SECTION
1
Page
1
PHYSICAL
CHEMISTRY
CHAPTER
1
ATOMS, MOLECULES AND
STOICHIOMETRY
Page
2
Introduction
The English scientist John Dalton proposed the atomic theory in which he stated that matter is made up
of tiny repeating units known as atoms. The atomic theory has proved very useful in explaining the
chemical properties and behaviour of matter. For instance, chemical reactions involve rearrangement
of atoms. The atom itself consists of a nucleus, containing neutrons and protons, and energy levels
(shells) containing electrons. Arrangement of electrons in the outer shell of an atom, as well as the
distance of the electrons from the nucleus, affects the chemical properties of an atom. Chemical
reactions involve sharing or total transfer of electrons between atoms. The rearrangement or transfer
of electrons during chemical reactions depend on how tightly they are held to the atom, and this in turn
depends on the number of protons (nuclear charge) in the nucleus as well as distance of the outer-most
shell from the nucleus. A large number of protons in the nucleus implies that the outer shell electrons
are firmly held to the atom and so they may not easily be transferred during chemical reactions. Take
sodium and chlorine as examples. The sodium atom is large and it has a small number of protons. The
outer electrons are therefore weakly held to the atom. This explains why sodium reacts by losing its
outer most-shell electron. On the other hand, the chlorine atom is smaller and has a larger nuclear
charge. The outer- most electrons are therefore firmly held to the atom. It is difficult for an atom of
chlorine to lose an electron, but it can easily accept an electron because the incoming electron is
strongly attracted by the large number of electrons in the nucleus.
Many substances are made up of molecules. A molecule is a particle made up of two or more atoms held
together by a covalent bond. This type of bond involves sharing of electrons between atoms . If the
molecules of a substance contain the same type of atom, the substance is known as an element, for
example, oxygen. On the other hand are compounds in which the molecules contain atoms of different
elements.
The particles encountered in chemistry, that is, the atoms, ions, electrons, molecules and protons, are
so tiny that chemists need a convenient method of counting them . Take hydrogen as an example. 4 g of
the gas contains a very large number of molecules, 12 x 1023. If we were to count these molecules, say,
in dozens, we would have an inconveniently large number, that is100 000 000 000 000 000
000 000 (1 x 1023) dozens. To avoid such cumbersome numbers, chemists have come up with the
concept of the mole. One mole is the quantity that contains 6 x 1023 particles. 4g of hydrogen gas
therefore contains 0.167 moles of hydrogen molecules. The concept of the mole is particularly useful
in stoichiometric calculations. These are calculations of amounts of products or reactants involved
during chemical reactions. When substances react, they combine in definite proportions, known as the
stoichiometric ratio, to form products which are also in a definite ratio. It is therefore possible to use a
balanced equation to determine the amounts of products or reactants involved in chemical reactions.
1.1
Relative masses of atoms and molecules
The masses of atoms and molecules are given relative to the 12C scale. On this scale, an atom of 12C atom
has a mass of 12 atomic mass units (a.m.u). This isotope of carbon was chosen because its atomic mass
could be determined accurately. Formally, the relative atomic mass, Ar, of an element is defined as
Ar =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
𝟏
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝑪−𝟏𝟐
𝟏𝟐
Similarly, the relative molecular mass, Mr, of a compound is defined as
Mr
=
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
𝟏
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝑪−𝟏𝟐
𝟏𝟐
The relative molecular mass of a compound is found by adding the Ar values for each atom in the molecule
of that compound. Ionic compounds such as sodium chloride do not contain molecules. They are made up
of formula units, for example , the formula unit for sodium chloride is NaCl. Instead of using the term
relative molecular mass for an ionic compound, the appropriate term is relative formula mass (RFM).
However, for convenience’s sake, chemists frequently use the term relative molecular mass for both ionic
and covalent compounds.
Relative formula mass
1.2
=
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒖𝒏𝒊𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
𝟏
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝑪−𝟏𝟐
𝟏𝟐
Isotopes
Isotopes are atoms of the same element that contain different numbers of neutrons.
In a sample of an element, some atoms are heavier than the others. The heavier atoms contain more
neutrons. However, the number of protons and electrons is the same in all the atoms.
Isotopes differ from each other in:
 Number of neutrons.
 Mass. Isotopes are identified by their mass numbers, for example, the isotopes of iron are
represented by the nuclide notation below.
54
Page
3
Fe
26
56
Fe
26
57
Fe
26
All the isotopes contain 26 electrons. The mass numbers 54, 56 and 57 distinguishes between the
isotopes (these numbers are also known as relative isotopic masses). Thus the isotopes can
simply be shown as Fe-54, Fe-56 and Fe-57 (or 54Fe, 56Fe, 57Fe). The heaviest isotope has the mass
number 57.
Recall the nuclide notation

nucleon number or mass number
= sum of protons and neutrons
Speed of movement in an electric field. The
heavier isotopes (with more neutrons) move
more slowly.
Isotopes are similar to each other in the following ways:


54
Fe
26

atomic number = number
They have the same number of electrons
Since the electronic structure is the same, they
have similar chemical properties.
They have the same number of protons.
of protons(= number of electrons)
Isotopes of hydrogen
Hydrogen has three isotopes which all have the atomic number 1 (that is, each isotope contains 1 electron
and 1 proton). The number of neutrons in each isotope is calculated from nucleon number = n + p, where
n = neutron number and p = proton number.
Thus
n = nucleon number – p
Isotope
atomic number
neutron number
(nucleon number - atomic number)
nucleon number
1
1 H
0
2
H
1
H
2
1
3
1
Table 1.0
The relative isotopic mass
This is defined as
Relative isotopic mass =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒂𝒏 𝒊𝒔𝒐𝒕𝒐𝒑𝒆 𝒐𝒇 𝒂𝒏 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
𝟏
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝑪−𝟏𝟐
𝟏𝟐
The relative isotopic mass coincides with the nucleon number (sum of protons and neutrons). This is
because a proton and a neutron each has a relative mass of 1. For example, an atom of 12C has 6 neutrons
and 6 protons. Its nucleon number = 12, which is equal to its relative isotopic mass.
Page
4
1.3
Relative masses of atoms and molecules from mass spectra
The technique of mass spectrometry, which uses a machine known as a mass spectrometer, is used to

Determine the structure of a molecular compound and its Mr

Find the number of isotopes in an element, their relative isotopic masses and their relative
abundances.
Instrumentation
Fig 1.1 is a block diagram illustrating the main features of the mass spectrometer
Fig 1.0
1. The sample is introduced as a vapour.
2. In the ionization chamber, fast moving electrons bombard atoms or molecules of the sample,
knocking out electrons from them. This results in the formation of positively charged ions.
3. The positive ions are accelerated through an electric field. It is the potential difference across
the field that causes the electrons to accelerate.
4. The beam of moving cations is deflected by a magnetic field. If the sample is an atomic
element such as a metal or a noble gas, the case is straightforward. The heavier isotopes will
be deflected less than the lighter isotopes. The result is that the isotopes are separated on the
basis of their mass. They are collected separately, and at a computer terminal the masses of
the different isotopes as well as their abundances are computed. The computer will
display this information on a mass spectrum.
The mass spectrum of iron is shown in Fig 1.0.

91.72%
abundance
(%)
Page
5
Fig 1.1

5.9%
54
2.1%

56 57
m/e
Number of peaks (vertical lines) = number of isotopes.
Thus iron has three isotopes.
The x-axis shows the relative isotopic masses of these
isotopes. The lightest isotope of iron is 54Fe, with an
isotopic mass of 54 units. It has the least number of
neutrons (54 - 26 = 28). The heaviest isotope is 57Fe, with
an isotopic mass of 57. It has the largest number of
neutrons (57-26 = 31).
The vertical axis gives the relative abundance of the
isotopes. The height of a peak is directly proportional to
the abundance of an isotope.
The most abundant isotope of iron is therefore 56Fe. We also conclude that it is the most stable
isotope. If it were not, it would be present in small amounts. The easiest way to understand these
isotopic abundances is this:
Consider a 100g sample of iron metal. Of this mass, 91.72 g would be due to 56Fe, 5.9 g to 54Fe . A
very small amount, 2.1 g, would be due to 57Fe.
1.3.1 Determination of relative atomic mass of an atomic element from
its mass spectrum
When the isotopic masses of all of the isotopes of an element, and their respective relative abundances are
known, the relative atomic mass of an element can be calculated using the formula
Ar =
( 𝐀𝐢 𝐱 𝐢𝐭𝐬 𝐚𝐛𝐮𝐧𝐝𝐚𝐧𝐜𝐞 ) + 𝐁𝐢 𝐱 𝐢𝐭𝐬 𝐚𝐛𝐮𝐧𝐝𝐚𝐧𝐜𝐞) +( 𝐂𝐢 𝐱 𝐢𝐭𝐬 𝐚𝐛𝐮𝐧𝐝𝐚𝐧𝐜𝐞)+ …
𝟏𝟎𝟎
Where Ai means the relative isotopic mass of isotope A and Bi means the relative isotopic mass of isotope
B etc. For iron,
Ar =
1.3.2
(54 x 5.9 )+ (56 x 91.72)+ (57 x 2.1)
100
= 55.74
Molecular elements and compounds
A molecule is a chemical entity made up of two or more atoms joined by a covalent bond. It is the smallest
unit of a compound or an element that preserves the chemical properties of that element or compound.
Some elements are molecular (made up of molecules), e.g. bromine, oxygen, phosphorous and sulphur (Fig
1.2).
sulphur atom
P
P
P
P
Fig 1.2 Molecular
structure of phosphorous
and sulphur.
a molecule of phosphorous( p )
4
a molecule of sulphur (S )
8
Mass spectrometry of compounds and molecular elements
Page
6
Mass spectrometry for a molecular element
We will use bromine as an example. Fragmentation of bromine molecules in the mass spectrometer
produces a mixture of positively charged ions. Ions of different masses are deflected to different extends, so
they can be collected separately. A computer terminal then calculates the relative abundances of these ions.
The relative abundances are expressed in terms of peak heights. The taller the peak, the larger the number
of ions giving rise to that peak.
Fragments of the same mass produces a single peak. Fragments are expected to differ in mass depending
on which isotope of bromine is present. In addition, there will be peaks caused by unfragmented bromine
molecules (molecular ions), Br2+. These are the molecular peaks.
Q
How many peaks would be expected on the mass spectrum of bromine and what would be
the ratio of peak height ( Bromine has two isotopes, 79Br and 81Br in the ratio 1:1).
A
5 peaks in the ratio 1:1:2:1:1
The following ions are formed in the mass spectrometer from bromine molecules:
1. ⁷⁹Br⁺
2.
81Br⁺
4. ⁷⁹Br-81Br⁺
5.
81Br-⁷⁹Br⁺
3. ⁷⁹Br-⁷⁹Br⁺
6.
81Br-81Br⁺
Species 1 and 2 are atomic ions formed by the fragmentation of bromine molecules. Since they have
different masses (79 and 81 respectively), they are collected separately and they form different peaks. The
ratio of peak heights (abundances) for these two species is 1:1.
The rest of the species are molecular ions, that is, they are unfragmented molecules which bear a positive
charge. Species 3 is the lightest of the molecular ions. It produces a peak at m/e 158 (79 + 79). Species 4
and 5 have the same mass (160), and so they produce a single peak. Species 6 is the heaviest of all the
species produced in the mass spectrometer (mass 162).
Note that the peak due to species 4 and 5 at mass 160 is two times taller than the rest of the peaks. This is
because there are two contributing species for this peak. A total of five peaks are therefore observed. These
peaks are in two groups. The first group, with two peaks, is made up of the peaks produced by the lightest
species, ⁷⁹Br⁺ and 81Br⁺. The second group, with three peaks, is made up of the peaks produced by the
heavier molecular ions (Fig 1.3).
Mass spectrum of a compound
Fig 1.3 The mass
spectrum of bromine
1
H
2
H
H
C
C
H H
3
O
4
H
Fig 1.4
Fragmentation of molecules of a compound forms a
mixture of positive ions with different masses, which are
collected separately in the massspectrometer, giving rise
to a number of peaks. The different ions are formed by
the breaking of different bonds in the molecule.
Different fragments (with different formulae) will give
rise to the same peak if they have the same relative
masses. Fig 1.4 shows the bonds that may break during
the fragmentation of ethanol in a mass spectrometer.
Different ions will be formed due to breaking of
different bonds as shown (Table 1.2).
The last species shown in the table is not a fragment. It
represents the whole unfragmented molecule (it is the
molecular ion). Using the Table 1.2 , we would expect six
peaks in the mass spectrum of ethanol.
The actual number of peaks depends on which bonds
actually break in the mass spectrometer.
Page
7
Note that fragments 2 and 3 have the same mass and will therefore form only one peak. In the mass
spectrum in Fig 1.5, peak heights (intensities) have been assigned arbitrarily.
The mass spectrum of ethanol (not to scale)
peak intensity
(abundance)
0
Bond broken
15 17
29
fragment formed
31
45 46
relative mass of fragment
CH3+
15
CH2OH+
31
2
CH3CHOH+
45
3
CH2CH2OH+
45
4
OH+
17
CH3CH2+
29
1
m/e
Fig 1.5 Mass spectrum
of ethanol
+
CH3CH2OH
46
Table 1.2 Ions formed
during the fragmentation of
ethanol
Of particular importance is the molecular ion peak at m/e 46. Since it represents the full molecule, its m/e
value is the same as the relative molecular mass of the compound being analyzed.
Example
An organic compound Q contains 40% C by mass and 53.3% O by mass. Its mass spectrum is shown
below.
peak
intensity
(height)
Page
8
15 17
43 59 60
m/e
(a)
(b)
Calculate the empirical formula of this compound.
Suggest an identity for Q and write down the formulae of the species responsible for
each peak in the spectrum.
Working
Since it is an organic compound, Q must also contain hydrogen.
%
Divide by Ar
C
40
O
53.3
H
6.7
40
53.3
6.7
12
16
3.333
Simplest ratio
3.33
1
1
6.7
1
2
(a) Therefore the empirical formula is CH 2O
From the mass spectrum, the compound has an Mr of 60.
(CH2O) n
= 60
30n
= 60
n = 2
Therefore the molecular formula of Q is C2H4O2.
b)
Q could be ethanoic acid, CH₃COOH
O
H
H C C
O H
H
The molecule may fragment as shown, giving rise to the fragments whose masses are given in the table.
Page
9
species
relative mass
CH3+
15
OH+
17
CH3CO+
43
CH3COO+
59
CH3COOH+
60
1.4
Chemical formulae
We have already used the empirical and the molecular formula in the foregoing calculation.
The empirical formula gives the simplest ratio of the different atoms present in the compound.
A molecular formula shows the actual number of the different atoms present in one molecule of a
compound.
Examples are given in Table 1.3.
compound
molecular formula empirical formula
Ethanoic acid
C2H4O2
CH2O
hydrogen peroxide
H2O2
HO
butane
C4H8
C2H4
ethyne
C2H2
CH
Table 1.3 Molecular and empirical
formulae of some compounds
Thus we know, for instance, that butane contains twice as much hydrogen as carbon.
Determination of formulae
Empirical formulae can be found from % composition data as in the previous example.
We find the ratio of atoms in the compound in terms of moles, which makes sense because when we write
a formula, say as C₂H₄O₂, we mean that in one mole of this substance, there are 2 moles of C, 4 moles of H
and 2 mol of O.
As already discussed, the molecular formula of a compound can be found by mass spectrometry. The peak
with the largest m/e value corresponds to the unfragmented molecular ion. This m/e value therefore gives
the Mr of the compound being analyzed.
Another method of determining Mr is by combustion analysis. This method is particularly applicable in
finding the molecular formula of hydrocarbons.
Finding molecular formula of a hydrocarbon by combustion analysis
Assuming complete combustion of the hydrocarbon, the only products would be water and carbon dioxide.
Let the hydrocarbon be CX H Y.
Then its combustion will be represented by the general equation (memorize it!) :
CxHy + x + y O2
4
y
xCO2 + 2 H2O
The usefulness of this formula becomes quite obvious if the values of x and y are large.
Consider the complete combustion of candle wax, a C30 alkane.
2C30H62 + 91O2 → 60CO2 + 62H2O
Moles of oxygen are found by (x + y/4) = 30 +
62
4
= 91/2.
Study the formula again:
Page
10
CxHy + (x + y/4)O2
xCO2 + y/2H2O
1 mol of the hydrocarbon combines with (x + y/4) moles of oxygen to form x moles of carbon dioxide and
y/2 moles of water. Now, this molar ratio is the same as the volume ratios of the reacting gases (number of
moles is directly proportional to volume of gas).
Since combustion is a gas phase reaction, the stoichiometry of the reaction can be expressed in terms of
volume ratios, for example if we are using cm³ as the unit for volume:
CxHy + (x + y/4)O2
xCO2 + y/2H2O
1 cm³: (x+ y/4) cm³
x cm³ : y/2 cm³
Suppose that instead of burning 1 cm³ of the hydrocarbon, we burn 10 cm³, then we have to multiply
throughout by 10 to maintain the ratio.
CxHy + (x + y/4)O2
xCO2 + y/2H2O
10 : 10 (x+ y/4)
10x
: 10(y/2)
Example
10cm³ of a hydrocarbon were completely burnt in 100 cm³ of oxygen. At the end of the reaction,
80cm³ of residual gas were collected, of which 40cm³ were absorbed in sodium hydroxide solution.
Find the molecular formula of the hydrocarbon.
Working
Let the hydrocarbon be CX HY
CxHy + (x + y/4)O2
xCO2 + y/2H2O
10cm³ : 10 (x+ y/4) cm³
10x
: 10 ( y/2)
The residual gas is the gaseous mixture that remains at the end of the reaction. It contains carbon dioxide
and unreacted (excess) oxygen.
The 40cm³ of gas which were absorbed in aqueous sodium hydroxide are carbon dioxide.
(carbon dioxide is an acidic gas and so dissolves in(reacts with) aqueous sodium hydroxide).
Excess volume of oxygen = total residual gas – volume of carbon dioxide
= (80 – 40) cm³
= 40 cm³.
Volume of oxygen that reacted = initial volume of oxygen – volume of excess oxygen
= (100-40) cm3
= 60cm3
From the general equation of combustion,
Volume of CO2 produced = 10x = 40
∴ x = 4
Volume of oxygen used = 10 (x+y/4) = 60
But x =4: substituting and solving for y
60 = 10 (4+y/4)
y=8
Page
11
The molecular formula of the hydrocarbon is C₄ H₈
1.5
1.5.1
Stoichiometry and the mole concept
The mole concept
The need to count things is part of our everyday lives. There are different units of counting, for example,
we often count bread and eggs in dozens.
Scientists usually deal with very large numbers of particles. For example, 2g of sodium metal contain a very
large number of sodium atoms. We need a convenient way of quoting such large figures. Towards this end,
chemists use a unit of counting known as the mole.
The mole is the amount of matter that contains the Avogadro number (L) of particles. The term
particle is general, for example, it could refer to atoms, molecules, ions or electrons.
The Avogadro number (constant) = 6 x 1023
Example
1(a)
(b)
2(a)
(b)
How many molecules are present in 0.35 mol of hydrogen gas?
How many moles are 1013 molecules of hydrogen?
Why is the statement ‘two moles of oxygen’ misleading?
How many moles of electrons are lost by half a mole of magnesium metal in forming the Mg 2+
ion?
Solutions
1(a) 1 mol
6.0 x 1023 molecules
→
0.35 mol →
less
Number of molecules =
0.35mol
1 mol
x 6.0 x 1023
= 2.1 x 1023
(b)
1 mol
less
→
6.0 x1023 molecules.
← 1013 molecules
number of moles =
1013 molecules
6.0 x 1023 molecules
= 1.67 x 10-11
2(a) The statement is ambiguous; it could refer to molecules of oxygen or to atoms of oxygen. For
example, 2 moles of oxygen contain 2 moles of oxygen molecules and 4 moles of oxygen atoms.
Page
12
(b)
Mg
→
Mg2+ + 2e-
1 mol of magnesium loses 2 mol of electrons. Therefore ½ a mol of magnesium would lose 1 mol of
electrons.
Inter- converting mass and moles
moles =
that is , n =
For example, 2g of sodium expressed in moles is 2/23 =0.087.
0.3 mol of hydrogen gas expressed in grams becomes Mr x
mols = 2 x 0.3= 0.6 g.
Similarly, 1.0 mole of water = 1.0 x 18 = 18 g.
From the last example, we make this general inference
mass in grams
Ar (or Mr)
m
Ar (or Mr)
thus
mass in grams = moles x molar mass
One mole of a substance has a mass equal to its Ar or
its Mr
Check using CO2. The Mr of CO2 is 46. 1.0 mol of CO2 should therefore have a mass of 46 g.
mass = mole x Mr
= 1.0 x 46 = 46 g
1.5.2
The molar gas volume
The volume occupied by 1.0 mole of a gas is constant, irrespective of the identity of the gas, provided
temperature and pressure are kept constant.
This volume is known as the molar gas volume, Vm .
Vm = 22.4dm3 at standard temperature and pressure (s.t.p)
OR
Vm = 24dm3 at room temperature and pressure(r.t.p)
1.0dm3 = 1000 cm3
Page
13
1.0cm3 = 10-3 dm3 = 10-6 m3
Example
All measurements are done at s.t.p
(a)
How many moles of gas are present in 15 dm³ of hydrogen?
(b)
What is the volume occupied by 0.13mol of ammonia gas?
(c)
What is the mass of 22.4 dm³ of water vapour?
(d)
How many atoms are present in 12 dm³ of oxygen gas?
(e)
What is the volume occupied by 1012 atoms of neon gas?
Solutions
(a)
mols = 15/22.4 = 0.670
(b)
1 mol  22.4dm3
0.13mol  0.13 mol x 22.4 dm3mol-1
= 2.91 dm³
(c)
22.4dm³ = 1.0 mol of water
= 18g (the mass of 1.0 mol of a substance = to its Ar or Mr)
Parts (d) and (e) are left as an exercise to the reader.
1.5.3 Stoichiometry
During a chemical reaction, the ratio in which reactants combine and products form is always constant for
that particular reaction. For instance, in the formation of ammonia, nitrogen and hydrogen combines in
the ratio 1:3.
N2(g) + 3H2(g)  2NH3(g)
The same result is obtained no matter how many times the experiment is carried out, and irrespective of
how many different people carry out the experiment. The reaction ratio (stoichiometry) for a reaction can
be deduced from the balanced equation of the reaction.
2 moles of hydrogen gas are mixed with 2 moles of oxygen and the mixture is sparked so
that the two gases react to form water. Which reactant is in excess and what would be the
molar composition of the residual gas?
Q
2H2(g) + O2(g)  2H2O(l)
According to the balanced equation, hydrogen and oxygen react in the ratio 2:1. In the given
situation, 2 moles of hydrogen require 1 mole of oxygen. Oxygen is therefore present in excess
by 1 mole. The residual gas contains 1 mole of oxygen (water forms as a liquid).
Page
14
A
In industrial processes, it is usually necessary to mix reactants in the exact ratio in which they react. This
minimizes wastage of raw materials and ensures that the product formed is not contaminated by excess
reactants. It would be time consuming, difficult and even expensive to separate the product from excess
reactants.
Reactions taking place in solution
Concentration
This is a measure of how much solute is dissolved in a given volume of solvent. In most reactions the solvent
is water. Concentration measures the extent of scattering of particles in a solution. If there is a very large
number of solute particles in a small volume of solution, then concentration of the solution is high. Adding
more solvent (dilution) will cause the particles to spread far, and so concentration is reduced.
amount of solute
Concentration(C) =
volume of solvent
The amount of solute is usually given in moles (n), and the volume (V) in dm³.
C= n
V
Often, concentration is expressed as mass m (in g) of solute dissolved in
a given volume of solvent (in dm3).
units = moles
dm3
That is, C =
m
V
, for which the units become g dm-3
= moldm-3
1 dm3 = 1000 cm3 = 1 litre
Example
1.
2.
Calculate the concentration of a 25.00 cm³ solution which contains 2g of sodium
hydroxide. Express your answer in moldm-3 and gdm-3.
What would be the final concentration when 25.00 cm³ of water are added to 100 cm³ of 0.5
mol/dm³ of sodium hydroxide solution?
Solutions
1.
Convert cm³ to dm³ and g to moles.
Moles of NaOH = g/Mr (NaOH)
= 2/40
= 0.05
Vol of solution = 25/1000
Page
15
= 0.025dm3
C = n/v
= 0.05/0.025
= 2.0 moldm-3
Converting to g/dm-3
C = 2g/0.025dm³
= 80gdm-3
2.
n
V₁
n = C1 x V1 … I
After dilution:
C2 =
0.5 x 0.1 = C2 x 0.125
C2 =
0.5 x 0.1
0.125
Concentration after dilution = 0.4 moldm-3
Before dilution:
C1 =
Since n remains constant after dilution,
we equate I and II.
C₁V₁ = C2V2
n
NB. The volumes were converted to dm³ because it is
a good habit to work in these units. The same answer
could have been obtained using volumes in cm³.Also
note that the final concentration is smaller. Dilution
always results in lowering of concentration.
V2
n = C2 x V2 ... II
Acid base titrations
It is possible to find the concentration of a solution of an acid or base by titration, as illustrated in this
example:
Example
25.0 cm³ of 0.1 moldm-³ sulphuric acid were required to exactly neutralize 20.0 cm³ of sodium
hydroxide solution. Calculate the concentration of the sodium hydroxide solution.
Solution
Page
16
Stoichiometry
H2SO₄ (aq) + 2NaOH → Na2SO4 (aq) + 2H2O (l)
1
:
2
= 0.0050
n(H2SO4) = C x V
C(NaOH) = mol/vol
= 0.1 x 0.025
= 0.0050/0.020
= 0.0025
n(NaOH) = 0.0025 x 2
= 0.0001 mol/dm³
Titration is a quantitative procedure, that is, it relies on the fact that reactants combine in a specific ratio
to form products. If any reagent is present in excess, only the amount required for reaction, as shown by
the balanced equation, will react, and the remainder will be present in a mixture with the products. During
titration, one solution has a known concentration. This is the standard solution which is placed in a burette.
The aim is to find the volume of the standard solution that would exactly react with a known
volume of the reagent whose concentration is not known.
A known volume of the second solution is transferred to a conical flask using a pipette. Most titrations
require the addition of a few drops of indicator into the flask. The indicator will change color at the end of
reaction, that is, when molar quantities of the reactants have reacted as shown by the balanced equation.
Commonly used indicators in acid-base titrations are phenolphthalein and methyl orange. Phenolphthalein
is pink in alkaline solution and colourless in acidic solution. The standard technique is to place the base in
the burette, though this is not a rule. The indicator is then added to the acid in the conical flask, giving a
colourless solution. The end point of the neutralization is when the first trace of pink appears.
Fig 1.0 shows the titration set-up.
Solution of known concentration
(titrant)
0
1
0
2
0
3
0
4
0
5
0
BURET T E ST AND
It should hold the burette in a
v ertical position (not tilted)
FUNNEL
Remov e soon after filling the
burette
BURET T E
Solution of know n concentration
(titrant) is placed here
BURET T E T AP
Check for leaks. Nev er use a leaking
burette. Also check that the tap can be
opened and closed without diffficulty
BURET T E T IP
placed just inside the burette
ANALY T E
This is the solution whose concentration is
to be determined. The analy te must hav e a
known v olume.
Fig 1.0 Set-up for a titration
Back titrations
Page
17
During a back titration, the amount of a reactant (analyte) is determined indirectly by titrating the
product of a reaction with another reagent. Two reactions are therefore involved. The first reaction
converts reactants to a product. In the second reaction, the product reacts quantitatively with another
reagent. In this way it is possible to determine the amount of product formed in the first reraction (see
example 1).
Another form of back titration involves putting a reactant (known amount) in the reaction vessel. Only a
certain amount will react according to the stiochiometry of the reaction. Another reagent is now added
into the reaction vessel to react with the excess reactant. It is then possible to determine the amount of
reactant that actually reacted. There are several reasons for carrying out a back titration, including

The analyte may be in solid form. In this case a liquid reactant is added in excess to react with the
solid. A back titration is then carried out to determine the amount of excess liquid that reacted. It
Page
18




then becomes possible to determine the amount of liquid reactant that reacted. By using the
stoichiometry of the reaction, the amount of solid that reacted can be determined.
If the analyte is contaminated by impurities, these may interfere with direct titration.
Direct (forward) titration may be too slow.
The direct reaction is reversible and does not go to completion.
The direct reaction does not have a clear end point.
Example
Cu2+(aq) reacts with iodine according to the following equation
2Cu2+(aq) + 4I-(aq)  I2(aq) + 2CuI(s) … (i)
In one experiment involving this reaction, the iodine produced reacted completely with 25.00 cm 3 of
0.10 moldm-3 sodium thiosulphate according to the equation
I2(aq) + 2S2O32-(aq)  2I –(aq) + S4O62-(aq) … (ii)
(a) Calculate the number of moles of Cu2+ that reacted with iodide ions reaction (i).
(b) Suggest an advantage of this method in the determination of the amount of Cu2+ that reacts with
iodide ions.
Solution
(a)
First calculate the number of moles of iodine that reacted with thiosulphate ions in reaction (ii).
This is the same as the number of moles of iodine produced in the first step.
25
Moles of S2O32- that reacted = C x V = 0.1 x
1 000
= 0.0025
Moles of iodine (reaction (ii)) = ½ moles of S2O32= ½ x 0.0025
= 0.00125
Reaction (i) therefore forms 0.00125 moles of iodine. According to the stoichiometry of this
reaction, moles of Cu2+ used = ½ moles of iodine used. That is,
Moles of Cu2+ = ½ x 0.00125 = 0.000 625
(b) The reaction of iodine with Cu2+ is easier to follow because starch can be added before the titration
to act as an indicator. When the reaction is complete, the blue-black colour of the iodine-starch
mixture disappears.
Example 2
Group II metal carbonates decompose when heated, according to the equation
MCO3(s)  MO(s) + CO2(g)
Page
19
4.08 g of a mixture of Bao and an unknown carbonate X was heated strongly. After heating, the mass
of the mixture decreased to 3.64 g. This residue was then dissolved in 100 cm3 0f 1.0 moldm-3 HCl.
The excess acid required 16 cm3 0f 2.5 moldm-3 NaOH for complete neutralization. Determine the
identity of metal X.
Solution
Moles of NaOH that reacted with excess acid
= C x V = 2.5 x
16
According to the stoichiometry of
this reaction,
1 000
= 0.040
moles of MCO3 = moles of CO2 = 0.01
HCl and NaOH react in the ratio 1: 1, so moles
of HCl that reacted = 0.032. This is the amount
of excess acid that was available to react with
NaOH (the rest of the acid had reacted with the
solid residue).
moles of acid added to residue = C x V
= 1.0 x
100
1000
MCO3(s)  MO(s) + CO2(g)
Also, moles of MO formed from decomposition
of MCO3 = moles of CO2 produced
= 0.01
MO reacts with acid according to the equation
MO(s) + 2H+  M2+ + H2O(l)
Therefore moles of acid that react = 2 x moles of MO
= 0.1
= 2 x 0.01 = 0.02
moles of acid that actually reacted with residue
= amount added – excess
= 0.1 – 0.04 = 0.060
The acid reacts with the residue in a neutralization reaction according to the equation
= moles of BaO + moles of MO from decomposition of MCO3.
Upon heating, mass of original mixture
decreased by
( 4.08 – 3.64) g = 0.44 g
This is the mass of CO2 lost from the metal
carbonate in the mixture.
m
Moles of CO2 lost =
Mr
=
0.44
44
= 0.010
The carbonate in the mixture is decomposed
by heat according to the equation
But total number of moles of acid that reacted with
residue = 0.060, so the remainder of the acid reacted
with BaO.
That is,
moles of acid that reacted with BaO = 0.060 -0.020
= 0.040
BaO + 2H+  Ba2+ + H2O
Moles of BaO = ½ moles of acid
= ½ x 0.040 = 0.020
Mass of BaO that reacted = moles x Mr
= 0.020 x 153 = 3.06 g
Mass of MO in the residue
= 3.64 g - 3.06
= 0.58 g
Now, mass of MO = moles x Mr, that is
0.58 = 0.01 x (M + 16),
where M is the Ar of the unknown metal.
Solving for M gives 42.
The metal M is therefore calcium.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
6.5in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text
1. box.]
The major chemicals present in the mineral dolomite are MgCO and CaCO . 2.75 g of a sample
Page
20
3
3
of dolomite was dissolved in 80 cm3 of 1.0 mol dm-3 HCl . The resulting solution was then diluted
to 250 cm3 with distilled water. 25 cm3 of the diluted solution required 20 cm3 of 0.1 moldm-3
NaOH solution for complete neutralization. Determine the percentage of MgCO 3 in the dolomite.
2. 4 g of contaminated limestone, CaCO3, is reacted with 200 cm3 of 0.5 moldm -3 HCl. The
resulting solution containing excess HCl required 50 cm3 of 0.5 moldm-3 NaOH for complete
neutralization. Determine the mass of impurities in the sample of CaCO3.
Page
21
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Questions, solutions and discussions
1(a) Define the term relative atomic mass.
(b) Chlorine has two isotopes, Cl-35 and Cl- 37.The relative atomic mass of chlorine is 35.5.
Calculate the isotopic abundances of the two isotopes of chlorine.
Solutions
(a) This is the ratio of the mass of an atom of an element to 1/12th the mass of one atom of
12C.
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝒂𝒏 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
Ar =
𝟏
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒐𝒏𝒆 𝒂𝒕𝒐𝒎 𝒐𝒇 𝑪−𝟏𝟐
𝟏𝟐
( b) Let the isotopic abundance of Cl -35 be y. The abundance of Cl- 37 becomes 100-y.
Using formula
Ar
=
35.5 =
( 𝐢𝐬𝐨𝐭𝐨𝐩𝐢𝐜 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂𝐥−𝟑𝟓 𝐱 𝐢𝐭𝐬 𝐚𝐛𝐮𝐧𝐝𝐚𝐧𝐜𝐞 ) + (𝐢𝐬𝐨𝐭𝐨𝐩𝐢𝐜 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂𝐥−𝟑𝟕 𝐱 𝐢𝐭𝐬 𝐚𝐛𝐮𝐧𝐝𝐚𝐧𝐜𝐞)
𝟏𝟎𝟎
35y +37(100−y)
100
y = 75
Therefore the abundance of Cl- 35 is 75% and that of Cl-37 is 25%
2
Boron has two isotopes, B-10 and B -11 in the ratio of 1:4. What is the relative atomic mass of boron?
Solution
Change the ratio to percentages.
B-10: % abundance = (1/5) x 100 = 20
B -11: % abundance = (4/5) x 100 =80.
By formula
Ar =
=
(10 x 20) + (11 x 80)
100
10.8
OR by using the ratio directly:
Page
22
Ar
=
(𝟏 x 10 ) + ( 𝟒 x 11)
𝟓
= 10.8 as before.
3
The mass spectrum of C3H7OH is as shown.
% abundance
M M+2
m/e
What is the most likely reason for the appearance of the (M + 2) peak?
A
presence of hydrogen bonds in the molecules
B
different fragmentations of the molecule
C
presence of oxygen isotopes
D
gain of two hydrogen atoms by C3H7OH
9189/3/O/N/2005
Solution C
Oxygen has two isotopes, O-16 and O-18. The M peak is caused by molecular ions, C3H7OH+
that contain the lighter O-16 isotope. Molecular ions that contain the heavier isotope O-18
would be two units heavier, giving rise to the M+2 peak.
4
Using a mass spectrometer, it is possible to determine
1 the number of protons in an atom
2 the relative molecular mass of an organic compound
3 number of isotopes of an element
Which of the numbered statements are correct?
9189/3/O/N/2005
Solution
2 and 3 are correct.
Notes


Page
23
5
The relative molecular mass of a compound corresponds with the highest m/e value obtained on the mass
spectrum of the compound. This peak is generated by the molecular (unfragmented molecule) ion.
The number of peaks generated on the mass spectrum of an element depends on the number of isotopes
present in that element.
The mass spectrum of an element X is as shown in the following diagram.
(a) Determine the formula of the element X.
(b) What are the values of p and y?
9 : 6:1
relative abundance
3 : 1
35 37
(c)
p
y
74
m/e
How many isotopes does the element X contain? Write down the formulae of the isotopes.
(d) Deduce the identity of the species responsible for each peak on the spectrum.
(e) Deduce the identity of element X, showing your reasoning clearly.
Solutions
(a)
X2 .The species at m/e = 74 has two times the mass of the species at m/e =37, suggesting
that the species at m/e = 74 has the formula 37X-37X.
(b)
p = 70 and y = 72
(c)
Two isotopes, 35X and 37X.
(d)
m/e
species
35
35
37
37
70
35
72
37
74
37
(e) The element has two isotopes, X-35 and X-37 in the ratio
3:1 (first two peaks on the mass spectrum).
X+
Ar of the element =
X+
35
+
X X
X35X+
X37X+
(35 x 3)+ (37+1)
4
= 142/4
= 35.5, so the element is chlorine.
[do not forget the positive charges on the ions]
6
How many peaks due to the Cl2+ ion are expected in the mass spectrum of chlorine, and in
what ratio would the heights of the peaks be?
Solution
Page
24
3 peaks (the last three peaks in the mass spectrum in question 5 above). These peaks
are due to the molecular ions shown in the following table.
ion
m/e
35
Cl35Cl
70
35
Cl37Cl
72
37
37
35
Cl Cl
72
Cl37Cl
74
relative abundance(peak height)
3/4 x 3/4 = 9/16
peak 1
[3/4 x 1/4] x 2 =*6/16
peak 2
1/4 x 1/4 = 1/16
peak 3
[* multiplied by 2 because there are two contributing species, doubling the
probability of finding such a species.]
The three species are therefore in the ratio of 9:6:1
7
The relative atomic mass of magnesium is 24.3. Which is the mass spectrum of
magnesium?
A
B
10
10
8
relative 8
intensity
relative
intensity 6
6
4
4
2
2
0
0
23 24 25 26
m/e
C
m/e
D
10
10
8
8
relative
intensity
relative 6
intensity 4
6
4
2
2
0
23 24 25 26
23 24 25 26
0
23 24 25 26
m/e
m/e
9189/3/O/N/2006
Solution C
Page
25
HINT: the most stable isotope of Mg is 24Mg (tallest peak).
Check by calculation
Ar =
(24 x7)+(25 x 1)+(26 x1)
9
= 24.3
8
What can be deduced from the following mass spectrum of a diatomic molecule X 2 showing the
X2+ peak heights?
peak height
1
6
5
4
3
2
1
0
158 160 162
m/e
Atoms in X2 have isotopic masses 79 and 81.
2 Relative abundance of each isotopes of X is 50%.
3 The Ar or x.
9189/3/O/N/2006 .
Solution: 1, 2 and 3 are correct
Atoms in X2 have isotopic masses 79 and 81. This explains the presence of the peak at
m/e = 160 (79 + 81).
Element X has two isotopes, X-79 and X-81. The peak at m/e = 158 is due to X2+ ions
containing X-79 atoms only(79 +79 = 158). The peak at m/e 162 is due to X2+ ions
containing X- 81 atoms only.
The two isotopes of element X are present in the ratio of 1:1, since the corresponding
peaks at 158 and 162 are equal in height. The Ar of element X can be
calculated using the formula
Ar =
=
(Abundance of isotope A x its isotopic mass)+ (Abundance of B x its isotopic mass)
sum of peak heights
(2 x 79)+ (2 x 81)
4
= 80
Element X is therefore Br2.
9
What is the mass of NaCl contained in 25cm3 of a of 0.01moldm-3 solution?
Page
26
Solution
moles of NaCl = C x V =
0.01 x
25
1000
=
2.5 x 10-4
mass of NaCl = moles x Mr(NaCl)
= 2.5 x 10-4 x 58.5
= 0.0146g
10
20cm3 of a 0.01 moldm-3 sodium sulphate are accidentally spilled from a 1dm 3 solution.
What mass of NaOH remains in the container?
Solution
Tip: In a question where you are given concentration and volume of solution, the first thing
that should come to mind is finding number of moles, even though you might not be sure about
the next step.
moles of Na2SO4 lost = C x V
= 0.01 x
=
mass of Na2SO4 lost =
20
1000
0.0002
n x Mr
= 0.0002 x 142
= 0.0284g
moles of solute in the 1 dm3 solution = 0.01
mass of solute in 1 dm3 solution = n x Mr
= 0.01 x 142
= 1.42 g
mass of sodium sulphate left in the container = 1.4200g -0.0284g
= 1.39g
11
Calculate the volume of water that must be added to 250cm 3 of 0.1 mol/dm3 NaOH to
reduce its concentration by 25%?
Solution
After reducing the concentration by 25%, it becomes C 2 where
C2 = 0.1 – 25% of 0.1 = 0.1 -
25
100
Page
27
= 0.075
x 0.1
Using C1V1 = C2V2,
where V2, the new volume after dilution, is to be calculated.
V2 = (C1V1)/ C2
=
0.1 x 250
0.075
= 333.33 cm3
Volume of water that must be added = 333.33 – 250
= 83.33 cm3
[Type a quote from the document or the summary of an interesting point. You can position the text
Exercise
box anywhere
1.0 in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
1
100 cm3 of 0.2 moldm-3 NaCl is diluted by adding 50 cm3 of water. What is the new concentration
of the solution?
2
Oxides of nitrogen are pollutant gases which are emitted from car exhausts. In urban traffic when
a car travels 1Km, it releases 0.23 g of an oxide of nitrogen, NxOy, which occupies 120 cm3. What
are the values of x and y?
9701/1/O/N/2007
3
The amount of calcium ions in a sample of natural water can be determined by using an
ion - exchange column.
water sample
Page
28
ion - exchange column
A 50 cm3 sample of water containing dissolved CaSO4
was passed through the ion- exchange resin. Each
calcium ion in the sample was exchanged for two H+ ions.
The resulting acidic solution collected in the flask
required 25cm3 of 1.0 x 10-2 mol/dm3 KOH for complete
neutralization.
What was the concentration of calcium sulphate in the
original sample ?
9701/1/O/N/2006
4
Wines are made by the fermentation of glucose, C6H12O6. Ethanol and carbon dioxide
are the only products. The chief acid in an unopened bottle of wine is tartaric acid.
COOH
tartaric
acid
H
C
OH
H
C
OH
50 cm3 of a wine were exactly neutralized by 26 cm3 of 0.010moldm-3
KOH.
What was the concentration of tartaric acid in the wine?
COOH
Page
29
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
CHAPTER
2
ATOMIC STRUCTURE
Introduction
The atom is the basic unit of matter. In chemistry, emphasis is mainly on chemical properties and
reactions of matter. Chemical reactions occur at the atomic level, involving new arrangements of the
atoms in reactants. The atom can be studied at two levels
 Electrons, found in shells around the nucleus, directly determine the chemical properties of an
atom. It is therefore important to understand how electrons are arranged in atoms.
 The nucleus (protons and neutrons) affect chemical properties of an atom in an indirect way.
Protons provide the attractive force which keeps electrons bound to the atom. The strength of
this force therefore affects the reactivity and other chemical properties of an atom .The effect
of proton number and distance of electrons from the nucleus is a recurring theme in chemistry.
It helps to explain the chemical behaviour of atoms.
2.1
The nucleus
The nucleus is made up of protons and neutrons. Both particles have a relative mass of 1. The proton has a
relative charge of +1 whilst the neutron has no charge.



Protons affect chemical properties of an atom indirectly by attracting and holding electrons to the
atom.
The mass of an atom is mainly due to the nucleus, that is, protons and neutrons. This is because the
mass of an electron is so small that its mass contribution is negligible.
Isotopes are atoms of the elements which have the same number of protons (that is, same atomic
number) and electrons, but different numbers of neutrons. Isotopes have the same chemical
properties but they have different masses. The isotopes with more neutrons are heavier. Table 2.0
compares properties of the nuclear particles (protons and neutrons) to those of electrons. Protons,
neutrons and electrons are collectively known as the sub-atomic particles.
Particle
Page
30
proton
Relative charge
Relative mass
Location
+1
1
nucleus
neutron
0
1
electron
-1
nucleus
shells around
nucleus
0.000 5 (1/1840)
Table 2.0 properties of the sub-atomic particles
2.2
Electronic structure
Electrons are involved in chemical reactions, but protons and neutrons are not. The arrangement of
electrons (electronic configuration) of an atom is important in determining the chemical behaviour of that
atom. Of particular importance are

Structure of the valence shell. The valence (outer-most shell) is the one which directly
participates in chemical reactions, either by gaining, losing or sharing electrons with other atoms.
The number of electrons in the outer-shell shell is important in determining the type of reaction
that the atom undergoes. Metals, with a very small number of electrons in the valence shell, usually
react by losing the valence electrons (oxidation). Non-metal atoms, with a large number of
valence electrons (close to, but not equal to eight) usually react by gaining a few electrons into the
valence shell (reduction).
Number of shells. As the number of shells increases, so does atomic size (atomic
radius). The valence shell is then a large distance from the nucleus. Valence electrons in such an
atom are loosely held to the atom and so can easily be lost. In other words, when atoms become
large, they begin to show some metallic properties (tendency to lose
electrons). In fact, elements become more metal-like on going down the group. This behaviour is
well illustrated by the Group (IV) elements. Increasing number of shells also increases the shielding
effect. This is the situation whereby inner shells prevent the valence shell from feeling the full
attractive force of the nucleus. The valence shell electrons are therefore weakly held to the nucleus,
causing them to be lost easily in chemical reactions.

2.2.1 Arrangement of electrons in shells and orbitals
Electrons are found in orbitals. In turn, orbitals of the same energy level belong to the same shell. An
orbital is not an object with physical boundaries. It is simply a region of space in which the probability
of finding an electron is very close to 100%. Orbitals are designated with the letters s, p, d and f. In any
given shell, the energy of the orbitals increases in the order shown in Fig 2.0 below.
s
Fig
2.0
p
d
f
least energy
m ost
energy
most stable
(closest to
nucleus)
least stable
(furthest
from
nucleus)
Note that this is a comparison of orbitals in the same shell.
A shell can be thought of as a discrete energy level, the level
being measured relative to the nucleus. One way to understand
this is to think of the nucleus as the ground level. The shell
which is closest to the nucleus is like an object which is closest
to ground level. Such an object has the least (potential) energy
and it is the most stable. A shell which is farther from the
nucleus is like an object well raised above ground level. Such an
object has a large potential energy and is relatively unstable.
The principal quantum number (n = 1, 2, 3 e.t.c) describes the energy of a shell and its orbitals. The
energy of a shell and its orbitals increases as n increases, for example, the shell n= 3 is at a higher energy
level (farther from the nucleus) than the shell n=2.
Page
31
A shell which is closer to the nucleus has less energy and is more stable than a shell which
is farther from the nucleus.
If an orbital is close to the nucleus its electrons are tightly held by the attractive force of the protons. Such
electrons are not easily removed or transferred during chemical reactions (they are stable). Shells which are
a large distance from the nucleus feel a small attractive force of the nucleus. Such electrons are easily lost
or transferred during reactions, that is, they are relatively unstable.
From this discussion, it can be deduced that it is the outer-most electrons (valence electrons) which
participate in chemical reactions. They are the farthest from the nucleus and the most weakly held to the
atom (least stable).
Nature of the orbitals
The s orbitals
The s orbitals are spherical. They have the least energy in any given shell. In other words, they are closest
to the nucleus and they are the most stable.
The designation 1s, 2s, 3s... is used to represent s orbitals in different shells (different energy levels). The
coefficients 1, 2, 3 e.t.c represent the principal quantum number, n which defines the shell (energies
level) in which the orbital is found. The larger the value of n, the higher in energy and the more unstable
the orbital. Even though the s orbitals are the closest to the nucleus in any given shell, they have different
energies depending on the value of n. For instance, the 2s orbital is higher in energy and more unstable
than the 1s.
Apart from affecting the relative energies of the orbitals, the principal quantum number n also determines
the size of the orbital. The smaller the value of n, the smaller the orbital. When n is small, the orbital is
close to the nucleus and so it is strongly pulled inwards by the attractive force of the protons, shrinking its
size.
For example, comparing the 1s and 2s orbitals (Fig 2.1), the 2s orbital is:

larger
 higher in energy
 less stable
 further from the nucleus
The p orbitals
Fig 2.1 The 1s and 2s orbitals have
similar shapes (spherical) but differ
in distance from the nucleus, energy
and size.
These dumb - bell shaped orbitals are higher in energy than
the s orbitals in any given shell.
The p sub-shell is made up of three orbitals, the px, the py and
the pz. They have the same shape and same energy (that is,
they are degenerate) but they differ in orientation. The px has
lobes along the x - axis, the pz along the z - axis and the py
along the y- axis (Fig 2.2). Note the use of the term subshell.
Sub-shells are orbitals that share the same shell
The d orbitals
Page
32
In a particular shell, the d - orbitals are higher in energy than the p - orbitals. There are five orbitals in the
d sub- shell, and in the neutral atom they are degenerate, that is, they are at the same energy level. These
five orbitals are designated
dxy, dxz, dzy, dz2 and dx2- y2
The shapes and orientations of the d- orbitals are shown in Fig 2.3.
Fig 2.1 Shapes and orientation of the p orbitals. (a) The direction convention used in
chemistry. (b), (c) and (d) Orientation of the pz, py and px orbitals. (e) Degenerate p orbitals
form the p sub - shell. These three orbitals are mutually perpendicular and their nodes coincide
with the nucleus. The nodal plane is a region in which the probability of finding an electron is
zero. The central position of the nodal plane is known as the node, and this
coincides with the nucleus.
Fig 2.2 Shapes of
the d-orbitals.
Which orbital is not
shown here?
Page
33
Note the following about the d – orbitals:
 The first three have lobes between the axes, for example, the dxy orbital has lobes between the x
and the y-axes, as shown in Fig 2.3. You can easily remember the names of these three orbitals by
noticing that the names take all possible combinations of two letters from the letters x, y and z .


Fig 2.3
The last two d-orbitals (dz2 and dx2 – dy2 ) have
lobes along the axes. The dz2 has two lobes which
lie along the vertical (z) axis. The dx2 –y2 orbital
has four lobes which lie along the x and y axis.
Of the five d-orbitals, four have four lobes and
have the same general shape. The dz2 has a
unique shape. It has two lobes and a ring around
the nodal region (Fig 2.4).
Electrons can be found in lobes and the ring, but not at the node
(nucleus) and nodal plane.
An important note
Fig 2.4
orbital
The dz2
The diagrams of orbitals given so far should be understood only as models. Real orbitals
do not have clearly cut boundaries as depicted in the diagrams. Further more, the
diagrams give the impression that orbitals are solid structures in which electrons reside.
This is misleading. An orbital has no structure but is simply the volume of space in which
electrons spend most of their time, that is, the probability of finding an electron in that
region is close to 100%. In advanced work, the orbitals are interpreted as mathematical
functions.
Consider the hydrogen atom which has only one electron. This electron has kinetic
energy and is free to move within a volume of space whose boundaries are approximately
spherical. Since the electron moves very fast, if we were able to see it, it would appear as
if it were everywhere at the same time, and it would appear like an electron cloud
occupying a roughly spherical volume.
An orbital is the volume of space in which an electron spends most of its time. The probability of
finding an electron within this volume of space is approximately 100%.
2.2.2
Electronic configurations: The Aufbau method
The term aufbau is a German word, meaning to “build up”. The Aufbau method gives rules about how
electrons enter the atom. These guidelines should be followed when writing electronic configurations for
atoms.
An electronic configuration shows how electrons are arranged in an atom.
Rules for filling electrons in an atom by the aufbau method
1. Electrons will enter the lowest energy orbital first.
The 1s orbital, which is closest to the nucleus, and which feels the greatest pull by the nucleus, will
always be filled first.
Page
34
2. Hund’s rule of maximum multiplicity.
A maximum of only two electrons may occupy an orbital, and these two electrons must be spin paired.
3. Pauli’s exclusion principle
Electrons will first occupy orbitals singly, thereafter pairing may occur according to Hund’s rule.
Since there is only one orbital in the s sub-shell, according to rule 2, this sub shell can carry a maximum of
only two electrons. The p sub-shell has three orbitals, so it can contain a maximum of six electrons. The d
sub-shell, with five orbitals, will carry a maximum of ten electrons (Table 2.1).
Sub-shell
Number of orbitals
in the sub-shell
Maximum number of
electrons that may enter each
sub-shell
s
1
2
p
3
6
d
5
10
f
7
14
Table 2.1
Consider the p sub - shell, with three degenerate orbitals. Also suppose that we wish to fill them with five
electrons. Rule 3 says we should first make sure that each orbital has a single electron; only then can we
start pairing them (Fig 2.5).
Fig 2.5
The concept of spin pairing
The electron is always spinning about an axis. When ever a charged particle moves, it generates a magnetic
field around it. When two electrons co-exist in the same orbital, they must spin in opposite directions, that
is, one should have a clockwise spin and the other should have an anticlockwise spin. By spinning in
opposite directions, the magnetic moments of the two electrons cancel each other (magnetic moments are
vector quantities, so they amplify or deplete each other depending on their size and the directions in which
they are operating). The canceling out of magnetic moments means that there is no net magnetic field in
the orbital and this tends to stabilize both electrons. The phenomenon in which two electrons in an orbital
spin in opposite directions is known as spin pairing.
By convention, the up half -arrow (↿) is used to represent the clockwise spin and the down arrow (⇂) the
anticlockwise spin.
The electronic structure framework
Page
35
Electrons will enter orbitals in the order given below:
In so doing, electrons enter the lower energy orbitals first. The electronic configuration of a hydrogen
atom (atomic number 1) is the simplest, having only one electron to account for:
H: 1s1
The next atom, He, has two electrons, giving the electronic configuration
He: 1s2
With two electrons, the 1s orbital is now full. Any subsequent electrons will have to enter the next shell.
Helium, with an exactly filled shell, is very stable and is classified among the stable Group 0 (noble)
elements. Table 2.2 below shows the electronic configurations of the rest of the elements up to Ca.
The reader should complete the table by filling in the last column (from Al to Ca).
Page
36
Table 2.2
Points to note
1. Certain configurations are associated with stability. There is stability associated with an exactly half filled
or exactly filled sub-shell or shell. The p3 configuration (half filled), as in nitrogen, is relatively stable.
This stability is explained in terms of the symmetric occupancy of orbitals that occurs when subshells or shells are exactly half - filled or exactly filled.
2. Whenever the outer-most (valence) shell is filled with eight electrons (an octet), the element is very
stable and is classified as a noble element (Group 0). An exception to the octet rule is Helium. It is in Group
0 even though it has two electrons (a duet) in the outer-most shell. However, the outer- most shell in helium
is exactly filled, so helium is as stable as the other noble gases.
The concept of shells
You may remember showing the electron arrangement in an atom (such as sodium) diagrammatically as
shown in Fig 2.6.
Na: proton (atomic)
number 11
This diagram shows that the atom has three shells of electrons around the
nucleus. But what exactly is a shell? Can we deduce the number of shells from
the electronic configuration?
Study the electronic configuration of Na:
1s2 2s22p⁶ 3s1
Fig 2.6
Recall that the coefficients (principal quantum number) of each orbital
represent the energy of that orbital. The obvious conclusion is that orbitals
with the same principal quantum number have similar energies. Such orbitals
form a shell.
A shell is an energy level made up of orbitals (sub shells) with the same quantum number, and
hence with similar energies.
However, this generalization fails when the distance of the shells from the nucleus increases. In such
cases, it is found that a shell may contain sub-shells that have different principal quantum numbers.
Page
37
Since there are three values of n in the configuration of sodium, that is, 1, 2 and 3, we conclude that a
sodium atom has 3 shells, the shell n=1 ( K shell), n=2 (L Shell) and n=3 (M shell).
The K shell always has one sub shell ( the 1s) . This shell is the lowest in energy. The L shell always has two
sub shells; the 2s and the 2p. The M shell always has two sub shells; the 3s and the 3p, as illustrated for
the sodium atom in Fig 2.7.
 The s orbital is always lower in energy (closer to the nucleus) than the p orbital, even when they
belong to the same shell (same principal quantum number n).
 The 1s orbital (K shell) is the nearest to the nucleus. It is therefore the most stable because it feels
the strongest attraction by the nucleus. It is also the smallest.
 The L shell has two sub-shells, the 2s and the 2p. The total number of electrons in the two subshells (in this case, 6+2) gives the total number of electrons in the shell. The two sub - shells in
the L shell are equivalent (not equal) in energy; the 2s orbital is slightly lower. However, the
energy difference between the two sub- shells is very small. In fact, it is so small that the two subshells behave as a single shell.
 As the value of n increases, the shells move outwards from the nucleus and become larger. The
largest value of n corresponds to the outer-most shell. Thus in Na the outer - most shell
is the 3s (n=3, M). The outer- most shell is also referred to as the valence shell. It is the least
stable. Being farthest from the nucleus, the electrons in the shell are relatively weakly held to the
atom. The electrons therefore easily participate in chemical reactions.
Fig 2.3 The three shells
(energy levels) in an atom of
Na. Notice that the second
shell is made up of two subshells, the 2s and the 2p, giving
a total of eight electrons.
Relative energies of the orbitals
Imagine stacking empty orbitals one above the other, starting with the least energy orbital. The order
obtained would be as shown in Fig 2.4 below.
Electrons enter the orbitals in the order shown on the left, up
to calcium. Calcium, with 20 electrons has the configuration.
1s22s22p63s23p64s2
Fig 2.4 Energy level diagram
for the empty orbitals up to the
3d.
The 3d-orbitals, though present, remain empty because they
are higher in energy than the 4s shell. The last shell (valence)
shell in Ca is therefore the 4s, which contains two electrons.
This explains why Ca achieves the +2 oxidation state in all of
its compounds, corresponding to loss of the 2 valence
electrons.
Take note of the following:
The energy levels of the 4s and the 3d are reversed when
electrons start to enter the 3d orbitals, as shown in Fig 2.5
Page
38
This is often a source of confusion among students. The order
given in Fig 2.4 is what we would expect if the 3d orbitals were
not occupied. However, as soon as electrons begin to enter the
3d orbitals, they repel electrons in the 4s orbital. The effect is
that the 4s electrons move farther from the nucleus, that is, they
become higher in energy than the 3d orbitals.
This is illustrated below for scandium, titanium and bromine.
Fig 2.5 The relative energies of
the 4s and the 3d orbitals are
reversed when electrons start
entering the 3d orbitals
21 Sc
: 1s22s22p63s23p63d14s2
22 Ti
: 1s22s22p63s23p63d24s2
35 Br
: 1s22s22p63s23p63d104s24p5
By writing these configurations in this way, it becomes clear that the valence shell in the transition
elements, in this case, Sc and Ti, is the 4s. We therefore expect the transition elements to lose electrons
from the 4s orbital first during chemical reactions. This is what is actually observed. The transition
elements have a +2 oxidation state, corresponding to the loss of the outer-most 4s electrons. However, in
some books these configurations are written so that the 4s comes before the 3d. This is misleading
because it gives the wrong impression that during reactions, transition elements lose electrons first from
the 3d sub-shell.
It should also be mentioned at this stage that
as distance from the nucleus increases (increasing values of n), the difference in energy between
consecutive shells becomes so small that they practically behave as one shell.
1 Thus the 3d and the 4s shells behave as one shell in the chemistry of the transition elements.
The 4s is slightly higher than the 3d, so electrons are lost from the 4s orbital first. However, electrons are
also lost from the 3d orbitals due to the closeness in energy to the 4s shell. This explains why transition
elements exhibit variable oxidation states in their compounds. Different numbers of electrons can be lost
in addition to the two 2s electrons, thus giving rise to different oxidation states. Consider calcium and
managanese as examples. Ca, a non-transition element, has only one oxidation state in all of its
compounds, corresponding to the loss of the two valence electrons in the 4s shell. Mn, a first row
transition element, has oxidation states from +2 to +7 in its compounds. The +2 state corresponds to
loss of the two outer-most 4s electrons. The +3 to +7 states correspond to further loss of electrons from
the 3d orbitals.
2.
The 4s, 3d and 4p sub-shells behave as a single shell, which can hold up to 18 electrons.
Consider the electronic configuration of Br (atomic
number
35)
Sub-shell
maximum occupancy
35 Br
: 1s22s22p63s23p63d104s24p5
3d
10
The outer-most shell is the shell n=4, made up of the
4s
2
4s and the 4p orbitals. This shell therefore contains 7
4p However, the 3d shell
6
electrons.
is so close in
energy to the 4s and the 4p 18
sub-shells that it behaves
TOTAL
as if it were part of the valence shell. This explains
why the fourth shell in Br is said to have 17
electrons (10 from 3d, 2 from 4s and 5 from 4p). The configuration for Br can be written in terms of shell
occupancy as 2.8.8.17 (Fig 2.6).
Page
39
Fig 2.6 Electronic structure of a bromine atom.
There are four shells, which is equal to period number. The fourth
shell, containing 17 electrons, is made up of the 3d, 4s and 4p sub
shells. When electrons are lost from the bromine atom, they come
from the outer-most sub shells first, the 4s and the 4p, which
contain a total of seven electrons. Similarly, when a bromine atom
gains an electron, it enters the highest sub shell, which is the 4p.
This gives the bromide ion the stable 2.8.8.18 configuration of
argon.
The next element after bromine is Ar, which has a total of 36 electrons. The fourth shell in Ar is exactly
filled with 18 electrons, explaining its position in Group 0.
Ar: 2.8.8.18
If we decide to treat the valence shell in Ar as being made up of the 4s and the 4p sub shells only, as is
often the case, then the number of valence electrons would be 8.
2.2.3




Relating electronic structure to the Periodic Table
The number of shells in an atom of an element is equal to the period number for that element.
The highest value of n in the electronic configuration corresponds to the outer-most shell, which
is the highest in energy. This is the first shell to participate in chemical reactions.
The number of electrons in the outer-most shell corresponds to the group number for that
element.
As shells move further and further from the nucleus, their energies become similar, and they may
be regarded as a single shell. This explains why the fourth shell and the fifth shells are said to take
a maximum of 18 electrons. Fig 2.7 shows at a glance the relationship between electronic
structure and position of an element in the periodic table.
Fig 2.7

Page
40

Fig 2.7 also emphasizes the fact that across a period, electrons enter the same shell, and for the
first three periods, the term same shell actually means same principal quantum number n.
However, from the fourth shell onwards, we see that the same shell can have sub-shells with
different principal quantum numbers.
Going down groups, new shells are opened, represented by larger values of n. The opening of new
shells has the effect of increasing the size of atoms. For instance, the potassium atom, with four
shells, is larger than the sodium atom which has three shells. We also infer that potassium is more
reactive than sodium. Because of its large size, the outer-most electron in potassium is a large
distance from the nucleus. Consequently it is relatively weakly attracted to the nucleus. The
electron is therefore quite easily lost during reactions.
2.2 .4
The transition elements
The elements after calcium, that is scandium to zinc, form the first row of transition elements. Across this
row, electrons enter the d sub-shell. Here is the electronic configuration of calcium, which just precedes
scandium.
Ca (atomic number 20): 1s2 2s2 2p6 3s2 3p6 4s2
It has already been mentioned that in the transition elements, the 4s sub-shell is higher in energy than the
3d sub-shell. In the transition elements, electrons therefore enter a lower energy (penultimate) 3d subshell. Thus in writing the electronic configurations of the transition elements, we show the 3d sub-shell
before the 4s sub-shell (Fig 2.8).
Fig 2.8
By the time electrons enter the 3d sub shell, the 4s sub shell is already
occupied. As electrons enter the 3d sub shell, they repel electrons in the
4s sub shell, causing it to rise in energy. The 4s therefore becomes the
outer most sub shell, as shown in the diagram. However, the 4s and the
3d are so close in energy that in chemical reactions, they behave as one
shell.
Table 2.3 gives the electronic configurations of the elements Sc to Zn.




Table 2.3 Electronic configurations of the
transition elements from Sc to Zn
Page
41

In the transition elements, electrons enter the
3d sub-shell. We might have expected them
to enter the 4p sub- shell. However, the 4p
sub-shell is higher in energy and will only be
occupied after the 3d sub-shell is fully
occupied by 10 electrons.
There are five orbitals in the d sub-shell.
These five orbitals, designated dxy , dxz , dzy ,
dz2 and dx2-y2 are at the same energy level in
the neutral atom ( that is, they are
degenerate).
According to the rules of electron filling up,
these five d orbitals are filled singly first up to
manganese. At manganese the d sub-shell is
now exactly half filled. There is an extra
stability associated with an exactly half filled
sub-shell; this stability is a result of the
symmetric distribution of charge (electrons).
After manganese, that is, from iron and
onwards, electron pairing begins until the
orbitals are fully occupied at copper and zinc.
There is also extra stability associated with a
fully occupied sub-shell.
The configurations of chromium and copper are unexpected. We would have expected the
following configuration for copper (atomic number 29)
1s22s22p63s23p63d94s2.
What actually happens is that an electron is unpaired from the 4s sub-shell and enters the 3d subshell. This results in the formation of an exactly filled d sub-shell (with 10 electrons), which is a
stable configuration. The same thing happens at chromium, atomic number 24. The expected
configuration for Cr is 1s22s22p63s23p63d44s2. The actual configuration is 1s22s22p63s23p63d54s1
As in copper, an electron leaves the 4s sub-shell and enters the 3d sub-shell. This creates
an electronic configuration in which the d sub- shell is exactly half filled. Such a configuration is
relatively stable.
The d4 and d9 configurations, expected in Cr and Cu respectively, do not exist. They are replaced by the
relatively stable d5 and d10 configurations respectively. The 4s sub shell in Cr and Cu is therefore occupied
by a single electron. The rest of the first row transition elements have the general configuration 4s 23dn.
2.2.5
The short-hand notation
The electronic configuration of an element can be abbreviated as shown below.
The rest of the
configuration ( the noble elements)
do not have this part.
Core: noble gas
configuration
The table 2.4 shows the short-hand notation for some elements.
Page
42
element number of
electrons
full configuration
abbreviated configuration
Li
3
1s22s1
He 2s1
Be
4
1s22s2
He 2s2
B
5
1s22s22p1
He 2s22p1
Na
11
1s22s22p63s1
Ne 3s1
P
15
1s22s22p63s2 3p3
Ne 3s2 3p3
K
19
1s22s22p63s2 3p64s1
Ar 4s1
Ca
20
1s22s22p63s2 3p64s2
Ar 4s2
Sc
21
V
23
Cr
21
1s22s22p63s2 3p63d54s1
Cu
21
1s22s22p63s2 3p63d94s1
1s22s22p63s2 3p63d14s2
1s22s22p63s2 3p63d34s2
1
2
Ar 3d 4s
Ar 3d34s2
Ar
3d54s1
9
1
Ar 3d 4s
Table 2.3 Short hand
configuration for some
elements
Summary






2.2.6
Electrons enter the orbitals starting with those of the lowest energy. Only two electrons may enter
an orbital and they must be spin paired. The orbitals in any shell are filled in the order s, p, d, f.
The electronic configuration framework gives the order in which electrons enter the atom: 1s 2s
2p 3s 3p 4s 3d 4p 5s …
The coefficients in this framework represent the energies of the orbitals. The coefficient is known
as the principal quantum number. Orbitals of the same principal quantum number belong to
the same shell. However, as distance of the shells from the nucleus increases, the shells become
closer and closer to each other in energy, so that they may behave as a single shell.
A notation like s or p shows an orbital. If the notation includes a principal quantum number, e.g.
1s and 2p, it represents an energy level or a sub-shell.
In noble elements the outer most p sub-shell is exactly filled with the maximum possible number
of electrons, that is, 6. All in all, the valence shell (ns, np) has a total of 8 electrons (an octet). This
octet configuration renders stability to the noble gases.
In the transition elements, electrons enter the penultimate 3d sub-shell. Penultimate here means
‘underlying’, because in terms of energy, this shell is lower in energy than the occupied 4s subshell. This is only true when the d-orbitals are occupied. When the d-orbitals are unoccupied, they
are higher in energy than the 4s, for example in Ca.
Evidence for electronic configurations
1. Formation of ions
Consider the elements in Period 3, Na to Cl. The first three elements, Na to Al are clearly metallic in nature.
In their reactions, they tend to form positive ions (cations) by losing valence electrons. But why is it that
sodium will form +1 and not + 2 ions, and magnesium will form + 2 ions, not +3 ions?
It is only the valence electrons that participate in chemical reactions. Being the farthest from the nucleus,
they are the least stable. They feel only a weak attraction by the nucleus and so they are easily lost.
Consider the configuration of a sodium atom
1s22s22p63s1
There is only one electron in the valence 3s shell. When sodium reacts, it loses this electron to form a
positive ion with a charge of +1. Note that formation of the ion exposes a shell filled with eight electrons
(stable octet). The cation formed therefore has the same stability observed in the noble gases. There is no
further tendency to lose an electron from the stable outer shell of the ion (Fig 2.9).
Na (Z=11)
electron lost
from valence shell
Na atom with one
electron in valence
Fig 2.9 Formation
shell(unstable)
of the Na+ ion.
Na+ion with a full
octet in the outer-most
shell(stable)
Page
43
Similarly, from the configurations of Mg and Al we deduce that only two and three electrons will be lost
respectively, forming the same stable core as in sodium (Fig 2.10).
Na
Mg
1s22s22p63s1
los
1s22s22p63s2
es
1e-
loses 2 e -
Al
2
2
6
2
1s 2s 2p 3s 3p
1
loses
3e
1s22s22p6
n+
stable neon core formed.
n= 1,2,3 respectively.
Fig 2.10
Note that the Na⁺, Mg2+ and Al3+ ions contain the same number of electrons ( they are isoelectronic).
2. Change in atomic radii across a period and down a group.
Atoms tend to shrink in size across the Periodic Table, but they increase in size down groups. Once more,
these observations can be explained by studying the pattern in electronic configurations down groups and
across periods.
Going across the periodic table, electrons enter the same shell. No new shells are opened, and so there is
no increase in the shielding effect. However, the number of protons in the nucleus increases across a
period. This increases the ability of the nucleus to attract shells inwards towards itself, resulting in a
decrease in atomic radii. Thus the sodium atom is much larger than the chlorine atom which is in the same
period. The change in atomic radii also influences the chemical properties of the elements. The sodium
atom, being large in size, will react by losing electrons because the outer-most shell, apart from being
sparsely populated, feels a weak attraction by the nucleus. On the other hand, chlorine, and indeed, non
metals, tend to react by gaining electrons, because the atoms are small so that the outer shell is closer to the
nucleus. Electrons in the outer-most shell feel a strong attraction by the nucleus. It is therefore difficult to
lose an electron from the outer shell. However, it is easy to gain an electron or electrons in the outer shell
because the incoming electron(s) feel a strong attraction by the nucleus.
Down the group, atomic radii tend to increase. This is because in moving from one element to the next down
a group, an additional shell is opened. The consequence of increasing atomic radii includes
 the tendency for elements to become more metallic down the group. It becomes easier for the
atoms to lose the valence electrons, which is what characterizes metals. A very good illustration of
this observation is found in the Group IV elements. Carbon is a non metal. Silicon and germanium
are metalloids, tin and leads are metals.
 Increasing reactivity of the metals down their respective groups. Metals react by losing valence
electrons. A metallic atom becomes more reactive as its size increases. This is because the valence
electrons are then a large distance from the nucleus and so they are only weakly held to the atom
by the attractive influence of the nucleus. T
 Increasing of the shielding effect down the Groups. The electrons in an atom are preveneted from
leaving the atom by the attractive influence of the ncuclues. Inner shells have a tendency of
preventing outer shell electrons from feeling the full attractive influence of the nucleus. This phenomenon is known as shielding. Going down a group, the number of shells increases, and so does
the shielding effect. The valence shell electrons are therefore more weakly held to the atom as the
group is descended.
Page
44
3. Ionization energies.
A formal definition of ionization energy will be given later on. For the time being, we will define it as a
measure of the amount of energy needed to knock out a valence electron from the atom of an element.
 In general, ionization energy increases across the periodic table. This is because proton number is
increasing and yet electrons are entering the same shell. The increase in proton number implies
that the outer shell electrons feel an increasing attractive force of the nucleus. The electrons become
more and more firmly held to the atom because the increase in the shielding effect is very small

when electrons enter the same shell. Increasing amounts of energy are therefore required to remove
an electron from an atom as proton number increases across the period.
However, the increase in ionization energy across the periods is not so straightforward. There are
some anomalies (irregularities) in this general trend, as illustrated for Period 3 in Fig 2.11.
I.E1
Mg
Na
Si
P
Cl
Ar
S
Al
Ar
Trend in first ionisation energy , IE 1 across the periodic table. IE1
is a measure of the energy required to remove an electron from
the outer - shell of an atom.
Fig 2.11 Trend in first
ionization energy across
Period 3.
(i) The ionization energies of aluminium and sulphur are lower than expected. These anomalies can be
explained in terms of the electronic configurations of the elements.
Mg
1s22s22p63s2
Al
1s22s22p63s23p1
Page
45
In Al, the electron to be removed comes from a p sub-shell which is higher in energy (farther from the
nucleus) than the 3s sub-shell from which an electron would be removed from P. It is therefore easier to
remove an electron from Al than from Mg, that is, Al has the lower ionization energy. Now compare the
electronic configurations of phosphorous and sulphur.
P
1s22s22p63s23p3
S
1s22s22p63s23p4
In sulphur, the electron to be removed comes from a p-orbital which is occupied by two electrons. The two
electrons in this orbital repel each other so that it becomes relatively easy to remove any one of them.
Compare with nitrogen, in which all 3p orbitals are singly occupied. Further more, the exactly half filled 3p
sub-shell in a phosphorous atom leads to stability of the sub-shell, so that it is not so easy to lose an electron
from it. Sulphur therefore has a lower ionization energy than nitrogen.
Ionization energy decreases down a group. As additional shells are opened, the outer shell electrons
become further and further from the nucleus. They feel a decreasing attractive force of the nucleus and so
can more easily be removed.
 Ionization energy decreases down a group. As already explained, atomic sizes increases down a
group due to increasing number of shells. This has two effects which both lead to a reduction in
ionization energy.
(i)
As the number of shells increases, the valence electrons experience more shielding from
the inner shells. The valence electrons therefore feel a weaker attraction from the nucleus.
(ii)
As the atom becomes larger, so does the distance of the valence electrons from the nucleus.
Once more, the valence electrons feel a weaker attraction from the nucleus. It is therefore
easier to remove an electron from the outer most shell as a group is descended.
The graphic in Fig 2.11 illustrates the fact that ionization energies increase across a period but decrease
down a group (or increase up a group).
Fig 2.11 Trend in ionization
energy across periods and down
groups.
Successive ionization energies
Removal of an electron from an atom results in the formation of a +1 cation. The energy required for this
conversion is measured by the first ionization energy (IE1). If a second electron is removed from the +1 ion,
a +2 ion is formed. The energy required to convert a +1 ion into a +2 ion is measured by the second
ionization energy (IE2) for that element. Similarly, the energy required to convert a +2 ion into a +3 ion is
the third ionization energy (IE3). IE1, IE2, IE₃ e.t.c are known as successive ionization energies for the
element under consideration. It is possible to deduce the position of an element in the Periodic Table by
studying its consecutive ionization energies.
Example 1
The successive ionization energies of an element Q are given below. Deduce, as far as possible, the
position of the element in the periodic table.
1st
Ionization energy/ Kjmol-1
2nd
3rd
4th
5th
6th
7th
8th
736 1450 7740 10 500 13 600 18 000 21 700 25 600
Solution and discussion
First notice that successive ionization energies have an increasing trend. This is expected. Suppose that an
atom X loses an electron to form the ion X⁺. It is going to be harder for the ion to lose another electron
to form the +2 ion.
Page
46
X⁺ → X2+ + e
This is because the second electron to be lost must come from a positively charged particle to which it is
naturally attracted. An electron naturally resists being removed from a centre of positive charge. For this
reason, the second ionization energy of an element is higher than the first.
Similarly, the third, fourth (e.tc.) ionization energies increase in that order because as the size of the positive
charge on the ion Xn+ increases, the electron to be lost becomes more and more strongly held to the ion. For
example, the third
ionization energy involves the process
X2+ → x3+ + e
and a very large amount of energy is required to knock out an electron from the +2 ion.
However, if you study the values given in the table above, you will find that there are sudden ‘jumps’ in the
ionization energies at some points. These sudden increases in energy can not be explained simply in terms
of an electron finding it harder to leave an ion with a large positive charge.
A change of the shell from which the electron is removed causes such abrupt changes in energy.
Successive ionization energies can therefore be used to deduce the energy levels (shells and sub-shells) in
an atom, and in turn this information can be used to deduce the position of the element in the periodic table.
In the given table, the sudden increase in energy is at the third ionization energy. This means that the first
two electrons removed came from the valence shell, but the third came from a new shell, which is closer to
the nucleus. Since the element has two valence electrons, we conclude that it is in group II.
If the first sudden increase in ionization energy is at the nth ionization energy, then the element must
be in Group (n-1).
In the given example, the first sudden increase in ionization energy was at the 3 rd ionization energy, so the
element must be in group (3-1) =2.
The element can not be Beryllium due to the fact that Be has too few electrons compared to the number of
electrons lost from the atom Q.
Example 2
Successive ionization energies are given below for an element M. Deduce, as far as possible, the
position of the element in the periodic table.
1st
-1
Ionization energy/ Kjmol
2nd 3rd 4th
1310 3390 5320 7450
8th
5th
6th
7th
11 000 13 300 71 000 84 100
Page
47
Solution and discussion
The first large ‘jump’ in ionization energy is at the 7th value. The element must Be in group VI. However,
there is evidence of the existence of sub-shells. The fifth ionization energy shows a relatively large jump
compared to the values preceding it. However, it is not large enough to lead us to a conclusion that the 5 th
electron is coming from a new shell. Assuming that the element is indeed in Group VI, and that it has six
electrons in the outer- most shell, then four of these six valence electrons must be in the p sub-shell.
When all these four electrons have been lost, the next two electrons must come from a new sub- shell (the
s), as shown by the relatively large jump in the fifth ionization energy. However, the fifth and sixth electrons
are still coming from the valence shell (a different sub shell of the same shell), so the change in ionization
energy is not too high.
[Type a quote from the document or the summary of an interesting point. You can position the text
Exercise
box anywhere
2 : in
miscellaneous
the document. Useproblems
the Text Box Tools tab to change the formatting of the pull quote
text box.]
1
Sir James Jeans, who was a great populariser of science, once described an atom of carbon as being
like six bees buzzing around a space the size of a football stadium.
(a)
(i)
Suggest what were represented by the six bees in this description.
(ii)
Explain (in terms of an atom of carbon) what stopped the bees from flying away from the
space of the football stadium.
(iii)
What is missing from Jeans’ description when applied to an atom of carbon?
(b)
The diagram below represents the energy levels of the orbitals in atoms of the secondperiod,
lithium to neon.
(i)
(ii)
In the space below, sketch the shapes of the two types of orbital.
(iii)
Complete the electron configurations of nitrogen and oxygen on the energy level diagrams
below, using arrows to represent electrons.
(iv)
Explain, with reference to your answer to (iii), the relative values of the first ionization
energies of nitrogen and oxygen. The values are given in the Data Booklet and should be
quoted in your answer.
(c) (i)
(ii)
2
Label the energy levels to indicate the principal quantum number and the type of orbital at
each energy level.
State the formulae of the negatively charged ions formed by these elements in simple binary
compounds (nitrides and oxides).
Why do nitrogen and oxygen form negative ions, but not positive ions, in simple binary
compounds?
9701/04/M/J/07
Which isotope of an element in the third period of the Periodic Table contains the same
number of neutrons as S?
Page
48
A Na
3
B Mg
C Si
D P
The successive ionisation energies, in kJ mol–1, of an element X are given below.
870
What is X?
1800
3000
3600
5800
7000
13 200
A
4
33As
B
40Zr
C
52Te
D
53I
Unnilpentium is an artificial element. One of its isotopes is 105Unp262
Which of the following statements is correct?
5
A
262
105Unp
has a nucleon number of 105.
B
The atom 105Unp260 is an isotope of 105Unp262.
C
There are 262 neutrons in 105Unp262.
D
The proton number of 105Unp 262 is 262.
The table gives the successive ionisation energies for an element X.
What could be the formula of the chloride of X?
A XCl
6
B XCl2
B 3s 3p 4s 3d
C 3s 4s 3p 3d
D 4s 3s 3p 3d
Which of the following particles would, on losing an electron, have a half-filled set of p orbitals?
A C–
8
D XCl4
What is the order of increasing energy of the listed orbitals in the atom of titanium?
A 3s 3p 3d 4s
7
C XCl3
B N
C N–
D O+
9701/04/O/N/2003
Use of the Data Booklet is relevant to this question.
It is now thought that where an element exists as several isotopes, the stable ones usually
contain a ‘magic number’ of neutrons. One of these magic numbers is 126.
Which isotope is unstable?
A
B
208Pb
C
210Po
D
208Tl
9701/04/O/N/2003
An atom has eight electrons. Which diagram shows the electronic configuration of this atom in
its lowest energy state?
Page
49
9
209Bi
9701/04/O/N/2005
10
The first six ionisation energies of four elements, A to D, are given.
Which element is most likely to be in Group IV of the Periodic Table?
9701/04/M/J/2005
11
What is the electronic configuration of an element with a second ionisation energy higher than
that of each of its neighbours in the Periodic Table?
A 1s22s22p63s2
B 1s22s22p63s23p1
C 1s22s22p63s23p2
D 1s22s22p63s23p3
9701/04/O/N/2005
12
Gallium nitride, GaN, could revolutionise the design of electric light bulbs because only a small
length used as a filament gives excellent light at low cost. Gallium nitride is an ionic compound
containing the Ga3+ ion.
What is the electron arrangement of the nitrogen ion in gallium nitride?
A 1s2 2s2
13
B 1s2 2s2 2p3
C 1s2 2s2 2p4
D 1s2 2s2 2p6
9701/04/M/J/2006
Use of the Data Booklet is relevant to this question.
The electronic structures of calcium, krypton, phosphorus and an element X are shown.
Which electronic structure is that of element X?
Page
50
14
A 1s22s22p63s23p3
B 1s22s22p63s23p64s2
C 1s22s22p63s23p63d64s2
D 1s22s22p63s23p63d104s24p6
9701/04/O/N/2006
The graph shows the first thirteen ionisation energies for element X.
What can be deduced about element X from the graph?
A
B
C
D
15
It is in the second period (Li to Ne) of the Periodic Table.
It is a d-block element.
It is in Group II of the Periodic Table.
It is in Group III of the Periodic Table.
9701/04/O/N/2007
The first seven ionisation energies of an element between lithium and neon in the Periodic Table
are as follows.
1310
3390
5320
7450
11 000
13 300
71 000 kJ mol–1
What is the outer electronic configuration of the element?
A 2s2
16
B 2s22p1
C 2s22p4
D 2s22p6
9701/04/M/J/2009
Use of the Data Booklet is relevant to this question.
The elements radon (Rn), francium (Fr) and radium (Ra) have consecutive proton numbers in
the Periodic Table.
What is the order of their first ionisation energies?
9701/04/M/J/2009
17
A simple ion X+ contains eight protons.
What is the electronic configuration of X+?
A 1s2 2s1 2p6
B 1s2 2s2 2p3
C 1s2 2s2 2p5
D 1s2 2s2 2p7
9701/04/O/N/2010
Page
51
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
CHAPTER
3
CHEMICAL BONDING
Introduction
We live in a very stable universe. Matter does not just start falling apart around us. From this
observation alone we might begin to suspect that particles of matter are held together by strong forces
which prevent them from falling away from each other.
Indeed, these forces exist, and they are known as chemical bonds. The concept of chemical bonding in
chemistry is important as it helps to predict the relative stability, and hence reactivity, of matter. If the
atoms of a substance are held together by very strong chemical bonds, the substance is relatively stable
and is not easily converted to other substances.
There are three ways in which atoms may bond to each other.
(i) Covalent bonding
(ii) Ionic (electrovalent) bonding
(iii) Metallic bonding
A recurring concept in this topic is this:
All bonding (not just ionic) is electrostatic in nature
Type of bonding
Covalent
Nature of electrostatic
attraction
Attraction between electrons
located between two atoms of
a bond and the nuclei of both
Page
52
atoms of that bond.
Metallic
Attraction between a sea
of delocialised electrons and
positive metal atoms.
Ionic
Attraction between positively
and negatively charged ions.
3.1 Covalent bonding
A covalent bond is formed by sharing of one or more electron pairs between two atoms. The simplest
example we can study is bonding in the hydrogen molecule (Fig 3.1). For a covalent bond to form between
two atoms, the appropriate orbitals from both atoms must approach each other and overlap. The electron
pair(s) contained in the overlap region is the covalent bond.
Why bonding occurs
By bonding, atoms achieve stability. For instance, a hydrogen atom is very unstable and does not have an
independent existence because of the single electron present in the outer shell. By bonding with another
atom, two things are achieved:


The electron is paired up with another. Electrons tend to be more stable when they are paired,
because by pairing their magnetic moments cancel each other out (remember that when electrons
pair up they do so with opposite spins).
Each atom then has the stable configuration of a noble gas in its outer-most shell. In hydrogen,
each atom now has two electrons (a duplet) in the outer shell, just as in helium. In other molecules,
each atom of a bond strives to achieve an octet configuration in its outer-most shell, as illustrated
for the chlorine molecule in fig 3.2.
Bonding in hydrogen
Hydrogen has a single electron in an s orbital. The s orbitals of two atoms approach each other and continue
to do so until they have sufficiently overlapped, as illustrated in Fig 3.1 below. This results in the formation
of a σ bond.
Page
53
Fig 3.1 sigma bonding
in the hydrogen
molecule
Bonding in the chlorine molecule
Each chlorine atom has seven electrons in the valence shell and is very unstable. As in hydrogen, a chlorine
atom has no stable independent existence. It tends to pair up with another chlorine atom, forming a
diatomic molecule, Cl2.
..
.. .
..
XX
X
X
X
XX
Cl [ 3s23p5]
Cl [ 3s23p5]
. ..
.. .



XX
X
XX
X
X
The valence shell configuration of each Cl atom is 3s23p5, giving
a total of 7 valence electrons. Each atom therefore has an odd (un
paired) electron.
When the Cl-Cl bond forms, the odd electrons are paired up.
Each atom now contains a stable octet of electrons in the outermost shell.
In the Cl2 molecule, each atom has three non-bonding pairs of
electrons (lone pairs). The two electrons shared between the two
atoms is known as a bonding pair.
Fig 3.2 Sigma bonding in
chlorine
The following account explains the bonding in Cl 2 in terms of the valence shell electronic configuration of
each atom of the bond.
Page
54
3.2 Sigma (σ) and pi (π) bonding
Formation of a covalent bond involves overlap of atomic orbitals. There are two possible orientations of
overlap.
 In sigma(𝜎) bonding, orbitals over-lap head-on.

In pi(𝜋) bonding, orbitals overlap side-ways.
Sigma overlap has already been illustrated above for chlorine. Fig 3.3 illustrates some orbital combinations
that may lead to the formation of σ and π bonds.
 bonding is possible between ...
nucleus
...two px orbitals,
...or between a p orbital
(px, Py , pz) and an s orbital.
 bonding is possible between ...
...two pz orbitals
Fig 3.3
... or between two p y orbitals.
Formation of pi and sigma bonds
Bonding in oxygen (O2)
O : 1s22s22p4
The 2s and the 2p sub-shells make up the valence shell. There are 6 electrons in this shell, two short of an
octet. Each atom therefore needs two more electrons in its valence shell, and this can be achieved through
the sharing of two valence electrons between the two atoms. In each O atom, there are two p orbitals which
are singly occupied. It is these orbitals which overlap, forming a double bond between the atoms. Two
orbitals overlap sideways (π) whilst the other two overlap head-on (σ) as illustrated in Fig 3.4(a) and (b).
Page
55
Fig 3.4 (a)
Fig 3.4 (b)
The bond between the two oxygen atoms therefore contains four electrons. It is a double bond, in which one
bond is σ and the other is π. Fig 3.4 below illustrates formation of a π bond by the overlap of two pz orbitals.
Fig 3.4
The formation of a π bond in O2 by the overlap of two pz orbitals
How does a π bond differ from a σ bond?
On average, and for any given bond, π electrons are a greater distance from the nucleus than σ electrons. π
electrons are therefore diffuse(spread out) and relatively weakly held to the nucleus. The π bond is therefore
weaker, longer and is more easily broken than the σ bond. When a double bond is subjected to a strain
strong enough to cause bond breaking, it will always be the π bond that breaks first.
Organic compounds which contain the carbon to carbon double bond (alkenes) are quite reactive due to the
presence of the π electrons. The diffuse nature of the π bond and the fact that the π bond is relatively weakly
held by the nucleus makes the C=C bond an electron rich centre. Such a centre reacts by attracting electron
deficient species known as electrophiles. Because it is relatively weak, a π bond can not form a single
bond. Consider a chlorine molecule which contains a single bond. This bond can not be π; it would be weak
and will cause the molecule to be unstable.
Page
56
All single bonds are sigma
Bonding in nitrogen (N2)
N: 1s22s22p3
In N, there are 5 valence electrons, contained in the 2s and the 2p sub-shells. The electrons in the 2s subshell are already paired up, so they will not be involved in bonding. They will remain as a non bonding
(lone) pair of electrons. Each of the three 2p orbitals is singly occupied. Thus all three will overlap and share
their single electron with the corresponding orbital on another atom. That is, the p x of one atom overlaps
with the px of the other atom (σ), the py of one atom overlaps with a py of the other atom (π). The pz orbitals
will do the same (π). The result is that each atom shares three electrons with the other atom, forming three
shared pairs of electrons, that is, a triple bond (Fig 3.5).
Fig 3.5
bonding in
nitrogen, N2
Important points to remember about π and σ bond formation

If a bond is single, as in Cl2, then it is sigma.

In a multiple bond (double or triple), only one of the bonds can be sigma. The rest are π.
3.3
The concept of hybridization
The methane molecule is drawn showing four covalent bonds around the C atom:
H
H C H
H
However, this does not agree with what we would expect from the electronic configuration of C:
Page
57
C: 1s22s22p2
The configuration shows that the valence shell of C has two sub-shells, the 2s and the 2p. The 2s, being fully
occupied, may not participate in bonding. This leaves the 2p orbitals. But there are only two occupied p
orbitals which may participate in bonding, so we would expect C to form only two covalent bonds in its
compounds, for example, CH2 would be formed instead of CH4. The concept of hybridization has been used
to explain why carbon forms four covalent bonds in its compounds, not two.
3.3.1 Hybridization in methane
Here is the ground state configuration of a C atom:
Before hybridization occurs, an electron is excited from the filled 2s sub-shell to the vacant 2pz orbital. This
creates four singly occupied orbitals, which may participate in bonding.
But before they overlap with orbitals from other atoms during bond formation, these four singly occupied
orbitals mix mathematically to form four equivalent hybrid orbitals. The hybrid orbitals have shapes which
ensure maximum overlap. The hybrids formed here are known as sp³ hybrid orbitals because they are
formed from the mixing of one s orbital and three p orbitals. Each hybrid orbital is occupied by a single
electron (Fig 3.6).
In methane, each hybrid orbital on the C atom overlaps
head-on with the 1s orbital of a hydrogen atom. Since
there are four hybrid orbitals, the carbon atom is able to
bond with four hydrogen atoms. Hybridization moulds
the hybrid orbitals so that they can overlap efficiently
with the 1s orbital of hydrogen, forming four sigma bonds
around the C atom.
Fig 3.6
Page
58
Why hybridization occurs
Hybridization is the mathematical combination of orbitals to form hybrid orbitals. Before this happens, an
electron must be excited from the 2s to the the pz orbital. This requires an expenditure of energy since the
2s sub-shell is lower in energy than the 2p sub-shell. However, the energy used is small since the 2s and the
2p sub-shells belong to the same shell (energy level) and have similar energies. The energy used is more
than compensated for by the formation of a larger number of covalent bonds, for example, in CH4. Bond
formation has the tendency of stabilizing molecules, so the more bonds that can be formed, the more stable
the compound becomes. Thus methane is more stable than the compound CH 2 that would be formed if
hybridization did not take place. Hybridization also modifies orbitals, creating hybrids whose shape, energy
and orientation allows efficient overlap with orbitals on other atoms.
NB The compound CH2 (methylene) does exist as an intermediate in certain types of reactions, but it is very unstable.
Its structure is shown below.
H
C
H
3.3.2 The bonding in ethene
A molecule of ethene is usually displayed as shown below.
Once more, we have to reconcile this structure with the electronic configuration of
a C atom. In ethene, there is a carbon to carbon double bond, made up of a π bond
and a σ bond (recall that in a multiple bond, there is always one σ bond). When C
forms four sigma bonds as in methane, it requires the use of four hybrid orbitals.
To form three sigma bonds as in ethene, three hybrid orbitals are required.
This implies that each carbon atom must use three bonding orbitals to make these hybrid orbitals. (Number
of hybrid orbitals used by an atom in bonding is equal to number of atomic orbitals that mix to form the
hybrid orbitals). In ethene, each of the two carbon atoms is hybridized as shown below.
Formation of ethene in terms of orbital overlap takes place in this way:
Page
59



Each carbon atom has three sp2 hybrid orbitals and an unhybridized pz orbitals for use in bonding.
A hybrid orbital on one C atom overlaps head-on with a hybrid orbital on the other C, forming a C
to C sigma bond.
A second hybrid orbital on each carbon atom overlaps head-on with the 1s orbital of H, forming a
C to H sigma bond.
The third hybrid orbital on each C overlaps head-on with the 1s orbital of another H forming
another C to H sigma bond. The three hybrid orbitals on each carbon atom are therefore used in σ
bond formation, resulting in the formation of a σ skeleton. The sigma skeleton only shows sigma
bonds (Fig 3.7).
H 1s
H 1s
sigma skeleton
sp2
C
2
sp2
C
H 1s
H
C
H
...resulting in the formation of a
sigma skeleton. The three sigma
covalent bonds are spaced out at an
angle of 1200 relative to each other
to minimize repulsion between
them.
Fig 3.7 Formation of
C-C and C-H sigma bonds
in ethene.
The pz orbital on the carbon atom is unhybridized and perpendicular to the pz orbital on the other
carbon atom. The two pz orbitals therefore overlap sideways to form a C to C π bond. The C to C
double bond in ethene therefore contains one sigma and one pi bond, as illustrated in Fig 3.8.
Fig 3.5
pi bond formation in ethene
3.3.3 Bonding in ethyne, C2H2
Ethyne has the structure shown below.
60
H
H 1s
The three hybrid orbitals on each
carbon atom overlap in sigma bond
formation ...
Page
1200
C
2
sp
sp2

H
sp
sp2
H C
C H
It contains a C to C triple bond. In this multiple bond, only one is sigma. The other two are pi bonds. In
methane, carbon forms four sigma bonds and it is sp3 hybridized. In ethene, carbon forms three sigma
bonds and it is sp2 hybridized. We can guess that in ethyne, where each carbon atom forms only one sigma
bond, there is sp hybridization as illustrated below.
Page
61

HC  CH

3.3.4 Bonding in carbon dioxide
Carbon has the valence structure:
… 2s22p2,
whilst oxygen has
… 2s22p4
There are therefore two singly occupied p orbitals in oxygen which can be used in bonding.
The two electrons in the 2s orbital of carbon are paired and can not participate in bonding unless they are
separated. The separation does happen, followed by hybridization, as explained before. There is sp
hybridization, as in ethyne. This results in the formation of two hybrid orbitals, leaving two orbitals on C
unhybridized (the py and the pz). The two unhybridized orbitals are at right angles to each other. There are
therefore a total of 4 orbitals to be used in bonding by the carbon atom, two sp hybrids and two unhybridized
orbitals.




One hybrid orbital overlaps head-on with a bonding p-orbital on one oxygen atom, forming a C-O
sigma bond.
The other hybrid orbital overlaps head - on with a p - orbital on the second oxygen atom,
forming the second C-O sigma bond. All hybrid orbitals on carbon are thus accounted for. This
leaves the two unhybridized py and pz orbitals.
One unhybridized orbital, the pz, overlaps sideways with the remaining bonding p-orbital on an
oxygen atom, forming a π bond.
The second unhybridized orbital on carbon overlaps sideways with the remaining p-orbital on the
second oxygen atom, forming another π bond. Each of the two C to O bonds therefore contains a
π and a σ bond (Fig 3.5).
Fig 3.5
in CO2
3.4
Bonding
Dot and cross diagrams
Page
62
We have described the bonding in some simple covalent molecules, including their dot and cross
diagrams. From now and onwards we put emphasis on the dot and cross diagrams, and hope that you can
 work out and describe the bonding in terms of the overlap of appropriate
atomic orbitals.
 deduce the numbers and nature of bonds (single ,multiple, dative) present in a molecule.
 determine the total number of electrons used by atoms in a molecule for bonding.
Guidelines for drawing dot and cross diagrams
1. Identify the central atom, for example, in methane, CH₄, there is a single atom of C. This becomes the
central atom.
2.Deduce the number of valence electrons for the central atom.
The number of valence electrons is equal to group number
for example, C in group (IV) has 4 valence electrons.
3. Not all of the valence electrons on the central atom may be used in bonding. You have to decide how
many electrons will be used for bonding. To do this, you need to find out how much short the valence shell
is of an octet, for example, carbon, with a valence of 4 electrons, is short by four electrons.
4. Repeat steps 2 and 3 for the other atoms present in the molecule. Consider the carbon dioxide molecule.
The central atom, C, will use all four of its electrons in bonding, as already explained. O (group VI) has six
valence electrons, two short of an octet. It will therefore use two of its valence electrons for bonding. This
will leave out four electrons unused on O. These four electrons will form two pairs of non bonding
electrons, also known as lone pairs.
5. Draw the dot and cross diagram, making sure each atom is using the correct number of electrons to pair
with electrons on the central atom, as worked out in 2 and 3 above.
You may need to do some ‘electron book-keeping’ to make sure you have used the correct number of
electrons. The total number of electrons shown in the dot and cross diagram should be equal to the sum of
the valence electrons of all the atoms used in bonding. For example, book-keeping for methane:
number of
valence electrons
C
H
4
1x4 =
Total
Check: there should be a total of 8 electrons in the dot and
cross diagram for methane.
‘Electron book-keeping’ becomes particularly useful when drawing
the dot and cross structures for ions.
4
8
Book-keeping for methane
Important exceptions to the guidelines given so far
Page
63
1. Some elements do not have enough valence electrons to satisfy an octet in their valence shell, even
when they share their valence electrons with other atoms in the formation of covalent bonds. Common
elements in this category are Be, B and Al. With only three valence electrons, the maximum number of
bonds that Al can form is three, as in AlCl3, which gives a sextet (set of six electrons) around the central
atom.
2. Some non- metallic elements can form more than four bonds around the central atom, that is, they are
able to exceed or expand the octet. Examples in this category include P, S, Cl, Br and I. Note that all
of them are in period 3 or beyond. Elements in periods 1 and 2 do not have the ability to exceed an octet,
as explained later on.
3. In odd electron systems it is not possible to pair up all electrons without leaving a single unpaired
(odd) electron in the outer most shell. An example is NO which has a total of 11 valence electrons.
Examples: dot and cross diagrams
1. CH4 (methane)

Central atom is C. It has 4 valence electrons and will use them all in bonding.

Hydrogen has one valence electron. It does not have to satisfy an octet. It only has to satisfy a
duet of electrons in its outer shell, because this will give it a stable Helium- like structure.
Hydrogen therefore uses the one available 1s electron, sharing it with an electron from C.


H
X
H X C XH
X
H
2.
The structure therefore has four single C-H (sigma) bonds.
Similar compounds formed by the group IV elements (such as
SiCl4) will have the same structure, because all of them, like C,
have 4 valence electrons.
H2O
O
X
H
The oxygen atom needs two electrons more to compete an octet. It will therefore
use two valence electrons for bonding. The water molecule has two bond pairs and
two lone pairs (non-bonding pairs) of electrons.
XH
Note
If asked in an exam to draw the dot and cross diagram of a molecule, you should not leave out lone
pairs, if any.
3.
NH3 (ammonia)

N, in group five, has five valence electrons. It needs three more electrons.
N
X
H
4.
XH
XH
Aluminium, in group (III), has 3 valence electrons. It needs five more electrons to satisfy an octet,
but there is no way it can obtain 5 electrons from 3 Cl atoms.
.. .
..Cl
..
x
64

Therefore it uses 3 of the valence electrons, sharing each of them with an
electron on H. This leaves a lone pair of electrons on the N atom.
The molecule PH3 is analogous to NH3. They are both group V hydrides.
AlCl3

Page

..Cl.x
.. ..
..
..
Al x.Cl..
Al therefore uses all three valence electrons, each being paired with an electron
from Cl. Chlorine, in group (VII), is one electron short of an octet. It will
therefore use one electron for bonding, pairing it with an electron on Al.
NB In the following examples we will omit lone pairs (non-bonding pairs) but
remember that they should always be included in dot and cross diagrams.


5.
This leaves six electrons (3 lone pairs) unused on each Cl. Three Covalent bonds are formed around
Al. This means that even after forming covalent bonds, Al still has not satisfied an octet. It lacks
two electrons to fill an octet.
AlCl3 is an example of an electron deficient molecule. This case is peculiar to molecules formed by
the group (III) elements, for example, boron triflouride (BF3) is also electron
deficient.
AlF3 is the halide analogue of AlCl3. However, it is predominantly ionic, so it is outside the scope of
our current discussion on covalent bonding.
BeCl2


Being in group two, Be needs 6 electrons to satisfy an octet. It therefore must share 6 electrons,
which is clearly impossible, since it has only two valence electrons. All it can do is form the
maximum possible number of bonds (two), by sharing each of its valence electrons with an electron
from Cl.
Beryllium chloride therefore has two bonds, that is, four electrons around beryllium, four short of
a valence. This is another example of an electron deficient molecule.
Cl X Be
6.
Sulphur hexafluoride (SF6)

Sulphur, like O, is in Group VI. It is therefore expected to use only two electrons for bonding,
pairing each of them with a fluorine atom. If this was the case, then the compound formed will be
SF2. In practice, S is able to use all of its 6 valence electrons for bonding. F, in group 7, needs one
more electron to satisfy an octet. Each of the six F atoms will therefore provide an electron to share
with each of the six valence electrons of S. This forms 6 bonds, or a total of 12 electrons around S.
This does happen because S is able to expand the octet. It has accessible d orbitals which it can use
to accommodate some of the excess electron density.
F
FX
X
XF
S X
FX X F
F



65
7
Page
X Cl
The term ‘accessible’ means that the vacant d orbitals are close in energy to the
occupied valence shell of Sulphur, so it is possible to involve them in bonding.
Electrons can be promoted into the d orbitals with a little expenditure of energy.
In fact the energy used to promote electrons to the 3d orbitals is so small that it
is more than compensated for by the extra stability attained by the formation of
extra bonds.
Ability to expand the octet starts from P in Period 3. Other non- metal elements after P can also
expand the octet, e.g. in the molecule ICl6.
Expansion of an octet results in the formation of more bonds than would be possible if an octet was
satisfied. The formation of extra bonds gives extra stability to a molecule. Remember that it is the
bonds that hold atoms together in a substance. The more bonds there are, the more stable the
substance.
The earlier members of the periodic table, that is, the non- metals in Period I and II can not expand
the octet because the d orbitals are not accessible to accommodate extra electron density.
PCl3

Phosphorous, in group five, needs three more electrons to fill the octet. P will therefore use three
of its valence electrons, leaving out two electrons, that is, one lone pair, unused. Cl needs one more
electron, so each chlorine atom will pair with an electron from S.
The molecule nitrogen trichloride (NCl3) is analogous to PCl3. Both are the trichlorides of a group
V element.
Cl
X
P X
Cl
X
Cl
Cl
X
N X
Cl
X
Cl
An octet is satisfied for both the P and the Cl atoms.
8.
PCl₅ (Phosphorous pentachloride)
Cl
X X
PX
X
X
Cl
Cl
Cl
Cl
As already explained, P and the latter non- metallic members of the
periodic table can expand the octet .This occurs, for example, in PCl₅. In
forming an octet, the phosporous atom uses all of its valence electrons so
that no lone pair is left on it.
3.4.1 Odd electrons and dative bonding
X
XNX O
OX X
Consider the dot and cross diagram for NO2 shown on the left.
The total number of electrons to be accounted for is 17.
In compound formation, electrons always seek to be paired, because
this makes them stable. However, if the total number of valence
electrons in the molecule is odd, as in NO2, one electron in the
molecule is bound to be left unpaired. Such an electron is known as
an odd electron. In NO2, the odd electron is on the nitrogen atom,
valence electrons
N
5
O 6x2
12
17
since it is the one which contributes an odd number of valence electrons.
A chemical species which bears an unpaired (odd) electron is known as a
free radical.
Nitrogen monoxide (NO) and halogen atoms are other examples of free radicals. Check that the total
number of valence electrons is odd in each species.
In all the examples given so far, it has been assumed that the covalent bond is formed by an equal
contribution by both atoms of the bond, that is, each contributes an electron to the bond. In NO 2, we see an
exception. In one of the N-O bonds, both electrons are contributed by one atom, that is, N. A bond formed
in this way is said to be dative or co-ordinate and its presence can be emphasized using the notation →.
The direction of the arrow shows which atom has contributed the electrons of the bond.
dative bond
odd electron
Page
66
X
O N XO
N
O
O
The dative bond avoids having to include a second
N to O double bond. The following structure (I),
containing two double bonds and no dative bond,
could have been drawn for NO2.
The structure might look appealing at first sight because it agrees with ‘electron bookkeeping’ considerations. 17 electrons are used as before. More-over, there are two
double bonds, so the structure is likely to be stable, since double bonds are stronger
than single bonds.
X
XNX
O
O X X
However, notice that the structure can not be correct because now nitrogen has too many electrons around
it. By including two double bonds, nitrogen has exceeded its octet, which is not possible for elements in
periods 1 and 2. To avoid this, one of the bonds should be dative.
Draw the dot and cross diagram for the CO molecule
Q
A
Carbon monoxide presents a rare example of a multiple bond containing a dative bond.
Remember that a dative bond is a way of preventing the exceeding of an octet by Period 2 elements. In CO,
oxygen donates a pair of electrons into a vacant p orbital on the carbon atom to form a dative bond. This
involves overlap of a fully occupied orbital (that is, containing two electrons) on the oxygen atom with the
vacant orbital on C.
Important note
Once formed, the dative bond becomes indistinguishable from all the other covalent bonds in a molecule;
it has no special properties, and so in many cases when we write down the structures of molecules, we do
not need to show dative bonds, but you need to be aware of their existence.
Consequence of an odd electron in a chemical species
An unpaired electron is highly unstable and it will strive to achieve stability by pairing up with another
electron from a suitable reagent. This makes free radicals quite reactive.
The dimerization of NO2
Dimerization is a chemical reaction in which two identical molecules join chemically to form
one molecule whose Mr is twice the Mr of each reactant molecule.
An NO2 molecule tends to dimer with another, because in so doing, the odd electron on the nitrogen atom
of both molecules becomes paired up, forming a more stable compound (Fig 3.6). Thermodynamic data
show that the product of this reaction, dinitrogen tetraoxide, is more stable than NO2.
O
N
Page
67
O
O
O
+
N
N
O
2NO2
O
N2O4
O
N
O
Fig 3.6
of NO2
dimerization
Free radicals such as the oxides of nitrogen have been largely blamed for the atmospheric reactions that
lead to the depletion of ozone. Another important free radical in the dynamic processes that destroy the
ozone layer is the chlorine atom. Since chlorine has an odd number of electrons in its outer-most shell, it
contains an odd electron in the valence shell.
Isoelectronic molecules
Molecules are said to be isoelectronic if they have the same total number of electrons.
Examples are N2 and CO. Both contain a total of 14 electrons. Isoelectronic molecules also contain the
same total number of valence electrons. Note that both CO and N2 contain a total of 10 valence electrons.
It is therefore not surprising that the bonding in isoelectronic molecules often shows some similarities.
Compare the bonding in N2 with that in CO and note that both contain a triple bond. However, there is no
dative bond in the nitrogen molecule.
3.4.2
Consequence of multiple bonds
Multiple bonds are stronger than single bonds, as illustrated in the Table 3.1.
Bond
Q
A
bond energy/Kj mol-1
C
C
348
C
C
C
C
612
837
Bond energy is a measure of the amount of energy
required to break a bond. In other words, it gives
information about the strength of the bond. The larger
the bond energy, the stronger the bond.
Table 3.1
Use Table 3.1 above to estimate the bond energy of the π bond in the C to C double bond.
Since the single C to C bond energy is 348Kj/mol. We would expect the C to C double
bond to have a bond energy of 348 x 2 = 969Kj/mol, assuming that the σ bond and the π
bond have equal bond energies. The actual value of the C to C double bond is smaller,
showing that the π bond has a smaller bond energy than the σ bond. The actual value of
the C to C double bond is smaller than the expected value by 696-612 = 84Kj. The π
bond is therefore weaker than the σ bond by 84Kj/mol, that is, it has a bond energy of
348 – 84 = 264Kj/mol.
Page
68
The calculation above shows that a π bond is weaker than the corresponding σ bond. This is expected.
Sideways overlap (π) is not as efficient as head-on overlap. Consequently, π electrons are, on average, a
larger distance from the nucleus than σ electrons. This implies that electrons of the π bond are relatively
weakly attracted to the nuclei of both atoms of the bond. This makes the bond relatively weak. When a
multiple bond participates in a reaction in which bond breaking has to take place, it is always a π bond that
breaks first. This happens, for example, in the reaction between bromine and ethene.
This reaction, known as electrophilic addition, involves breaking of the C to C π bond.
3.4.3
Dative bonding in addition compounds
An addition compound is one formed by the combination of two molecules to form only one product. If
the two molecules are identical, the addition reaction is known as dimerization. The dimerization of two
molecules of NO2 to form N2O4 has already been discussed. Here we look at two more compounds that are
formed via an addition reaction and that contain dative bonds.
BF3.NH3
As the formula suggests, this compound is an addition product between BF3 (boron triflouride) and NH3.
It is completely different from either of the reactants, having no chemical or physical resemblance to
them. In the previous discussion it was shown that NH 3 has a non- bonding pair of electrons (lone pair)
on the nitrogen atom. Ammonia can therefore act as a good electron donor. On the other hand, BF3 is
electron deficient, having six electrons only in the valence shell of boron. A dative bond can therefore
form between the B atom in BF3 and the N atom in NH3:
pz
F
F
B
dative bond
BF3
F
NH3
(BF 3.NH 3 - white solid)
Fig 3.6 illustrates the formation of a dative bond in
BF3.NH3. The 2s orbital on N of the ammonia
molecule, containing a lone pair of electrons,
overlaps (σ) with a vacant pz orbital on the boron
atom of BF3. The dative bond, when formed, may be
shown as below (a). However it should be
remembered that once formed, the dative bond
becomes indistinguishable from any other covalent
bond (b):
..
N
H
H
H
Fig 3.6 formation of a dative bond in
BF3.NH3.
Al2Cl6
This compound is an addition product (dimer) of two molecules of AlCl 3.
2AlCl3 → Al2Cl6
The dimerization also involves dative bonding. The structure of Al2Cl6 is shown below
Page
69
In this case, two dative bonds are formed. The reaction occurs because AlCl 3
is an electron deficient molecule. By pairing up with another molecule
through dative bonding, each Al atom achieves an octet and the dimer would
be more stable than the individual molecules .
3.5
Molecular geometry
The shape (geometry) of a molecule is important, for example in determining how it packs with other molecules, how it forms molecular interactions, and in predicting polarity.
To determine the shape of a molecule, we use a method known as the Valence Shell Electron Pair
Repulsion (VSEPAR) theory. This theory has two important statements.
1. Electron pairs around the central atom in a molecule repel and reposition themselves as far
from each other as possible to minimize any further repulsion.
2. Lone pair-lone pair repulsion is greater than bond pair-lone pair repulsion which in turn is
greater than bond pair- bond pair repulsion.
Types of electron pairs in a molecule
Consider a hypothetical molecule XP2 that contains two lone pairs and two bond pairs.
Steps in determining the shape of a molecule
1. Draw a dot and cross diagram for the molecule.
2. Determine, from the dot and cross structure, the total number of electron groups/pairs around the
central atom.
3. Determine the best way to arrange these electron groups so that they are as far from each other as
possible. At this stage it is not necessary to distinguish between lone pairs and bond pairs.
4. To obtain the final shape, ignore the lone pairs. If lone pairs are present, they would have been used in
stage 3 to determine the best possible geometry for all the electron groups present. To obtain the final
bond angle, you have to take into consideration repulsions involving lone-pairs, if any.
The shape of a molecule refers to the geometry of its bonds (bond pairs), but the bond angle is determined by the electron pair geometry – both bond pairs and lone pairs.
3.5.1
Shapes of neutral molecules
Note: The dot and cross diagrams for all molecules in these examples will be reproduced from the
previous discussions. Make sure you can draw them on your own.
Page
70
Important note
There is a strong temptation among students to simply memorize the shapes and bond angles from textbooks without being able to work them out. This is risky because examiners are rarely limited to the
examples given in text books, neither are they limited to the molecules and ions specified in the syllabus.
The following examples show you how to work out the shapes and bond angles in molecules and ions.
1. BeCl2
i. Dot and cross diagram:
Cl X Be X Cl
Notice that lone-pairs around peripheral atoms (chlorine in this case)
are not relevant in determination of shape, so they have been ignored.
Emphasis is on electron groups around the central atom, which is Beryllium in this case.
ii.There are two electron groups (pairs) around the central atom. Both are bonding pairs.
iii.What is the best way to position two electron groups around the central atom so that they are as far
from each other as possible? Placing them at 1800 with respect to each other is the best way to minimize
repulsion between them.
1800
Cl
Be
Cl
So the two covalent bonds of this molecule are at 1800 from each other. The shape is linear,
with a bond angle of 1800.
2. CO2
O XXC XX O
There are two electron groups around the central atom, giving a bond angle of 180 0, as in
BeCl2. The shape is linear.
O
C
O
3. AlCl3
There are three bond pairs around Al. The best geometry to minimize repulsion between these bond
pairs is to place them in a triangular plane; at an angle of 1200 to each other. The shape is therefore
trigonal planar, bond angle 1200 (𝑡ℎ𝑎𝑡 𝑖𝑠,
360
)
3
4. CH4
Page
71
There are four bond pairs around the carbon atom. They should be placed at the corners of a regular
tetrahedron, giving a bond angle of 109.50. This is the optimum angle needed to ensure minimum
repulsion among four bonding pairs of electrons.
The tetrahedron is a three dimensional shape. The perfect tetrahedral bond angle is 109.50. CCl4 and
CH3Cl are further examples of molecules which show the tetrahedral geometry. However, in CH 3Cl, the
tetrahedron is distorted since the peripheral atoms are different.
5. SF6
Page
72
There are six bonding pairs around S. This gives an octahedral geometry (a 3 dimensional shape with 8
faces). We may visualize an octahedron by noticing that it is in fact a square bipyramid, that is, a bipyramid
whose two apices are symmetrically situated opposite a central square plane. In a perfect octahedron, as in
SF₆, all bond angles are equal to 900 (of course this is not apparent on paper). The reader is advised to use
solid models to get a clearer picture of the three dimensional shapes discussed so far (tetrahedron,
triangular bipyramid, octahedron).
All the examples we have considered so far included only bond pairs around the central atom of the
molecule. The following examples include one or more lone pairs around the central atom.
6. NH3
The best way to arrange the four pairs of electrons around the N atom so that they are as far from each as
possible is to place them at the corners of a regular tetrahedron, as already discussed in the previous
example (at this stage we do not need to distinguish between lone pairs and bond pairs). The final shape
is of course not tetrahedral, but it is worked out from the tetrahedral geometry. The final bond angle will
also be estimated from the perfect tetrahedral angle of 109.50.
Page
73
To obtain the correct shape, remember that the shape of a molecule refers to the spatial arrangement of
bonds around the central atom. Since a lone-pair is not a bond, we ignore it. Imagine ignoring the apical
lone pair in the diagram above. The shape becomes trigonal pyramidal, as shown below.
Notice that the bond angle has been reduced from the initial angle of 109.5 0, which is what we would have
expected had there been no lone pair on the nitrogen atom. The effect of the lone pair is to repel the bond
pairs, pushing them closer, so that the angle between them becomes smaller.
Analogous compounds of elements in the same group of the periodic table will have the same geometry.
This is because the elements have the same number of valence electrons, for example, PH3 is analogous to
NH3. Both are the hydrides of group V elements and they share the same geometry.
7. H2O
Shape : bent/v-shaped/angular
Bond angle : 104.50
The shape of the water molecule is worked out in Fig 3.7.
CH4, NH3 and H2O illustrate the VSEPAR theory well. Study the table below.
Molecule
Page
74

bond angle.
NH3
tetrahedral
trigonal pyramidal
109.50
107.50
H2O
bent/angular/ v- shaped
104.50
CH4

shape
The bond angles in the three molecules are close to each other because they are all derived from
the arrangement of four electron groups (lone pairs and bond pairs) around the central atom.
This arrangement is tetrahedral, with a bond angle of 109.50
NH3 and water show a reduction in bond angle from the tetrahedral angle. CH 4 has a perfect
tetrahedral angle. To explain the reduction in bond angles in H 2O and NH3, we invoke the VSEPAR
theory. The reduction is smaller in NH3 because there is only one lone pair which repels the bond
pairs. The bond angle in water is smallest due to significant repulsion between two lone pairs on
the oxygen atom.
Fig 3.7 The electron pair geometry in water is tetrahedral, but the shape of the molecule (bond pair
geometry), after ignoring the lone pairs is angular, with a bond angle of 104.50.
Q
Predict the shape and bond angle in H2S.
A
H2S is the hydride analogue of water. The two molecules have the same shape
(V shaped) and bond angle (1040/104.50).
Q
Predict the shape of the NO2 molecule.
A
The molecule is V-shaped, as in water, but the bond angle is different. Any value slightly
greater than 1200 is a good estimation for the bond angle in NO2.
The dot and cross diagram for the nitrogen dioxide molecule was given in a previous discussion.
Page
75
X
XNX
O
O X X
There are three electron groups around the central N atom. These three groups, which include an odd
electron, are arranged in a trigonal geometry, at an angle of 1200 relative to each other. In the final
analysis, we ignore the odd electron on nitrogen, and we consider its repulsive action on the two N-O
bonds. At this moment we can say that the bond angle is close to 120 0. The odd electron does not cause
any significant repulsion to the bond pairs (compare with the effect of a lone pair). The two N to O bonds
are therefore able to space out until the bond angle is greater than 1200. The actual bond angle is quoted
in chemical literature as being approximately equal to 1340. The shape, after ignoring the odd electron, is
angular.
Important note
It is possible for two molecules to have the same shape but have different bond angles. This is illustrated by H 2O and
NO2. The two molecules have the same shape (angular) and yet they have completely different bond angles. The bond
angle for water is estimated from a tetrahedral angle of 109.50 whereas that of NO2 is estimated from a trigonal planar
angle of 1200.
The shape of a molecule refers to the geometry of its bonds (bond pairs), but the bond angle is determined by the
electron pair geometry – both bond pairs and lone pairs, around the central atom.
Water and NO2 have the same shape because they have the same number of bond pairs. They have different bond angles
because they have different numbers and nature of electron pairs around the central atom.
3.5.2
Shapes of ions
‘Electron book-keeping’ is very important in the determination of shapes and bond angles in ions. This
will be illustrated using the following ions.
Nitronium (NO2⁺) , nitrite (NO2⁻ ) and nitrate (NO3-)
1. The nitronium ion (NO2⁺)
This ion is so unstable that it has no independence existence. It is generated insitu from nitric acid during
some reactions, for example, the nitration of benzene (conversion of benzene to nitrobenzene). This is an
important step during the synthesis of explosives.
The first step in the determination of the shape of an ion is to find out how many (valence) electrons
should be used in the dot and cross diagram. For the nitronium ion:
Page
76
The dot and cross diagram should use 16 electrons. Notice that one electron had to be deducted during
the electron book-keeping. Being positively charged by one unit, the species has one less electron.
Notice that in the dot and cross diagram above, nitrogen is using a valency of four, not five, as would have
been expected. This is because the dot and cross structure has one less electron. The electron is removed
from the nitrogen atom, because it would have been the fifth and odd electron. An odd electron renders a
molecule unstable, and if the molecule is to form an ion by losing an electron, then it would be the odd
electron that is lost.
2. The nitrite ion (NO2⁻)
You will notice that in the dot and cross diagram of the nitrite ion, nitrogen has six electrons, instead of the
expected 5. This is because the book-keeping shows one more electron due to the presence of the -1 charge.
The extra electron is placed on N, because this would make the total number of electrons around it even.
That is, there is no odd electron on the nitrogen atom, which would otherwise cause instability.
3. The nitrate ion (NO3-)
Page
77
The dot and cross diagram should make use of 24 electrons. Remember that though double bonds stabilize
a molecule, too many double bonds around the nitrogen atom results in too many electrons around it
(nitrogen can not exceed the octet). To avoid exceeding the octet around nitrogen, remember to include a
dative bond.
You will have noticed that one of the oxygen atoms has
seven electrons around. You might not have expected
this because oxygen has six valence electrons, not
seven. The electron book-keeping shows that there is
an extra electron from the charge on the ion. The question is where to put this extra electron, on the nitrogen
atom or on the oxygen atom. The electron is
accommodated on the oxygen atom because oxygen is
able to accept and stabilize a negative charge.
You may recall several other species in which an oxygen atom bears a negative charge, including the
hydroxide ion (- OH) and the carboxylate ion (COO⁻). In the NO3⁻ ion, the extra electron is on an oxygen
atom, and it has been shown as a small triangle. It is this electron that gives a charge of -1 to the ion.
Further examples: the importance of intuition and common sense
What is the shape and bond angle in nitric acid, HNO3? Here is where common sense and intuition
becomes an asset. A question which comes to mind is
‘Where is the hydrogen atom bonded, to the central nitrogen atom or to an oxygen atom?’
To answer the question we have to remember the behaviour of an acid such as nitric acid in water:
It releases a proton (a hydrogen ion).
That means the hydrogen in nitric acid is bonded to an oxygen atom. If it were bonded to a nitrogen atom,
it could not be released in solution (think of ammonia, NH3, in which hydrogen atoms are directly bonded
to a nitrogen atom. None of these hydrogen atoms are ionizable in aqueous solution).
The dot and cross diagram and shape of the nitric acid molecule is shown in Fig 3.8. Of course, the nitrate
ion would have the same geometry. It is formed by loss of the hydrogen atom. This does not change the
shape, but it leaves an extra electron (shown as a small triangle) on the oxygen atom which bore the
hydrogen atom. This electron is responsible for the -1 charge on the nitrate ion. Thus in the nitrate ion, one
oxygen atom will appear to have contributed seven electrons to the structure, not the expected six.
The same type of reasoning can be used to deduce the
structures of other oxyacids, for example, sulphuric acid,
H2SO4, shown below.
Fig 3.8
Page
78
H2SO4 is diprotic, that is, an acid which is capable of liberating two protons into solution. From this it can
be deduced that both hydrogen atoms on sulphuric acid are bonded to oxygen atoms. The HSO 4⁻ and SO4⁻
ions, which are formed when sulphuric acid loses one and two protons respectively, have the same
geometry.
The electron book-keeping for the sulphate ion and the corresponding geometry is shown below.
Q
State the change in shape and bond angle that occurs in the following conversions.
Illustrate your answer with suitable diagrams.
(i)
A
CO2 to CO3-
(ii) H2O to H3O+ (oxonium)
(iii) NH3 to NH4+
(i) Please see main text.
(ii) From angular to trigonal pyramidal (as in ammonia). Bond angle changes from 104.50 to
1070.
(When you draw the structure of H3O+ , note that it contains an O-H dative bond and one
lone pair).
(iii) Left as an exerciser to the reader. Note that the amminoium ion, NH 4+ contains a N-H
dative bond.
3.6
Features of the covalent bond
Two important features of the covalent bond are bond strength and bond polarity. Bond strength is
purely a thermodynamic concept, so it will be discussed in detail under the relevant topic.
3.6.1 Bond polarity
Electrons that make up the covalent bond are under the attractive influence of the nuclei of two atoms. But
will the two atoms attract the electron pair to the same extend, or one of the atoms will have a greater pull
on the electrons? It all depends on the relative electronegativities of the atoms that make up that bond.
Electronegativity
This is a measure of the relative tendency of an atom to withdraw electrons of a bond towards itself.
Page
79
Linus Pauling invented an arbitrary scale of relative electronegativities which we use today. The scale
ranges between 0 and 4. The larger the Pauling value, the more electronegative an atom is. An extract of
the scale is shown in fig 3.9.
Fig 3.9 The Linus scale of electronegativities


Electronegativity increases across a period and up a group.
The most electronegative elements are therefore situated at the top of the far right side of the
Periodic Table. The three most electronegative elements are N, O and F. The most electronegative
element is fluorine.
The noble elements have zero electronegativity.
This is because they do not have a tendency to form covalent bonds, at least in most of the
situations. A discussion of electronegativities is therefore irrelevant for the noble elements.
Explaining the trend in electronegativities
The trend in electronegativity across a period and down a group is expected.
Going across the period, the number of protons in the nucleus (nuclear charge) increases. The attractive
power of the nucleus for the bonding electrons therefore becomes stronger.
Going down a group, additional shells are opened. The effect is to increase the size of the atom, and to hide
the nucleus deeper and deeper inside the atom. The distance of the nucleus from the shared pair of electrons
between the two atoms increases, and attraction for these electrons becomes correspondingly weak, that is,
electronegativity decreases.
Polar bonds
If the two atoms of a bond have different electronegativities, they will attract electrons of the bond to
different extends.
The result is that the electron cloud of shared electrons is skewed towards the more electronegative atom.
This atom therefore becomes relatively negatively charged and the less electronegative atom becomes
relatively positively charged. The less electronegative atom will have a smaller electron density around it.
Page
80
The electron rich side of the bond is represented by δ- and the electron deficient side by δ+.
This notation shows that the molecule now has two poles as far as electron distribution is concerned; an
electron rich pole and an electron deficient pole. Such a bond is therefore said to be polar. It is also said to
possess a dipole.
What exactly should be the difference in electronegativity of the two atoms for a bond to be regarded as
being polar?

An electronegativity difference of zero implies that there is perfect sharing of the bonding
electrons. The bond is therefore perfectly non-polar.

If the difference in electronegativity is greater than zero, but is vey small, the bond is regarded as
being approximately non-polar. This is because neither of the two atoms has the ability to strongly
attract the bonding electrons to itself. However, it should be noted that such bonds do contain a
degree of polarity. The C-H bond is a good example. The difference in electronegativity between C
and H is 0.4 (refer to Fig 3.9). This is a very small difference, and so the bond is said to be
approximately non-polar.

Bonds in which the difference in electronegativity is greater than 0.4 should be considered as
being polar. However, if the difference in electronegativity exceeds about 1.7, the bond becomes
ionic, that is, one atom will pull electrons of the bond (almost) completely to itself. It then becomes
a negative ion and the other atom becomes a positive ion. In LiF, the difference is
4.0-1.0 = 3, making the compound predominantly ionic.
Covalent or ionic?
It must be clear from the foregoing discussion that it is not always easy to put a clear cut boundary
between an ionic and a covalent bond. It is easy to say that the Cl-Cl bond is purely covalent because the
difference in electronegativity between the two atoms is zero. But what about a compound such as AlCl3?
One might expect it to be ionic from a consideration of what is taught in juniour chemistry: a bond
between a metal and a non-metal is ionic. This statement is not always true. In fact, AlCl3 is a covalent
substance. The difference in electronegativity between the Al and the Cl atom is 1.5, which is very close to
the threshold value of about 1.7 required to make a bond ionic. We therefore describe AlCl 3 as being covalent
but with a significant degree of ionic character. In fact, according to the calculations done by Linus Pauling,
AlCl3 has about 43% ionic character and about 57% covalent character.
Also consider magnesium chloride, MgCl2. The difference in electronegativity between Mg and Cl is 1.8.
This is greater than the threshold value of 1.7. MgCl 2 is therefore ionic. However, the difference in electronegativity between Mg and Cl is very close to 1.7, and so we should expect a high degree of covalent
character in MgCl2. The chemical properties of MgCl 2 show that it is ionic with a high degree of covalent
character. Using Pauling’s method, MgCl2 would have about 55% ionic character and 46% covalent
character.
The reader is advised to use Pauling’s values with great care. They only give a qualitative method of estimating covalent and ionic character. It is not always possible to classify a bond as being ionic or being
covalent simply by carrying out calculations using Pauling’s values. For example, one might expect HF to
be ionic with a high degree of covalent character. In fact, it is covalent with a high degree of ionic character.
Page
81
3.6.2 Interaction of dipole moments
The forgoing discussion on bond polarity was concerned with an isolated bond.
But what happens if two or more polar bonds are present in a molecule? The polarity of a bond is measured
by a quantity known as the dipole moment, µ .The Debye, D, is used as a unit for this physical quantity.
The dipole moment is actually a vector quantity, that is, it is defined by its size and by its
direction. For instance, we may show the dipole moment of the HF bond as a vector quantity, in the
following way.
Now, if there is more than one polar bond in the
molecule, the dipole moments will add up in the
same way as vectors do. The result is that there may
be one large net (resultant) dipole moment working
in a specific direction.
In this case, the dipole moments have amplified each other. Another situation arises when the dipole
moments are oriented in such a way that they diminish each other. In fact, if the molecule is symmetric, the
dipole moments cancel each other out and the molecule will have no net dipole moment, even though
it contains polar bonds. For a molecule to be symmetric

it must have a shape which has a centre of symmetry.

all bonds must be identical, or if there are different bonds, they should be in the
ratio of 1:1, and they should be positioned so that the molecule has a centre of symmetry.
There are therefore two reasons why a molecule might be non- polar:
(i) either it has no polar bonds at all (for example chlorine),
(ii) or it has two or more polar bonds whose dipole moments cancel each other out by reason of the
molecule being symmetric.
Examples
The following molecules all contain polar bonds, but they are non-polar because, being symmetric, the
dipole moments cancel each other out.
1. AlCl3
The geometry of the molecule (triangular planar) is symmetric.
Page
82
2. CO2
In the two examples above, all bonds are identical, so the dipole moments are equal to each other. CCl 4,
with four identical bonds and a perfect tetrahedral shape, falls into the same category. It is non-polar.
3. 3,4-dinitrobenezene
4. [Pt(NH3)2Cl2]
This molecule has a square planar geometry. There are two possible isomers, shown below.
In each isomer, there are four polar bonds, two Pt-N bonds
and two Pt-Cl bonds. There are different bonds, but in the
ratio of 1:1. In trans-platin, identical bonds are
symmetrically arranged relative to each other, so their
dipole moments cancel out and the molecule is overally
non-polar.
In cis- platin, identical bonds are asymmetrically arranged relative to each other, so their dipole moments
do not cancel each out.
Cis-platin is a well known anti-cancer drug but Trans-platin has no medicinal value. Because of its
polarity, cis- platin is able to bind to the polar DNA molecules in cancerous cells. This prevents the DNA
from multiplying. When DNA of cancerous cells fails to multiply, the cells are not able to reproduce
themselves.
The following molecules are not symmetrical, and they have a net dipole moment, that is, they are polar.
5. CHCl3 (trichloromethane)
In this molecule, there are four bond pairs around the central carbon atom. The shape is therefore
tetrahedral as in CCl4 (tetrachloromethane).
Page
83
However, CCl4 is non-polar, whereas CHCl3 is polar. This is because in CHCl3, there are two types of bonds,
the C-Cl and the C-H bonds. The shape is therefore a distorted tetrahedron, which is asymmetric. The
dipole moments cancel out to some extent, but they do not completely nullify each other.
6. 1,3 - dinitrobenzene
In 1,3-dinitrobenzene , the two polar bonds are asymmetrically
arranged with respect to each other. Consequently, their dipole
moments do not cancel each other. Rather, they add up, to give a net
dipole moment operating in the shown direction.
3.7
Intermolecular forces
Forces of attraction exist between the molecules of a substance. These are termed intermolecular forces to
distinguish them from intramolecular forces (Fig 3.9), which are the covalent bonds holding atoms together.
It should be mentioned here that the term intermolecular force applies as well to substances which do not
contain molecules, for example neon. ‘Intermolecular’ forces of attraction exist between the individual neon
atoms.
Fig 3.9
Intermolecular forces are much weaker than covalent bonds. When a substance is heated, it is always the
intermolecular forces that break first. Covalent bonds, being stronger, will only start to break at higher
temperatures.
Breaking of intermolecular forces does not change the chemical identity of a substance. It simply pulls the
molecules apart; since no covalent bonds are broken, the molecules remain with the same formula as
before. Breaking of intermolecular forces, for example, during boiling, is therefore a physical process. However, breaking of covalent bonds is a chemical process because it changes the identity of the substance.
Intermolecular forces are important in explaining the physical properties of compounds, such as boiling
points, volatility, melting points, densities, colour etc.
Boiling points and melting points
Page
84
When a substance boils, its molecules are separated from each other and escape into the gaseous state.
Separation of these particles requires the breaking of intermolecular forces that exist between the
molecules. Different substances have different boiling points, reflecting the different strength of the
intermolecular forces present in them. The stronger the intermolecular forces, the more energy required to
separate the molecules. The boiling point of the substance would be correspondingly high.
Similarly, when a substance melts, intermolecular forces between particles of a solid are broken. The
molecules therefore spread and are able to flow, that is, the liquid state is formed.
Volatility
This is a measure of the ease with which a substance can turn to a vapour. A substance does not necessarily
have to be heated to its boiling point for its particles to evaporate. Try asking this question to a
number of people,
‘At what temperature does water evaporate?’
You will be surprised by the number of people who think that water evaporates at 100⁰C. The fact is that
liquid water can evaporate at any temperature. However, at low temperatures, say 20⁰C, the rate of
evaporation of the water molecules is very small. This shows that the forces of attraction between water
molecules are relatively strong. A relatively large amount of heat energy is required to separate the
molecules from each other. In other words, water is involatile. Now take a substance like ether, CH3OCH3.
The forces of attraction between its molecules are so weak that at a relatively low temperature, say 20 0C, a
large number of the particles escape and enter the vapour state. Ether is therefore said to be a volatile liquid.
Density
Density is a quantity which gives an idea of how tightly or loosely packed the particles are in a substance.
If the particles are tightly packed, then the substance has a high density. Intermolecular forces tend to hold
together molecules of a substance. If the intermolecular forces are relatively strong, then the particles are
held relatively tightly to each other. Density would be correspondingly high.
Types of intermolecular forces
Intermolecular forces have been classified as:



3.7.1
permanent dipole - dipole attractions
Van der Waals forces
hydrogen bonds
Permanent dipole - dipole attractions
These arise between molecules which bear a permanent dipole.
Consider the HCl molecule, which is permanently polar


H Cl
In a container of HCl, the molecules attract each other electrostatically; the Cl δ¯ end of one molecule will
attract the Hδ+ end of a neighbouring molecule.

H
 
Cl
H

Cl
permanent
dipole-dipole attraction.
The same forces of attraction are also present in molecules which
contain more than one polar bonds. However, the molecules must be
asymmetric; otherwise the dipole moments will cancel each other
out, making the molecule non-polar.
Page
85
The following compounds have permanent dipole - dipole attractions between their molecules because the
molecules have polar bonds and do not have a centre of symmetry.


Trichloromethane, CHCl3 (there are three polar C-Cl bonds which are asymmetrically arranged).
Chloromethane, CH3Cl.
δ+
CH3
δ-
Cl
δ+
CH3
δ-
Cl
Q
Explain the difference in the boiling points of the two isomers below.
NO2
NO2
NO2
NO2
1,3 - dinitrobenzene
A
1,4 - dinitrobenzene
The first isomer has the higher boiling point because it has a net dipole moment (it is
asymmetric). The net dipole moment acts in the direction shown by the arrow in the
diagram below.
NO2
NO2
1,3 - dinitrobenzene
The 1, 4-isomer is non-polar. Even though it has two polar bonds, their dipole moments
cancel out because the molecule is symmetric. Being non-polar, there are only weak
forces of attraction between the molecules. In the 1, 3- isomer, relatively strong
permanent dipole-dipole forces exist between the molecules. More energy is required to
break these forces, that is, the isomer has a higher boiling point.
3.7.2 Van der Waals forces (temporary dipole-induced dipole attractions)
This type of intermolecular force arises between non polar molecules or atoms.
Examples are methane (containing molecules) and neon (containing atoms). The C-H bond in methane is
non- polar because the difference in the electronegativities of C and H is very small.
The fact that in a non-polar substance there are forces of attraction between the molecules is demonstrated
by the observation that such substances can be liquefied or frozen. During these processes, the molecules
or atoms get closer together and they become less mobile. This allows the intermolecular forces to become
stronger and more stable.
Page
86
The origin of Van der Waals forces
Page
87
Features of Van der Waals forces

They are very weak, much weaker than permanent dipole - dipole attractions.

They are extremely short lived (temporary). The Van der Waals attraction between any two given
molecules is instantaneous. However, since there will be a large number of molecules in the
container , on average, there is always some forces present between some molecules.

The strength of Van der Waals forces increases with increasing number of electrons
(increasing Mr). The more electrons there are, the greater the distortion of molecular
orbitals and the stronger the Van der Waals forces.

Van der Waals forces are found in all molecular or atomic substances. However, they are only
important in the absence of stronger intermolecular forces (hydrogen bonds or permanent
dipole-dipole attractions). Consider a molecule such as HCl. There are two types of
intermolecular forces between the molecules, that is, Van der Waals and permanent dipoledipole attractions. However, the latter forces are much stronger and they make the Van der
Waals forces insignificant.
The effect of number of electrons on strength of Van der Waals forces is clearly illustrated by the group VII
elements. Chlorine is a pale greenish-yellow gas, bromine is a red-brown liquid and iodine is a shiny black
solid at room temperature. How can this trend in physical properties be explained?
As the number of electrons increases down the group, so does the strength of the Van der Waals forces.
More energy is required to break these forces. In chlorine the forces are so weak that they can not hold the
molecules together at room temperature. In bromine the forces are strong enough to hold the molecules
loosely together in the liquid state. The forces are strongest in iodine. They are strong enough to hold the
molecules closely and tightly in the solid state.
The same observation is made for the hydrocarbons. Low Mr hydrocarbons such as methane and ethane
tend to be gaseous at room temperature. Higher Mr hydrocarbons such as pentane are volatile liquids. At
even higher Mr values the alkanes become solid, e.g. candle wax, which has, on average, 30 carbons per
molecule.
3.7.3
Hydrogen bonds
These are the strongest of the intermolecular forces. A hydrogen bond is a special type of a permanent
dipole-dipole attraction. It is formed between molecules which contain the N-H, O-H or F- H bonds.
N, O and F are the three most electronegative atoms. On the other hand, the hydrogen atom has no
intervening electrons between the nucleus and the valence shell. The nucleus of the hydrogen atom can
therefore easily be exposed when the hydrogen atom is bonded to N, O or F. The proton in the nucleus of
the hydrogen atom becomes available for the formation of hydrogen bonds.
The origin of hydrogen bonds
Consider a hydrogen atom bonded to one of the three most electronegative atoms, N, O and F. We will use
HF for our illustration.
The illustration given above applies not only to the H-F bond, but to the H-O and H-N bonds as well.
Requirements for a hydrogen bond

A hydrogen atom bonded to one of the three most electronegative atoms, N, O and F

A lone pair of electrons on the more electronegative atom (N, O, F). It so happens that in their
compounds, these atoms possess one or more lone pairs. A hydrogen bond specifically involves
attraction between a lone pair on N, O or F and an electron deficient hydrogen atom, as illustrated
for HF.

H

F

H

F
hydrogen bond
The fact that a hydrogen bond is always directed towards a lone pair implies
that hydrogen bonds are directional, that is, they operate in a specific
direction. Consequently, orientation is very important for their formation. Two
molecules must be oriented favourably with respect to each other for a
hydrogen bond to be formed between them.
Page
88
The same directionality exists for permanent dipole- dipole attractions. Compare with Van der Waals
forces which are non- directional. Orientation factors are therefore not important in the formation of Van
der Waals forces.
Hydrogen bonding in water
Water contains O-H bonds, so hydrogen bonding is possible between its molecules.
One question which immediately comes to mind is this,
“How many lone pairs does each oxygen atom use to form
hydrogen bonds, and how many hydrogen atoms does each
water molecule use?”
The answer to this question depends on whether we are discussing liquid or solid water (ice). In liquid
water, the molecules are always moving around. Consequently, there is a continuous process of hydrogen
bond formation and hydrogen bond breaking. A water molecule may use one or both lone pairs for hydrogen
bonding, or no lone-pair is used at all. All that can be said is that on average, the probability of there being
a hydrogen bond between two water molecules is high. This probability decreases with increasing
temperature. Hydrogen bonds, though the strongest of the intermolecular forces, are still many times
weaker than covalent bonds. They are therefore easily broken when temperature and kinetic energy of the
molecules increase. Conversely, the probability of a water molecule forming one or more hydrogen bonds
with neighbouring water molecules increases with decreasing temperature. As temperature drops, molecules lose kinetic energy and come closer together. This favours the formation of hydrogen bonds. Each
water molecule has the capacity to hydrogen bond with four other water molecules. The two lone pairs on
the oxygen atom form hydrogen bonds with two neighbouring water molecules. The two hydrogen atoms
make hydrogen bonds with two other neighbouring water molecules. All four possible hydrogen bonds are
formed (per molecule of water) in ice and this has important consequences on its properties.
Hydrogen bonding in ammonia
Ammonia, which contains N-H bonds, is also capable of forming hydrogen bonds (Fig 3.10). Once more,
the extent of hydrogen bonding depends on the temperature.
Hydrogen bonding in a mixture of water and ammonia
Page
89
Fig 3.10 hydrogen
bonding in pure ammonia
Ammonia dissolves well in water, and there are two reasons for this
observation.
 Water and ammonia molecules are capable of associating
through hydrogen bonds (Fig 3.11). This allows the two substances to form a single phase mixture (solution).
 Some ammonia molecules react with water, to form a mixture
containing NH4+ and OH- ions, in addition to unreacted water
and ammonia molecules.
Fig 3.11 hydrogen
bonding in a mixture of
ammonia and water.
Intermolecular versus intramolecular hydrogen bonds
Compounds A and B below are isomers. Isomer A has the higher boiling. How can this observation be
explained?
Isomer B has a hydrogen bond between the NH2 and the OH group
because of the close proximity of these two groups to each other.
This type of hydrogen bond, formed between groups on the same molecule, is known as an
intramolecular hydrogen bond. By forming this bond, the molecule uses up its capacity to hydrogen bond
with neighbouring molecules. Only weak Van der Waals forces are left to operate between the molecules.
Because of their separation in space, the NH2 and the OH groups on isomer A can not form intramolecular
hydrogen bonds. Intermolecular hydrogen bonds can therefore form between neighbouring molecules (Fig
3.12). More energy is required to break the hydrogen bonds between neighbouring molecules. Thus the first
isomer has the higher boiling point.
Fig 3.12
Properties of hydrogen bonds


They are the strongest of the intermolecular forces. However, they are much weaker than covalent
bonds. They are highly directional. A hydrogen bond acts only in a certain direction. It is always
from a H atom on one molecule to a lone pair of electrons on N, O or F on the other molecule.
They are a special type of permanent dipole - permanent attractions.
Consequences of hydrogen bonding





Anomalous behavior of water.
Anomalous boiling points.
Viscosity.
Anomalous Mr values.
Solubility.
Page
90
Anomalous behaviour of water
Water expands as it cools, so its solid state, ice, is less dense and tends to float on the liquid state. This is
quite unexpected. The general observation is that when matter cools it contracts, and the solid state
therefore becomes denser than the liquid state, that is, the solid state should sink into the liquid state.
This anomalous behaviour of water can be explained in terms of hydrogen bonding. As water is cooled, its
particles lose kinetic energy and come closer together. This allows more and more hydrogen bonds to form.
Each water molecule can form a maximum of four hydrogen bonds to four neighbouring water
molecules; two hydrogen bonds via its two lone pairs and two hydrogen bonds via its hydrogen atoms. In
liquid water, not all four hydrogen bonds may form because of the high kinetic energy of the molecules
which tends to interfere with formation of hydrogen bonds. In ice, all four hydrogen bonds are formed for
each water molecule. To minimize inter-electronic repulsions between these four water molecules, they
must be spaced out in a tetrahedral geometry. This spacing out implies an increase in volume, as it forms a
very open structure. In other words, water tends to expand as it freezes. This explains why a bottle of water
put in a fridge will burst when the water freezes. Fig 3.13 below illustrates hydrogen bonding in ice.
Fig 3.13 Hydrogen
bonding in ice.
Suppose that we freeze a fixed mass of water. In ice, the volume has increased, but mass has
remained the same. Using the relationship
D = m/V,
we can see that an increase in volume at constant mass results in a decrease in density, that is, density and
volume are inversely related. The decrease in density causes ice to float on liquid water. Most of us will
remember the story of the Titanic, the luxury ocean liner which collided with an iceberg on its maiden
voyage on Sunday, the 14th of April, 1912. An iceberg can be as big as a huge mountain. What causes it to
float is that its volume is much greater than its mass, giving it a lower density than liquid water.
Floating of ice on water causes lakes to freeze from the top. The layer of ice so formed then insulates the
liquid water underneath, keeping it relatively warm and unfrozen. This allows aquatic creatures to survive
at the bottom of frozen lakes.
Page
91
Anomalous boiling points
The boiling point of water is much higher than expected from its Mr. Water is a relatively small molecule
(small number of electrons), and yet it has a higher boiling point than many molecules which are heavier
than it. This is well illustrated by the group six hydrides, of which water is a member. The Group (VI)
hydrides are, from the top of the group, H 2O, H2S, H2Se and H2Te. Going down the group, the number of
electrons in the molecules increases, and so does the strength of the Van der Waals forces. Consequently,
boiling points are expected to increase as more and more energy is required to break the Van der Waals
forces. This trend is indeed observed, except for water, which is clearly out of trend. Instead of having the
lowest boiling point by virtue of it having the least number of electrons (Mr), it has the highest boiling point.
This anomalous boiling point of water is explained by the presence of hydrogen bonds in water and their
absence in its analogues. Hydrogen bonds are much stronger than the Van der Waals forces present in the
other Group (VI ) hydrides. A large amount of heat is required to break these forces and separate the water
molecules into the gaseous state.
It is important to note that Van Der Waals forces are present in all molecules. However, in hydrogen bonded
substances the important force is the hydrogen bond. In the absence of a stronger intermolecular force such
as the hydrogen bond and the permanent dipole-dipole attractions, Van der Waals forces become the
important force of attraction.
The same trend discussed for the Group (VI) hydrides is found among the Group (V) and (VII)
hydrides. In each case, the first member can form hydrogen bonds between its molecules, and so its boiling
pointer is much higher than expected. Compare with the group (IV) hydrides in which all members can not
form hydrogen bonds. Consequently, the first member, methane, which contains the least number of
electrons, has the lowest boiling point, as expected (Fig 3.14).
Fig 3.14 Trends in the boiling points of the hydrides of the Group (V), (VI) and (IV)
hydrides.
Viscosity
This is a measure of the ability of a liquid to resist flowing. Viscosity gives an idea of the size and strength
of intermolecular forces acting in a substance. Some substances are viscuous and so do not flow easily, for
example, honey, which is a rich mixture of concentrated glucose and fructose. Both glucose and fructose
have a large number of OH groups which can participate in hydrogen bonding. The molecules of water,
fructose and glucose are therefore closely associated to each other by hydrogen bonds and can not move
freely.
Anomalous Mr values
The apparent Mr of ethanoic acid in a solvent such as benzene is 120. The expected value is 60. This
observation can be explained in terms of hydrogen bonding. In benzene, molecules of ethanoic acid
associate to form dimers through hydrogen bonding, thus doubling the Mr (Fig 3.15).
Page
92
Fig 3.15
Dimerization in ethanoic acid
In water, the dimerization does not take place because
the ethanoic acid molecules are more likely to form
hydrogen bonds to water molecules than to each other.
Dimerization of the ethanoic acid molecules therefore
occurs in a solvent, such as benzene, which has no
capacity to form hydrogen bonds with the ethanoic
acid molecules.
The same observation is made for HF, whose apparent Mr in some solvents like benzene is also double the
expected value.
Solubility
The solubility of some substances in water can be explained in terms of hydrogen bonding. For example,
methane and ammonia have similar Mr values (16 and 17 respectively). Ammonia is soluble in water but
methane is not. Ammonia can form hydrogen bonds with water. In this way the water and ammonia
molecules can mix homogeneously. Similarly, alcohols such as methanol and ethanol dissolve in water
through hydrogen bond formation as shown in the following illustration.
C
H
Water and methanol (CH3OH) are miscible
in each other through formation of
hydrogen bonds
H
3.7.4
H
.. .O.
H
H
H
..O..
..O..
H
H
H
H
.O..
.
hydrogen bond
Consequences of intermolecular forces
Intermolecular forces do not affect the chemical properties of a substance. However, they are responsible
for the physical properties of substances, e.g. boiling points, melting points and colour. A discussion of
physical properties should centre on intermolecular forces, with one important exception:
The melting and boiling of giant covalent substances such as silicon dioxide and diamond requires the
breaking of covalent bonds.
Boiling and melting points
Boiling and melting requires the pulling apart of molecules. During melting, the particles are pulled apart
from the solid so that they become more loosely associated in the liquid state. Boiling requires a further
pulling apart of liquid particles so that they become free of each other in the gaseous state.
The boiling and melting point of a substance depends on the type and strength of intermolecular forces
present. It also depends on the number or extent of such forces. A comparison of boiling points of pairs of
substances is given below to illustrate this point.
1. Ammonia and methane
Page
93
Ammonia has a higher boiling point. There are hydrogen bonds holding the molecules of ammonia
together. These require a relatively large amount of heat to break. In methane there are only weak Van der
Waals forces, which require a small amount of energy to break. Note that the two molecules have similar
Mr values, so the difference in boiling point can not be explained in terms of strength of Van der Waals
forces.
2. CHCl3 and CCl4
CCl4 has a higher boiling point (76.80C) than CHCl3 (61.20C). CCl4 contains a larger number of electrons
(larger Mr), so there are relatively strong Van der Waals forces operating between its molecules.
3. CH3OH and CH3CH2OH
Both compounds have hydrogen bonds between their molecules, so the difference in boiling points must
be due to the difference in number of electrons. Ethanol (CH 3CH2OH), with the larger number of
electrons, has stronger Van der Waals forces between its molecules.
4. Ne and Ar
In both substances, there are weak Van der Waals forces between the atoms. However, argon has the
larger number of electrons. It therefore has stronger Van Der Waals forces. The boiling point is
correspondingly higher.
5. AlCl3 and AlF3
AlF3 has the higher boiling point because it is predominantly ionic, whereas, AlCl3 is predominantly
covalent.
NB Aluminium chloride does not boil but sublimes.
6. n-pentane and 2,2- dimethylpropane
The structures of these two compounds, which are isomeric to each other (same Mr), are shown in
Fig 3.16.
n-pentane has the higher boiling point because each molecule has a large surface area over which Van der
Waals forces can operate with a neighbouring molecule. The second compound has a roughly spherical
conformation. The contact area between two molecules is therefore much smaller (Fig 3.17). Since Van der
Waals forces act over a small area in 2,2-dimethylpropane, they are weaker and more easily broken.
Fig 3.17
Summary: intermolecular forces
Page
94


There are three types of intermolecular forces; permanent dipole-dipole attractions, Van der
Waals forces and hydrogen bonds.
These forces are too weak to affect the chemical properties of substances, but they are responsible
for physical properties, such as boiling points, melting points, color, etc.




Hydrogen bonds are the strongest, followed by permanent dipole-dipole attractions. Van der Waals
forces are the weakest (of course, we have to compare compounds of similar Mr values.)
Van der Waals forces are the predominant forces in non- polar substances such as the noble gases
and the diatomic elements such as O2.
Permanent dipole-dipole attractions occur in compounds which contain bonds with permanent
dipoles. However, if the molecule is symmetrical the dipole moments cancel out, and overally the
compound would be non-polar, e.g. AlCl3.
Hydrogen bonds occur if:
(i) a bond is made up of H bonded to one of the three most electronegative atoms, N, O or F.
(ii) A N , O or F atom is present in the neighbouring molecule to use its lone pair for forming the
hydrogen bond.
3.8
Ionic bonding
An ionic bond is a strong electrostatic attraction between a positively charged ion (cation) and a
negatively charged ion (anion).
Ionic bonding between a metal and a non-metal
When a metal reacts with a non- metal, an ionic compound is formed in which the metal exists as a cation
and the non-metal as an anion.There are exceptions to this generalization; for instance, AlCl 3 has already
been discussed under covalent substances, even though it is formed between a metal and a non-metal.
It has been noted in the previous discussion on covalent bonding that electronegativity of the elements
increases across the periodic table. Thus metals have very low electronegativities, that is, a metal atom has
very little tendency to accept electrons of a bond towards itself. In fact, the opposite is true for metals; they
have a strong tendency to lose their valence electrons. We say they are electropositive.
Page
95
Electropositivity is a measure of the tendency of an atom to lose an electron or electrons from the valence
shell in chemical reactions.
Electropositivity decreases across a period, that is, in any period; the metals are the most electropositive.
They have the greatest tendency to lose electrons.
This behavior of metals is expected. In any period, the metals have larger atoms than their non-metal
counterparts. Atomic radii decrease across a period. This has already been explained in terms of increasing
effective nuclear charge. Going across a period, the number of protons in the nucleus increases and yet
electrons are entering the same shell. The outermost electrons feel an increasing attractive force of the
nucleus (effective nuclear charge). The outermost shell is therefore pulled closer and closer to the nucleus,
resulting in a shrinking of atomic volume.
In metals, because of the large size of the atoms, the outermost electrons are weakly held to the atom.
These electrons are therefore easily lost. Further more, a loss of valence electrons exposes the underlying
shell, which has a compete octet of electrons. A metal therefore tends to be stabilized by the formation of
positive ions. On the other hand, to fill an octet in the outer most shell, a non-metal atom needs to accept
just a few electrons. Also, because of the small atomic size, the outer shell is closer to the nucleus, and it is
capable of accepting electrons, because the incoming electron(s) would be stabilized effectively by the
attractive force of the nucleus. Non- metals tend to be stabilized by accepting electrons to form negative
ions.
Consider what happens when chlorine and sodium are mixed. A sodium atom loses its valence electron,
forming a sodium ion, Na+. This ion is more stable than a sodium atom because it now has an outer shell
with a full octet of electrons. The electron lost by a sodium atom is gained by a chlorine atom. By accepting
this electron into its valence shell, Cl now has a full octet of electrons in the outer shell. The chloride ion is
therefore more stable than the chlorine atom.
The positive sodium ion and the negative chloride ion then attract to form sodium chloride, NaCl. It is this
attraction between a positive ion and a negative ion which is termed an ionic bond.
The bonding in sodium chloride is illustrated in Fig 3.18.
Fig 3.18
Ionic bonding in sodium chloride
Use of dot and cross diagrams
Page
96
In a strict sense, an ionic compound does not have a dot and cross diagram. Dot and cross diagrams are
mainly used to distinguish between shared electrons of a covalent bond because we need to know the origin
of each electron of the bond. In ionic compounds, there is no electron sharing, so it may not be necessary
to distinguish between electrons. However, examiners often ask candidates to draw the dot and cross
diagram for an ionic compound. The dot and cross diagram for NaCl is shown in Fig 3.19 below (only the
valence shells have been shown).
Fig 3.19
The formulae of ionic compounds
The formula of an ionic compound is written so that the total of positive charges balances with the total of
negative charges. An ionic compound, and indeed, all compounds, should be electrically neutral. It is
hoped that at this stage the reader has no difficulties in writing the formulae of ionic compounds. The
reader may do self-check exercise 3.1 below.
Exercise 3.1
Write down the formulae of the following compounds
1. Magnesium chloride.
2. Iron (III) chloride.
3. Aluminium oxide
4. Sodium nitride.
5. Magnesium nitride
6. Aluminium sulphate
7. Lithium oxide
8. Ammonium sulphate
9. Sodium nitrite
10. Barium sulphite
The lattice structure of ionic compounds
One important property of an ionic bond is that it is highly directional, that is, it will only operate in a fixed
direction. It therefore tends to be localized. This means that ions and anions can only be held together in a
specific way, that is, the ions are not randomly organized. This results in a regular three dimensional lattice
of ions, with well defined layers. Such a regular three dimensional arrangement of particles in space is
known as a crystal or lattice. This can be illustrated using sodium chloride (Fig 3.20). The co-ordination
number in the NaCl crystal is six. This means that each ion is surrounded by six ions of opposite charge.
Cl-
Page
97
Na+
Properties of ionic compounds
1. High melting and boiling points
Fig 3.20 The crystalline
nature of NaCl
The electrostatic attractions between positive and negative ions are very strong. A large amount of heat
energy is required to pull the ions apart and break down the crystal into a molten state. Because of their
high melting points, some ionic compounds are frequently used as refractory substances. These are
substances whose use depends on their ability to resist heat, for example, MgO which is used as a furnace
lining.
2. They are hard
The particles of the lattice (cations and anions) are held in fixed positions by strong directional
electrostatic attractions. It is difficult to displace the particles with respect to each other.
3. Non conductors of electricity in solid state.
There are no free ions or electrons to carry electric charge.
4. Conducts electricity when molten or in aqueous solution
In the liquid (molten) state or aqueous solution, there are mobile ions (cations and anions) to carry
electric charge.
5. Brittle.
Under strain, ionic compounds snap without bending first. A force may displace layers and align them in
such a way that positive ions are directly in line with positive ions, and negative ions are directly in line with
negative ions. This causes sudden repulsion between layers, causing a smart cleavage to occur. The cleavage
is clean (not jagged), because it occurs along layers.
6. Soluble in polar solvents but insoluble in non-polar solvents
The δ- end of a polar molecule such as water is attracted to the positive ions of the ionic compound. The
δ+ end of the solvent molecule is attracted to the negative ions of the solid lattice. This results in ions of
the crystal being pulled apart. This is illustrated for the dissolution of NaCl in water in Fig 3.21.
Page
98
Fig 3.21
This pulling apart of ions by solvent molecules uses energy. However, the energy used is more than
compensated for by the formation of solvated ions. Formation of ion-solvent electrostatic attractions
releases energy. It is this energy which drives further dissolution of the solid.
An ionic compound will not dissolve in a non-polar solvent such as benzene. Solvent- ion attractions
would be very weak, that is, very little energy would be produced by solvation to compensate for the
energy that is used to pull the ions apart from the solid crystal.
Ionic or covalent?
In discussions on bonding, an impression is sometimes created that there is a clear cut boundary between
covalent and ionic bonding. It should be mentioned here that in most of the cases, a covalent bond has some
ionic character, and an ionic bond has some covalent character. When we say a bond is covalent, we mean
that the percentage of ionic character is much smaller than its covalent character, and most of its properties
are those of a covalent substance.
However, we do have cases where, in a covalent compound, the ionic character is so high that it has an
important effect on the properties of the substance. Similarly, some ionic compounds have considerable
covalent character. A good example is magnesium chloride. We usually expect an ionic compound to form
a neutral solution when it dissolves in water. When MgCl 2 dissolves in water, the solution formed is not
neutral, but is acidic. This shows that MgCl2 has significant covalent character.
3.8
Metallic bonding
The bonding in metals is known as metallic bonding. It arises like this:

Metal atoms give up their valence electrons and contribute them into a common pool.

The metal atoms then become positively charged. These positive atomic cores are more stable
than the atoms, because by giving up the valence electrons they expose an underlying octet. The
term positive ion is sometimes used, but this is misleading because it gives the wrong impression
that the metal atoms completely lose their valence electrons. The metal atoms do not lose
electrons; they simply contribute them into a common pool. The term cation should particularly
be avoided.

The electrons then form a delocalized sea. The term ‘delocalized’ emphasizes that the electrons
are not fixed (localized) in one place as in a covalent bond. They are mobile and no single electron
belongs to any single atom. The term ‘sea’ emphasizes that all electrons form a common pool. It is
not possible to distinguish between the electrons and tell where each electron came from.

The positive atoms and the sea of delocalized electrons then attract electrostatically. This
attraction is very strong, and it is termed a metallic bond.
Metallic bonds are electrostatic attractions between positive metal atoms and a sea of delocalized
electrons.
Structure of metals
Page
99
The bonding in metals is illustrated in Fig 3.22 below.
The bonding in metals is giant metallic. There are strong forces of attraction permeating throughout the
structure in all directions. The solid state is crystalline, that is, it is characterized by a regular three
dimensional array of particles, just as in ionic compounds.
Properties of metals
As in covalent and ionic compounds, the properties of metals are closely related to the nature of bonding.
1. Most metals have high boiling and melting points
The metallic bond is very strong. The electrostatic attraction between the positive atomic cores and the sea
of delocalized electrons is so strong that a large amount of energy is needed to pull atoms apart and break
down the crystal into its molten state.
Important note
The statement ‘metals have high melting points’ is a useful generalization but it has exceptions, for
example, sodium metal begins to melt at only 980C!
2. Malleable and ductile
Metals can be drawn into wires (they are ductile) and they can be moulded into different shapes
(they are malleable). The layers of atoms in metals are regularly arranged. It becomes possible for one layer
to slip over another. As the slipping occurs, the layers remain held together by the strong forces of attraction
between the positive atomic cores and the sea of delocalized electrons. This metallic attraction is nondirectional. It works in all directions and is therefore not broken when layers and atoms change positions.
The slipping of layers causes an increase in length and a decrease in width, and if this continues the metal
becomes a wire.
When a metal is hammered, the strain displaces and spreads out the atoms, but the structure does not easily
crumble because whatever new positions the atoms take, they are still held by the strong metallic forces.
Page
100
3. Good conductors of electricity
When a potential difference is applied between the ends of a metallic object, the mobile electrons in the sea
of delocalization are induced to move in one direction, thus allowing the transmission of electricity.
4. Can be alloyed
Alloying involves doping a metal with a small quantity of another metal, or other metals.
Metallic bonds are non-specific. If a different metal atom is introduced, it loses its valence electrons and it
is then attracted to the sea of delocalized electrons like any other atom in the structure. It is therefore the
non- specific attraction between electrons and positive atomic cores that allow atoms of other metals to be
integrated into the structure. There are several reasons for alloying metals, including:
(i) Increasing strength
The foreign atoms may introduce irregularities in the structure, especially if they have very different atomic
sizes to those of the metal being alloyed. Irregularity hinders the slip of layers and atoms when a
stress is applied. The metal is then stronger and it can not easily be deformed.
(ii) Increasing electrical conductivity
The added atoms may be able to lose a larger number of electrons into the lattice. This is particularly true
if the added metal is a transition element. There will be more electrons to carry charge than in the unalloyed metal.
(iii) Increasing resistance to corrosion
Stainless steel is an alloy of iron. It is well known for its resistance to rusting. The presence of other elements
such as chromium helps to prevent corrosion. For example, when Cr atoms react with atmospheric oxygen,
a tough layer of impermeable chromium (III) oxide is formed, which prevents further entry of oxygen and
water.
(iv) Enhancing magnetic properties
A metal is magnetic if it has unpaired electrons in its valence shell. The more unpaired electrons there are,
the stronger the magnetic properties of the metal. An un-paired electron behaves as a tiny magnet. This is
a property of every charged particle; when it moves, it becomes a tiny magnet, whose strength can be
measured in terms of the magnetic moment. If electrons are paired, their magnetic moments cancel each
other. This is because paired electrons spin in opposite directions. Their magnetic moments will have the
same magnitude, but they will be oriented in opposite directions, so they cancel each other out. Magnetism
is a property of the transition elements. Most of them have unpaired electrons in the outer-shell. Alloying a
non-magnetic metal with a transition element may give it magnetic properties.
5. Lustre in metals
Metals are lustrous (shiny). When electrons in the delocalized sea absorb quanta of light energy, they are
promoted to higher energy bands. They will not stay long in these unstable bands. They have a tendency to
fall back to their original energy levels. As they do so, they dissipate the same quanta of energy they had
gained. It is this emitted light that is picked by the eye as lustre.
A final note on bonding
Page
101
Bonding occurs to stabilize matter. What actually needs to be stabilized is the valence shell of atoms,
because they are the ones which are involved in chemical reactions. If the valence shell has an unstable
electronic configuration, it will tend to participate in reactions that ensure stabilization of the outer shell.
It is usually noticed that the valence shell becomes stable if it acquires the octet number of electrons. In
covalent substances, this happens by sharing of electrons between two atoms.
In ionic compounds, the metal atom achieves stability by shedding the unstable valence shell. This exposes
the underlying shell which has a full octet of electrons. The non-metallic atom attains stability by gaining a
few electrons into the valence shell. This gaining of electrons fills an octet and stabilizes the ion formed.
In metals, the metal atom also achieves an octet in the outer shell by shedding the unstable valence shell,
exposing the underlying shell, which has an octet of electrons.
Exercise 3.2
1
The elements of Group IV all form tetrachlorides with the general formula MCl4.
(a)
Draw a diagram of a molecule of SiCl4, stating bond angles.
(b)
Describe and explain how the volatilities of the Group IV chlorides vary down the group.
9701/04/M/J/2008
2 The molecule dichlorocarbene, CCl2, can be produced under certain conditions. It ishighly unstable,
reacting with water to produce carbon monoxide and a strongly acidic solution.
Suggest the electron arrangement in CCl2 and draw a dot-and-cross diagram showing this. Predict the
shape of the molecule.
9701/04/M/J/2010
3
The oxides of the third period include the following:
Na2O; MgO; Al2O3; SO2; SO3.
Showing outer electrons only, draw a dot-and-cross electron diagram for magnesium
oxide, MgO.
9701/02/O/N/2002
4 (a) Salt, sodium chloride, forms transparent colourless crystals. Describe the bonding insodium
chloride crystals, give the formula of each particle and sketch part of the crystal structure.
(b) Explain why crystals of sodium chloride do not conduct electricity, but molten sodium
chloride does.
9701/02/O/N/2003
5
Hydrogen sulphide, H2S, is a foul-smelling compound found in the gases from volcanoes.
Hydrogen sulphide is covalent, melting at –85 °C and boiling at –60 °C.
(i)
Draw a ‘dot-and-cross’ diagram to show the structure of the H2S molecule.
(ii)
Predict the shape of the H2S molecule.
(iii) Oxygen and sulphur are both in Group VI of the Periodic Table.
Suggest why the melting and boiling points of water, H 2O, are much higher than those of H2S.
9701/02/M/J/2005
6
Carbon disulphide, CS2, is a volatile, stinking liquid which is used to manufacture viscose, rayon
and cellophane.
Page
102
(a) The carbon atom is in the centre of the CS2 molecule.
Draw a ‘dot-and-cross’ diagram of the carbon disulphide molecule.Show outer electrons only.
(b) Suggest the shape of the molecule and give its bond angle and shape
9701/02/O/N/2002
7
Ethyne is a linear molecule with a triple bond, C≡C, between the two carbon atoms.
Draw a ‘dot-and-cross’ diagram of an ethyne molecule.
9701/02/M/J/2006
8
When separate samples of copper or iodine are heated to 50 °C, the copper remains as a solid while
the iodine turns into a vapour.
(i) Explain, in terms of the forces present in the solid structure, why copper remains a solid at
50°C.
(ii) Explain, in terms of the forces present in the solid structure, why iodine turns into a vapour
when heated to 50°C.
9701/02/O/N/2006
9.
Ethene, C2H4, and hydrazine, N2H4, are hydrides of elements which are adjacent in the Periodic
Table. Data about ethene and hydrazine are given in the table below.
(a)
Ethene and hydrazine have a similar
arrangement of atoms but differently shaped
molecules.
(i)
What is the H-C-H bond angle in ethene?
(ii)
Draw a ‘dot-and-cross’ diagram for
hydrazine.
(iii)
What is the H-N-H bond angle in
hydrazine?
(b) The melting and boiling points of hydrazine are much higher than those of ethene.
Suggest reasons for these differences in terms of the intermolecular forces each compound
possesses.
9701/02/M/J/2007
10
This question is about the bonding of covalent compounds.
(a) Sketch the shapes of a 1s, a 2s, and a 2px orbital.
(b) Covalent bonding occurs when two atoms share a pair of electrons. Covalent bonding may also be
described in terms of orbital overlap with the formation of σ bonds.
(i) How are the two atoms in a covalent bond held together? In your answer, state which particles
are attracted to one another and the nature of the force of attraction.
(ii) Draw sketches to show orbital overlap that produces the σ bonding in the H2 and
HCl molecules.
Page
103
(c)
The bond in the HCl molecule is said to be ‘polar’.
(i)
What is meant by the term bond polarity?
(ii) Explain why the HCl molecule is polar.
9701/04/O/N/2007
11
(a)
The structural formulae of water, methanol and methoxymethane, CH 3OCH3, are given below.
(i) How many lone pairs of electrons are there around the oxygen atom in methoxymethane?
(ii) Suggest the size of the C–O–C bond angle in methoxymethane.
The physical properties of a covalent compound, such as its melting point, boiling point,vapour
pressure, or solubility, are related to the strength of attractive forces between the molecules of that
compound.
These relatively weak attractive forces are called intermolecular forces. They differ in their strength
and include the following.
A
B
C
(b)
interactions involving permanent dipoles
interactions involving temporary or induced dipoles
hydrogen bonds
By using the letters A, B, or C, state the strongest intermolecular force present in each
of the following compounds.
ethanal CH3CHO
ethanol CH3CH2OH
methoxymethane CH3OCH3
2-methylpropane (CH3)2CHCH3
9701/02/M/J/2008
12 (a) Methanol and water are completely soluble in each other.
(i)
Which intermolecular force exists between methanol molecules and water molecules that
makes these two liquids soluble in each other?
(ii) Draw a diagram that clearly shows this intermolecular force. Your diagram should show any
lone pairs or dipoles present on either molecule that you consider to be important.
(b) When equal volumes of ethoxyethane, C2H5OC2H5, and water are mixed, shaken, and then
allowed to stand, two layers are formed.
Suggest why ethoxyethane does not fully dissolve in water. Explain your answer.
9701/02/M/J/2008
Page
104
13
Ketene, C2H2O, is a member of a class of unsaturated organic compounds that is widely used in
pharmaceutical research for the synthesis of organic compounds.
CH2=C=O
ketene
Suggest values for the H-C-H and C=C=O bond angles in ketene.
9701/02/O/N/2008
14
At low temperatures, aluminium chloride vapour has the formula Al2Cl6.
Draw a ‘dot-and-cross’ diagram to show the bonding in Al2Cl6. Show outer electrons only.
9701/02/M/J/2009
15
Elements and compounds which have small molecules usually exist as gases or liquids.
(a) Chlorine, Cl2, is a gas at room temperature whereas bromine, Br2, is a liquid under the
same conditions.
Explain these observations.
(b) The gases nitrogen, N2, and carbon monoxide, CO, are isoelectronic, that is they have the
same number of electrons in their molecules.
Suggest why N2 has a lower boiling point than CO.
(c)
A ‘dot-and-cross’ diagram of a CO molecule is shown below. Only electrons from outer shells
are represented.
In the table below, there are three copies of this structure.
On the structures, draw a circle round a pair of electrons that is associated with each of the
following.
(i) a co-ordinate bond (ii) a covalent bond (iii) a lone pair
9701/02/M/J/2010
16
Which statements are a true about the following molecules
Page
105
HCN
CO N2
1. They are isoelectronic
3. They all have a strong triple bond
2. N2 is the least reactive of the three compounds
CHAPTER
4
STATES OF MATTER
Page
106
Introduction
There are three states of matter, namely solids, liquids and gases. These states mainly differ in
 regularity in the packing of particles
 how loosely or tightly the particles are packed
 the kinetic energy of the particles
 shape
Bonding and structure in a substance is important in explaining the state at room temperature of that
substance, as well as its physical properties, such as melting points and boiling points. If particles of a
substance are held by very strong forces of attraction, then the substance is likely to be solid at room
temperature. At this temperature, there is no sufficient energy to pull apart the particles from each
other and form the liquid or gaseous state. Thus ionic solids and metals are solids at room temperature
because of the very strong electrostatic attractions that hold particles together. However, even covalent
substances which contain Van der Waals forces can be solid at room temperature if these forces are
strong enough. A good example is candle wax, in which the Van der Waals forces are strong enough to
hold alkane molecules together in the solid state. This is a result of the presence of a large number of
electrons in each molecule of candle wax.
Some substances exist as gases at room temperature. This shows that the intermolecular forces in the
substance are so weak that even at room temperature; there is enough energy to break the forces and
keep the molecules apart in the gaseous state. It is possible to liquefy and solidify any gaseous
substance, for example, neon and nitrogen. This can be achieved by lowering temperature, increasing
pressure, or both. The effect is to bring the molecules closer, reduce their kinetic energy and allow
stronger intermolecular forces to form.
4.1
The gaseous state
Gases are characterized by
 the random and loose ordering of particles. The particles are not arranged in any definite order,
and they are always rapidly shifting positions in an unpredictable way.
 high kinetic energy. The particles are in a state of random and rapid movement, frequently colliding
with each other and with the walls of the container.
 lack of shape. A gas has no particular shape because the gas particles always spread to fill he
available space.
4.1.1
Ideal versus real gas: What properties should a gas have?
If a gas meets all of the following criteria, then it is acting as expected of a gas. We say it is behaving as an
ideal gas. These criteria are referred to as the assumptions of an ideal gas.
Assumptions of an ideal gas
1. The assumption of zero volume
Each particle of a gas (it could be an atom as in the noble gases or a molecule as in oxygen) has a negligible
volume relative to the volume of the container. The ratio
Volume of molecule
Volume of container
is so small that it can be rounded off to zero.
Common error
Students commonly report that an ideal gas has no volume. This is not true. The gas as a whole certainly
has a volume. However, the volume of an individual particle is negligible.
2. Collisions are perfectly elastic
When particles collide, they bounce back without loss of energy. Thus collisions do not reduce the kinetic
energy of particles. Such collisions are said to be perfectly elastic.
3. There are no forces of attraction between particles
The gas particles are therefore able to move independently of each other because they are not associated
through intermolecular forces.
Page
107
4. An ideal gas obeys the ideal gas equation perfectly.
This equation is
PV = nRT
4.1.2 Validity of the assumptions
To what extend are the assumptions given above valid?
1. The assumption of zero volume.
It does not make sense to assign zero volume to a particle. If it exists, then it has a volume. Anything which
can be shown to exist is matter, and matter is that which occupies space, that is, has a volume. Each particle
of a gas has a volume. However, the volume is so small that for practical purposes, it may be rounded off to
zero.
2. Collisions are perfectly elastic
This is not practical. Whenever two particles collide, energy is lost as heat due to friction. However, the loss
of energy is very small owing to the small size of the particles.
3. There are no forces of attraction between particles
Electrostatic attractions between particles are inevitable. They will always exist. However, in gases, the
forces are very weak because of the large distance between particles. Further still, the particles are always
moving rapidly and randomly. This causes a continuous disruption of the intermolecular forces.
4. Ideal gases obey the ideal gas law.
The ideal gas equation is derived from assumptions 1, 2 and 3. Since these assumptions have their own
limitations, the fourth assumption is also limited. Real gases do not obey the ideal gas equation perfectly.
4.1.3 Real gases
In practice, ideal gases do not exist. Instead, we have real gases which approach (but will never quite
attain) the ideal state under certain conditions.
In a real gas
 each particle has a volume
 When particles collide, they lose some energy
 There are forces of attraction between particles
Because of these properties, a real gas does not perfectly obey the ideal gas equation.
Under what conditions does a real gas approach ideal gas behavior?
Page
108


At high temperature
At low pressure
At high temperature, the particles gain kinetic energy and spread away from each other, that is, the gas
expands. The volume of a particle then becomes negligible compared to the volume of the container. Also,
intermolecular forces become almost non-existent as distance between particles increases.
The reverse happens when a gas is cooled. The particles come closer and intermolecular forces begin to
form between the particles. The volume of a particle relative to the total volume occupied by the gas also
becomes important as the particles get closer and closer together. That is, a real gas deviates most from
ideal gas behavior at low temperatures.
At low pressure a gas is close to being ideal. When pressure is reduced, a gas expands (assuming constant
number of moles). The particles spread away and this favours the ideal behaviour. At high pressure a gas
deviates from ideality by forcing the particles into a smaller volume.
4.2
The ideal gas equation
The following observations (for a fixed mass of gas) lead to the ideal gas equation:
1. Volume varies directly as temperature, provided pressure is kept constant (Fig 4.1)
V ∝ T,
V = k1T
…
I
Increasing temperature causes a gas to
expand. The volume increases
proportionally with increasing
temperature. The reverse happens when
the gas is cooled.
Fig 4.1
2. Volume varies inversely as pressure, provided temperature is constant (Fig 4.2).
V ∝
𝟏
𝐩
,
V = k2/P
Increasing pressure results in compression of
the gas, that is, volume decreases, but of
course, it can not become zero, hence the
inverse nature of the graph.
Fig 4.2
3. Pressure varies directly as the number of moles of gas (Fig 4.3)
P ∝ n,
P = k3n … III
Increasing the number of moles of a gas results in a proportional increase in pressure, provided temperature and
pressure are kept constant.
Fig 4.3
4. Pressure varies directly as temperature (Fig 4.4)
Page
109
P = k4T ... (IV)
Pressure of a gas is a measure of the frequency with which
gas particles strike the walls of the
container. When
temperature increases, so does the kinetic energy of the
particles. They strike the walls of the container more
frequently, that is, pressure increases.
Fig 4.4
Combining the relationships (I) to (IV) gives the ideal gas equation
PV = nRT
Examiners frequently set questions that require use of the ideal gas equation. The problem students usually
face is in using the correct units.
Important derivations from the ideal gas equation
(i) Determination of Mr of an ideal
gas or a volatile liquid
PV = nRT ... (i)
But n =
m
Mr
substituting n in equation (i) gives
Page
110
PV =
Rearranging gives
, where m = mass in grams
m
RT
Mr
(ii) Determination of density of an
ideal gas
PV =
m
RT
Mr
Rearranging gives mass m in grams
Mr =
𝐦𝐑𝐓
𝐏𝐕
This form of the general gas equation is useful in the
determination of the Mr of a gas or a volatile liquid (the
liquid must be vapourised first).
That is, m = D x V ... (ii)
Equating equation (i) and equation (ii)
DxV=
PV x Mr
RT
and dividing both sides by V
D=
𝐏 𝐱 𝐌𝐫
𝐑𝐓
PV x Mr
m=
Now, D =
RT
m
V
... (i)
, Where D = density
(iii) Pressure of a combined system
In Fig 4.5 , two flasks A and B are arranged so that they can be combined into one system by opening the
tap. V1 and V2 represent the capacities, in m3 of the two flasks respectively. Before connection, the
pressure in flask A is P1 and that in Flask B is P2.
What would be the total pressure, PT in the combined arrangement after connection?
Fig 4.5
From PV = nRT,
we find n1 and n2, the number of moles of gas in flask A and B respectively.
n1 =
and
n2 =
P1V1
RT
P2V2
RT
Total number of moles, nT is given by
nT = n1 + n2
=
factorizing LHS
P1V1
RT
+
P2V2
RT
(P1V1 + P2V2) =
RT
Page
111
multiplying both sides by RT gives
rearranging gives
PTVT
RT
PTVT
RT
P1V1 + P2V2 =
PT =
=
PTVT
P1V1 + P2V2
VT
Where VT is the combined volume
(iv) The relationship P1V1 = P2V2
For n moles of an ideal gas at a pressure P1 and volume V1, the ideal gas equation is given by
P1V1 = nRT ... (i)
Now suppose that pressure and volume are increased to P 2 and V2 respectively at constant temperature.
The ideal gas equation becomes
P2V2 = nRT ... (ii)
Since the right hand-sides of equations (i) and (ii) are equal, it follows that the left hand sides are also equal,
that is,
P1V1 = P2V2.
This relationship is only true for an ideal gas at a constant temperature and constant number of moles.
Q
A gas occupies 87 cm3 at a pressure of 87.5 KPa. What pressure is required to reduce the
volume of the gas by 2% ?
A
Volume decreases by 87 cm3 x 2/100 = 1.74 cm3. The new volume (V2) is therefore
87.00 - 1.74 = 85.26 cm3. Let the new pressure be P2
P1V1 = P2V2
P2 =
P₁V₁
V₂
=
87.5 x 87
85.26
= 89.29 KPa
4.2.1
Deviations from ideality
You now know that a gas deviates the most from ideal behaviour when pressure is high and when temperature is low. Look again at the ideal gas equation, PV = nRT.
When temperature and number of moles are kept constant, we may write
PV = constant ... (i) (since R is also constant).
This outcome can be shown graphically. Suppose that volume is increased, that is, the gas expands (Fig
4.6(a)). This increase is countered by a proportional decrease in pressure, so the product PV remains constant. Similarly, when pressure is increased (Fig 4.6(b)), that is, the gas is compressed, volume decreases
proportionally, so the product PV also remains constant.
PV
PV
PV = constant
PV = constant
(a)
(b)
112
V
Page
Fig 4.6 For an ideal gas, the product PV
remains constant when pressure or
volume is altered, provided temperature
and amount of gas (moles) is kept
constant.
P
In practice, a plot of PV against V or P is not constant but becomes approximately constant at high
temperature and pressure (the gas approaches ideality under these conditions). In other words, for a real
gase, the graph of PV against P or V deviates from the horizontal line shown in Fig 4.6 above. This is
llustarted in Fig 4.7 below.
At low pressures (less than P2), a real gas shows negative deviation from ideal gas behaviour (the curve is
below the line for an ideal gas).
A real gas shows positive deviation from ideal gas behaviour when pressure is sufficiently high. The curve
is then above that of a real gas.
Fig 4.7 deviation of a real gas
from ideality
Q
Which of the following gases deviates the most from ideal gas behaviour. Explain your
answer.
A. NH3
A
B. HCl
C. O2
D. N2
NH3. Intermolecular forces are strongest in ammonia, where there is hydrogen bonding. The
assumption that an ideal gas possesses no forces of attraction between molecules is therefore the
least valid for ammonia.
Fig 4.8 below shows the relative deviations of four gases NH 3, HCl, O2 and H2 from ideality.
Fig 4.8


Page
113


NH3 contains relatively strong hydrogen bonds, so it deviates the most from ideal gas behaviour.
Comparing HCl, O2 and H2, HCl has the strongest intermolecular forces (permanent dipole-dipole
attractions), so it shows the greatest deviation.
Both H2 and O2 contain relatively weak Van der Waals forces. However, these forces are stronger
in oxygen which has the larger number of electrons. Oxygen therefore deviates more from ideal gas
behaviour.
In hydrogen, intermolecular forces are a very weak. Deviation from ideality is the smallest and it is
positive. Of all gaseous elements, hydrogen (Ar = 2) resembles an ideal gas the most, followed by
helium (Ar = 4). Note that each particle of hydrogen and helium contains two electrons. However,
helium atoms are heavier, so at any given temperature, they have the smaller kinetic energy. This
favours establishment of intermolecular forces.
Page
114
Worked examples
Answer D
1.
Using PV =
Use of the Data Booklet is relevant to this question.
Which expression gives the pressure exerted by
1.6 x 10–3 mol of N2 in a container of volume 3.0 dm3 at
273 oC?
A.
B.
C.
D.
1.6 x 10¯3 x 8.31 x 273
3.0 x 10¯⁶
3.0 x 10¯⁶
3.0 x 10¯³
Pa
1.6 x 10¯3 x 8.31 x (273+273)
3.0 x 10¯³
Pa
9701/01/O/N/2004
mRT
PV
83.1
10⁶
Pa
Pa
RT
Remember to convert temperature to K and volume to m3
(83.1 cm3 =
1.6 x 10¯3 x 8.31 x (273+273)
1.6 x 10¯3 x 8.31 x 273
Mr =
m
Mr
m3 = 83.1 x 10-6 m3)
3.
When an evacuated glass bulb of volume 63.8 cm3 is
filled with a gas at 24 °C and 99.5 kPa, the mass increases
by 0.103 g. Deduce whether the gas is ammonia,
nitrogen or argon.
9701/02/M/J/2004
Solution
Mr =
mRT
PV
=
0.103 x 8.31 x (24 + 273)
99.5 x 1000 x 63.8 x 10¯⁶
= 40.0
Answer D
The gas is therefore argon.
using PV = nRT
P=
nRT
4.
... (i)
V
First convert volume to m3 and temperature to K:
3.0 dm3 =
3
10³
m3 = 3.0 x 10-3 m3
2730C = (273 + 273) K
R is the ideal gas constant = 8.31 JK-1mol-1 (Data
Booklet)
then substitute in equation (i) above.
A small spacecraft of capacity 10m3 is connected to
another of capacity 30m3. Before connection, the pressure
inside the smaller craft is 50KPa and that inside the larger
is 100KPa. If all measurements are made at the same
temperature, what is the pressure in the combined
arrangement after connection?
9701/01/O/N/1990
Solution
Use PT =
2.
Use of the Data Booklet is relevant to this question.
=
In an experiment using a gas syringe, 0.10 g of a gas is
found to occupy 83.1 cm3, measured at standard pressure
(1.0 x 105 Pa) and 27°C.
𝐏₁𝐕₁ + 𝐏₂𝐕₂
𝐕𝐓
50 (10)+ 100(30)
30+10
= 87.5 KPa
What is the relative molecular mass of the gas?
Page
115
A.
C.
0.10 x 8.31 x 27
1.0 x 105 x 83.1
0.10 x 8.31 x 27
1.0 x 105 x 83.1 x 10¯⁶
B.
D.
0.10 x 8.31 x 300
1.0 x 105 x 83.1
0.10 x 8.31 x 300
1.0 x 105 x 83.1 x 10¯⁶
5.
When a 0.150g sample of substance X was vapourized
at 600C, the vapour occupied a volume of 70.0 cm3 at a
pressure of 101KPa. Calculate the Mr of X
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise 4.1
text box.]
1. When an evacuated glass bulb of volume 63.8 cm3 is filled with a gas at 24°C and 99.5 kPa, the mass
increases by 0.103 g. Deduce the density of the gas.
Hint
D=
𝐏 𝐱 𝐌𝐫
𝐑𝐓
(First find Mr of the gas)
2. The density of ice is 1.00 g cm–3.
What is the volume of steam produced when 1.00 cm3 of ice is heated to 323 °C (596 K) at a pressure
of one atmosphere (101 kPa)?
[1 mol of a gas occupies 24.0 dm3 at 25 °C (298 K) and one atmosphere.]
A 0.267 dm3
B 1.33 dm3
C 2.67 dm3
D 48.0 dm3
9701/01/M/J/2008
Hint
Since D = m/V, you can find the mass (m) of ice . This will also be the mass of steam. Then use the formula
𝐦
PV = RT
𝐌𝐫
3. When used for cutting or welding, ethyne is transported in cylinders which contain the gas under
pressure. A typical cylinder has a volume of 76 dm3 and contains ethyne gas at 1 515 kPa pressure at
a temperature of 25 °C. Use the general gas equation, pV = nRT, to calculate the amount, in moles,
of ethyne in this cylinder.
9701/02/M/J/2006
4. Flask X contains 1dm3 of helium at 2 kPa pressure and flask Y contains 2dm3 of neon at 1 kPa
pressure. If the flasks are connected at constant temperature, what is the final pressure?
1
A 1 kPa
3
5.
1
B 1 kPa
2
2
C 1 kPa
3
D 2kPa
9701/01/M/J/2002
A 2 g sample of hydrogen at temperature T and of volume V exerts a pressure p.
Deuterium, ≾ H, is an isotope of hydrogen.
Page
116
Which of the following would also exert a pressure p at the same temperature T ?
A
2g of deuterium of volume V
B
4g of deuterium of volume
C
a mixture of 1 g of hydrogen and 2 g of deuterium of total volume V
D a mixture of 2 g of hydrogen and 1 g of deuterium of total volume
9701/01/O/N/2002
4.3
The solid state
Crystalline versus amorphous solids
It is common for science students to refer to solids as having a regular arrangement of solid particles. This
is not always the case.
Solids with a regular three dimensional arrangement of particles are said to be crystalline. The system of
regularly repeating particles is known as a lattice. Some solids have particles which are irregularly arranged. Such solids are said to be amorphous (non-crystalline). Examples of amorphous solids include glass
and coal (a very impure form of carbon).
Crystalline solids
A crystalline solid can be a covalent, ionic or metallic substance. Ionic compounds are crystalline in nature.
These have already been discussed in Chapter 3, using NaCl as an example. Metals also have a crystalline
form.
When most elements and compounds turn to solid, they form a regular array of particles. Some substances
are gaseous or liquid at room temperature. When we discuss their solid states, we are referring to what we
would get after freezing the substance.
To describe the solid crystalline state of a given substance, reference should be made to:
1. Type of chemical bonding present
This should be described as simple or as giant. The bonding in metals is giant metallic. Bonding in ionic
solids is giant ionic. In covalent compounds there could be giant covalent bonding, as in silicon dioxide,
graphite and diamond, or simple covalent bonding, as in carbon dioxide, iodine and argon. In substances
with giant bonding, the bonding repeats and extends throughout the structure in all directions, without
breaks.
2. Nature of the lattice particles
These are the particles which build up the solid. Lattice particles can be positive and negative ions (ionic
compounds), positive atomic cores (metals), atoms (noble gases) or molecules as in ice and iodine.
3. Forces holding the lattice particles together.
These are intermolecular forces in simple covalent substances: Van der Waals forces, permanent dipoledipole attractions, or hydrogen bonds.
In ionic compounds these are ionic bonds (electrostatic attractions). In metals, the lattice particles are held
together by metallic bonds (electrostatic attractions)
4.3.1 Ionic solids



Page
117
There is giant ionic bonding.
Lattice particles are positive ions (cations) and negative ions (anions).
These ions are held together by strong electrostatic attractions extending throughout the structure
in all directions. Being directional, these forces hold the ions in a regular three dimensional array.
 NaCl has already been used to illustrate these points.
The properties of ionic compounds should be explained in terms of the structure and bonding present (see
Chapter 3).
4.3.2
Metallic solids

The bonding in metals is giant metallic bonding.


The lattice particles present are positive metal atoms (not cations)
These atoms are held together by the electrostatic attractions between them and a sea of delocalized electrons (Fig 4.6). These forces of attraction work in all directions (non-directional) and they
hold the atoms in a regular three dimensional array. The forces of attraction extend in all directions
throughout the structure without breaks.
The properties of metals are closely related to the structure and nature of bonding (See section 3).

Copper and aluminium
Aluminium


Fig 4.6

Aluminium has a high mass to strength ratio,
that is, it is very strong and yet very light. This
makes it useful in situations were lightness and
strength should be present together. For this
reason, it is used to make aircraft bodies.
Aluminium is also resistant to corrosion. This is
because it forms a thin superficial (on the
surface) layer of Al2O3. This layer is tough and
impermeable, so it protects the underlying metal
from further reaction with oxygen and water.
This, plus its excellent ability to conduct heat,
makes it useful in making cooking pots.
Aluminium is an excellent electrical conductor and is used to make electrical cables. However, it is
not used extensively for this purpose because it is expensive.
Copper

The uses of copper depend on its malleability and ductility. Whilst it is true that all metals are
generally ductile and malleable, some metals are more ductile and malleable than others.
 it is an excellent conductor of heat and electricity.
 it is inert and so resists attack by atmospheric agents( corrosion)
 it can be alloyed with many other metals ability .
Copper is thus used to make electrical cables and components, car radiators (it conducts heat from the
engine rapidly and loses it to the environment easily, it also resists corrosion) and in water pipes
(resistant to corrosion). Pure copper tends to be too soft and weak for some uses. Alloying produces
metals which are stronger and harder.
Page
118
Bronze (reddish-brown) is an alloy of copper and tin. Bronze has greater tensile strength than copper. It
can also be cast into moulds more easily. For this reason, it is used to make sculptures, such as the one
shown in Fig 4.7.
Other uses of bronze include :
1. For electrical connectors (it is a better conductor than most steels)
2. Boat and ship propellers (It is resistant to corrosion)
Brass (muted yellow, but duller than gold) is an alloy of copper with zinc. It is
harder and stronger than copper. Its colour makes it decorative. The uses of
brass include
1. Making musical instruments (brass has excellent acoustic properties)
2. In ornaments.
Fig 4.7 A bronze statue
4.3.3 Simple covalent solids



The bonding present is simple covalent bonding.
The lattice particles are molecules such as iodine and carbon dioxide. They can also be atoms e.g.
argon.
The particles are held together by weak intermolecular forces, e. g Van der Waals forces in solid
iodine, and hydrogen bonds in ice. However, in solids, these forces are strong enough to hold the
particles in fixed positions, forming a regular three dimensional structure.
The lattice structure of solid iodine
Fig 4.2 above is a sketch showing the lattice structure of iodine.
The lattice particles are iodine molecules (I2), which are held
together by weak Van der Waals forces.
 This explains why iodine easily sublimes to gas. A small
amount of heat energy is required to break the Van der
Waals forces between the molecules.
 Iodine is poorly soluble in water. For iodine molecules to
dissolve in water, the relatively strong hydrogen bonds
between water molecules must be broken first. When
new Iodine - water interactions form (Van Der Waals
Fig 4.8
forces), they are much weaker than the hydrogen bonds
found in pure water. Such a process in which stronger forces are replaced by weaker forces is not
energetically favourable.

The lattice structure of ice
Page
119
The lattice particles are water molecules, held together by hydrogen bonds. A peculiar property of ice is
that it is less dense that liquid water, and so tends to float on the surface of the liquid. This property is due
to the open structure of ice (Fig 4.9). Each water molecule is tetrahedrally bonded to four others through
hydrogen bonds. This geometry spaces out the water molecules, resulting in an open structure whose
volume is much higher than its mass, that is, water tends to expand as it freezes. This lowers its density.
4.3.4 Giant covalent substances
In giant covalent substances, there are a large
number of strong covalent bonds extending
throughout the structure in all directions.
The lattice particles are molecules or atoms, but
these are connected to each other by strong covalent bonds without breaks, forming a giant
structure. Examples are silicon dioxide, diamond
and graphite.
Bonding in graphite
Graphite is an allotrope of carbon. Carbon has
two other known allotropes, diamond and
Fig 4.9 The lattice structure of ice
buckminsterfullerenes.
2
2
2
The electronic configuration of C is 1s 2s 2p . With this configuration, C can form two bonds only, because
it has two singly occupied p orbitals. An electron is excited from the 2s sub-shell to a vacant p orbital. This
creates four singly occupied orbitals. There are now four orbitals which can be used for bonding. However,
these orbitals have to be hybridized first to make them equivalent to each other, and to ensure that they
have the correct shapes for overlap. sp2 hybridization takes place, that is, one s orbital mixes with two p
orbitals to form three hybrid orbitals (Fig 4.10). This leaves the pz orbital occupied but unhybridized. When
we discuss the bonding in graphite we should account for how each carbon atom uses its three singly
occupied hybrid orbitals, and its unhybridized pz orbital.
Fig 4.10
Page
120
Each carbon atom in graphite uses the three hybrid orbitals to overlap head-on with corresponding hybrid
orbitals from three neighbouring carbon atoms. This forms three sigma bonds around each carbon atom.
The geometry around each carbon atom is therefore trigonal planar. This overlap results in one carbon
atom being bonded to three other carbon atoms, which must therefore be spaced out at 1200 relative to each
other to minimize repulsion between the three bond pairs of electrons.
Fig 4.11
The perpendicular unhybridized pz orbitals on the carbon atoms overlap sideways to form a pi bond which
stretches throughout the layer (Fig 4.12).
Fig 4.12
Each carbon atom therefore contributes one electron to the pi bond. These electrons are delocalized
throughout the layers. In graphite, there are a large number of such layers, held together by Van der Waals
forces. Fig 4.13 shows the key features of graphite.
The C-C bond length in graphite is intermediate between a single C to C bond and a C to C double bond.
This shows that the covalent bond is reinforced to some extend by pi bonding. The bonds are therefore
stronger than C to C single bonds. This accounts for the high melting point of graphite (3 730° C). A very
large amount of heat energy is required to break the strong covalent bonds and separate the carbon atoms
from each other.
The uses of graphite are closely related to its structure and the nature of bonding present (table 4.2).
Three properties of graphite which makes it a very useful substance are



It is slippery
It is a good electrical conductor along layers (it does not conduct across layers, that is, at right
angles to the layers)
It has a very high melting point
Page
121
Fig 4.13
Key features of graphite
Table 4.2
Uses of graphite in relation to its properties.
Use
Explanation in terms of structure and bonding
Dry lubricant
Slippery due to the presence of weak Van der Waals forces between layers.
Being weak, these forces easily break, allowing layers to slide against each
other.
Graphite is very useful as a lubricant in industrial processes which involve
very high temperatures. It is not easily decomposed by heat because of its
very high melting point.
Electrodes in electrolytic
processes.
It is a good conductor of electricity along layers due to the presence of
delocalized π electrons.
NB Graphite does not conduct across layers.
Heat resistant crucibles for
molten metals
It has a very high melting point.
Bonding in diamond
As in graphite, the lattice particles in diamond are C atoms. These are also connected by strong covalent
bonds that extend in all directions throughout the structure. The atoms are arranged in a regular three
dimensional array, that is, diamond is a crystalline solid. Like graphite, diamond has a giant covalent
structure.
Page
122
In diamond there is sp³ hybridization. Excitation of an
electron from the 2s sub-shell to the vacant 2pz of a carbon
atom creates four singly occupied orbitals. All four of them mix
to form four hybrid orbitals. Each hybrid orbital is occupied
by a single electron. This is sp³ hybridization (as in methane),
because one s orbital mixes with three p orbitals. These four
hybrid orbitals are arranged tetrahedrally (bond angle=
109.50) around the C atom (Fig 4.14). This is the geometry
which minimizes inter-electronic repulsion between the hybrid
Fig 4.14
orbitals.
Each carbon atom bonds tetrahedrally to four others by sigma overlap. This forms C to C sigma bonds that
extend throughout the structure (Fig 4.15).
Properties and uses of diamond
Fig 4.15
Structure of diamond
1. Diamond is one of the hardest substances
known.
Its hardness is due to the presence of a large
number of very strong covalent bonds extending
in all directions.
Because of its hardness, it is used to make the
tips of drilling instrument, for example, in
mining equipment. Diamond tipped electrical
saws are also used to cut diamonds. (Diamond
cuts diamond).
2. Diamonds have a large refractive index. Refraction of light by diamonds gives them a sparkle. They are
therefore used in jewelry.
Table 4.3 gives a comparison of the two allotropes of carbon, diamond and graphite.
Silicon dioxide
Silicon dioxide is a giant covalent substance with a diamond like structure (Fig 4.16).
The simplest repeating unit (lattice particle) is SiO 2. Each Si atom is
tetrahedrally bonded to four O atoms, so the bond angle, as in diamond,
is 109.50. Each O atom is in turn bonded to a Si atom. This network of SiO covalent bonds extends throughout the structure in all directions. The
properties of silicon dioxide are similar to those of diamond. It has a very
high melting point because to melt it a large number of very strong
covalent bonds must be broken to form discrete SiO2 molecules.
The presence of a large number of very strong covalent bonds also makes
it hard and rigid.
Fig 4.16
4.3.5
Ceramics
Page
123
Ceramics are ionic or giant covalent materials whose uses depend on their strength, resistance to heat,
chemical inertness and insulating properties.
These properties are a result of strong and numerous bonds that stretch throughout the structure. The
bonds hold atoms rigidly in space and this explains their hardness. Ceramics with an ionic nature, for
example, magnesium oxide and aluminium oxide, find widespread use because of their very high melting
points. Magnesium oxide is used to line furnaces and as electrical insulators in equipment which are likely
to become very hot.
Aluminosilicates are clays based on the SiO2 structure. They have a sheet structure in which some silicon
atoms are replaced by Al atoms. When the clay is fired, water is driven from between the sheets and a three
dimensional network of bonds is established. Fired clay is used to make a wide range of useful products,
including china, bricks and crockery. The main disadvantage of ceramics is that they are brittle.
Table 4.3
Comparison of graphite and diamond
Question and answer section: states of matter
Q1.
Which substance has a diamond-like solid structure?
A. SiO2
A.
B. Graphite
C. Ice
D. Copper
SiO2
Q2. Which substance does not contain delocalized electrons
A. Silicon
A.
B. Graphite
C. Copper
D. Aluminium
Silicon.
Q3. Which gas deviates the least from ideal gas behaviour under the same conditions of temperature and pressure ?
A. Hydrogen
A.
Q4.
124
Page
C. Carbon dioxide
D. Neon
Hydrogen
Which substance(s) contain two types of chemical bonds?
(i) Ice
A.
B. Helium
(ii)
graphite
(iii) diamond
(i) and (ii) are correct. Both substances contain strong covalent bonds and weak Van der Waals forces.
Q5.
What happens when water freezes?
(i) Its volume decreases
A.
Q6.
Q7.
Which relationship(s) are correct for an ideal gas
(ii) P α T (iii) PV α T
(ii) and (iii) are correct. PV = nRT, but for any given ideal gas, nR is constant, so PV = kT where K = nR.
that is, PV α T. When temperature increases, both P and V increases , so the product PV also increases.
Which of the following solids has a simple molecular lattice?
A . magnesium oxide
A.
Q8.
(iii) Its molecules spread apart
(ii) and (iii) are correct.
(i) Vm α P
A.
(ii) Its density decreases
B. sodium
C. silicon (IV) oxide
D. sulphur
Sulphur. It exists as S8 molecules
For an ideal gas, the plot of pV against p is a straight line. For a real gas, such a plot shows a deviation from
ideal behaviour. The plots of pV against p for three real gases are shown below. The gases represented are
ammonia, hydrogen and nitrogen.
What are the identities of the gases X, Y and Z?
A
B
C
D
A.
Page
125
Q9.
X
ammonia
hydrogen
nitrogen
nitrogen
Y
nitrogen
nitrogen
ammonia
hydrogen
Z
hydrogen
ammonia
hyrogen
ammonia
D
Which diagram correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured
in K.)
9701/01/M/J/2008
APTER
5
Page
126
A.
D. Graph in A should be inverse, not linear. Recall that for an idea gas at constant temperature,
PV = constant. The graphs in B and C should therefore straight lines that are parallel to the x - axis.
CHEMICAL ENERGETICS
Chemistry is concerned, among other things, with studying the changes that matter may undergo. These
changes can be chemical (involving the formation of new substances) for example, the conversion of hydrogen and nitrogen to ammonia, or they can be physical, for example, the sublimation of iodine.
Chemical energetics (thermodynamics) is concerned with answering the question
‘Is the process feasible (possible) or not, and why?’
For example,
Is the conversion of graphite to diamond feasible (graphite and diamond are allotropes of carbon)?
Note the use of the term ‘feasible’. If a process is feasible, it means it is possible though in practice it may
not happen. For example, it is possible (feasible) for a boulder at the top of a hill to roll down, but whether
this actually happens or not depends on other factors. Chemical energetics simply tries to predict the
feasibility of a process. Whether in practice the process actually takes place or not is beyond the scope of
energetics, but is addressed by another branch of chemistry, known as reaction kinetics. For instance,
the conversion of graphite to diamond is energetically feasible. From an energetic point of view, it is
possible to convert the graphite (‘lead’) in your pencil to diamonds! In practice, this does not take place,
or takes place under conditions which are so stringent that the whole process is uneconomic. Chemical
energetics and reaction kinetics therefore represent two points of view in chemistry. A process might be
feasible from an energetic point of view, but from a kinetic point of view, it might not take place.
In this topic we explore why chemical processes occur and we get an idea of how to predict feasibility of
a reaction. We make this analysis by studying the transfer of heat energy during reactions.
5.1
Absorption and release of energy during chemical reactions
Page
127
Chemical energetics (thermodynamics) is the branch of chemistry that studies the flow of heat energy
during chemical processes.
In a chemical reaction, this flow of energy is in two parts
 Absorption of energy when bonds break. For a chemical reaction to take place, bonds in
reactants must first break, and this requires energy.
 Release of energy when new bonds are formed. The second process involved in a chemical
reaction is bond formation. This occurs when atoms of reactants rearrange and bond in new ways.
Bond breaking is said to be an endothermic process, that is, one which absorbs or requires heat energy
in order to take place. Bond formation is an exothermic process, that is, heat energy is released when a
bond is formed.
5.2
Energy content
Chemical reactions are usually accompanied by a change of energy content of the reacting substances. This
change in energy content manifests as a loss or gain of heat energy during the reaction. The energy content
of products and reactants can therefore be conveniently expressed in terms of heat energy,
referred to as enthalpy, H. Although it is impossible to actually determine the absolute energy content(enthalpy) of reactants and products, it is possible to determine the energy difference between them.
This energy difference, known as an enthalpy change, ΔH, manifests as net absorption or net release of
heat energy during the reaction.
You are probably familiar with reactions which give out heat energy. Such reactions are said to be
exothermic. In such reactions, the temperature of the contents of the reaction vessel increases. On the other
hand, endothermic reactions, which absorb energy, are accompanied by a decrease in temperature of the
products.
Understanding exothermic and endothermic reactions
Recall that a chemical reaction involves two processes, bond breaking (endothermic) and bond formation
(exothermic). In terms of energy flow, these two processes oppose each other. Bond breaking absorbs (uses)
energy whereas bond formation releases energy. The net energy flow during the reaction
(net absorption or net release) depends on which process has a numerically greater energy (enthalpy) term.
An endothermic process is assigned a positive enthalpy value. This is because the reactants gain
(absorb) energy which is then transferred to products during the reaction. In other words, the products are
now at a higher energy level than the reactants. In other words, the energy content of the products increases.
An exothermic process is assigned a negative enthalpy value. The negative sign denotes net loss of energy
to the environment. The energy content of the products therefore decreases.
We may represent an endothermic process with an arrow pointing up, showing an increase in energy
content of the products. An exothermic process may then be represented by an arrow pointing downward,
showing a net loss of energy.
Consider a hypothetical reaction in which 500Kj of energy are absorbed to break bonds and 800 Kj are
released when new bonds are formed in the products (Fig 5.1). The net flow of energy, ∆H, is obtained by
adding the two enthalpy terms. Thus
ΔH = 500 + (-800) = - 300Kj
The negative sign indicates that there is a net release of heat energy. In words, we say
Page
128
‘The enthalpy change of the reaction is -300Kj’ or “the process releases 300Kj of heat energy’.
Fig 5.1 Energy
‘transactions’ of a chemical
reaction
Note that it is not proper to say ‘the process releases -300Kj of energy’, because the use of the term
‘releases’ denotes a negative enthalpy.
The enthalpy change of the process can be shown on an energy level diagram (Fig 5.2).
In this reaction, more heat energy is given out when new
bonds in products form than is used when bonds in
reactants break. On the other hand, if a reaction is
endothermic, the process of bond breaking uses up more
energy than is released by the process of bond formation.
The net effect is therefore absorption of energy (Fig 5.3).
An example of an endothermic process: the
dissolution of ammonium nitrate in water
Fig 5.2 Energy level diagram of an
exothermic reaction which gives
out 300Kj of energy
thermometer
Page
129
ammonium
nitrate
Fig 5.3 In an endothermic reaction
there is a net absorption of energy, so the
products are higher in energy.
Fig 5.4
Ammonium
nitatre dissolving
in water
In this process, the thermometer registers a decrease in
temperature. The walls of the beaker will also feel cooler.
This is a property of an endothermic reaction: temperature
drops during the process.
Some readers may see a contradiction here and expect temperature to increase since heat energy is being
absorbed. As an example, consider an endothermic reaction which takes place in solution. There is a net
flow (absorption) of heat energy from the solution into the bonds of the products. Because heat energy has
moved from the solution to bonds of the product, it can be said that the solution has lost heat energy, not
to the environment, but to the bonds of the products, so its temperature should drop. During an
endothermic process, there is therefore a net transfer of energy to the bonds of the products. This energy
is not kept in the bonds as heat energy, but as chemical energy, otherwise the temperature would not
change. An endothermic process is therefore accompanied by an increase in the energy content (potential
energy) of the products, but temperature of the system drops.
A well known example of an endothermic reaction is photosynthesis. During this reaction, light energy is
absorbed and stored in the bonds of the product (carbohydrate) as chemical potential energy.
Now consider an exothermic process
Neutralization of a strong monoprotic acid by a strong base such as NaOH releases about 57 Kj of energy
per mole of the acid used (that is, ∆H = -57Kjmol-1). This is an exothermic process in which bond
formation in the products releases more energy than is used to break bonds in the reactants. The excess
energy enters the solution as heat, so a thermometer would record an increase in temperature.
Energy content of the products therefore decreases due to the conversion of potential chemical energy to
heat, but the temperature of the system increases because it has gained heat.
An important note on the +, - notation in chemical energetics
Page
130
Which enthalpy change is greater, -100Kj and -200Kj? Students need to be careful here. From a
consideration of directed numbers in mathematics, students might say that -100Kj is greater than -2ooKj, but this is
wrong. From a point of view of chemical energetics, -200Kj is the larger number. It is important to realize that the signs
+ and – are only symbolic, they simply tell us whether a process releases energy (-) or absorbs energy (+). A process
with an enthalpy change of -200Kj gives out 200Kj of energy, but a process with an enthalpy change of -100Kj gives out
only 100Kj. The second process clearly has the smaller enthalpy change (Fig 5.4). It is often more useful to use the
terms more exothermic/ less exothermic, for example, the process whose ∆H value is -200 Kjmol-1 is more exothermic
than the one with a value of -100Kjmol-1.
Fig 5.4 Process A has the greater enthalpy, disregarding the sign of
the value, since the sign is only symbolic.
5.3
Enthalpy of reaction and stability
The sign of the enthalpy change of a reaction, that is, whether it is positive or negative, helps in predicting
relative stability of products and reactants.
If a reaction is exothermic, there is a net loss of energy from the bonds of reactants, so the products should
be at a lower energy level. Since the products have less potential chemical energy, they are more stable,
relative to the reactants. From an energetic point of view, such a reaction is feasible (Fig 5.5).
Page
131
Fig 5.5 If a reaction is
exothermic, the products are
lower in energy, and therefore
more stable than the reactants.
The relative stability of the products can be explained in terms
of formation of bonds that are stronger than those that are
broken in the reactants. As you now know, bond formation
releases energy. Formation of a strong bond is accompanied
by the release of large amounts of heat energy. Combustion of
a fuel such as methane is exothermic because a product (CO 2)
of the reaction has bonds that are stronger than those in the
reactant molecules. On the other hand, if a reaction is
endothermic, energy is absorbed and so the products are
higher in energy. The products are therefore less stable than
the reactants. Such a reaction is not energetically favourable.
It is possible that it might not take place, but a large number
of endothermic processes actually take place.

If the magnitude of the enthalpy change is relatively small, enough energy might be found in the
immediate surroundings to drive the reaction.

To explain why some endothermic reactions take place, we use the concept of entropy, that is, the
tendency of a system to achieve disorder. If a process is accompanied by an increase in entropy
(disorder), then it may take place, because by attaining disorder, stability is also achieved. This
explains why some endothermic processes take place, for example, the dispersing of ions of
NH4NO3 in water leads to an increase in entropy. This, with the fact that the enthalpy change of the
process is a small positive number, explains why NH4NO3 dissolves in water even though the
process is endothermic.
From this discussion, it can be concluded that exothermic reactions are feasible, because from an
energetic point of view, they are favourable: the products formed are relatively stable. But this does not
necessarily mean that the reaction will take place. A good example is the reaction between oxygen and a
fuel such as petrol. The reaction is feasible, as it is accompanied by release of energy, implying that the
products would be lower in energy than the reactants. However, in practice, mixing oxygen and petrol does
not result in a reaction, until a spark or strong heating is applied. Such a reaction is said to be feasible but
non-spontaneous. It will not occur under ordinary conditions of temperature and pressure. Some exothermic reactions are spontaneous. Such reactions occur under ordinary conditions of temperature and
pressure; as soon as the reactants are mixed, they react. An example is the reaction between AgNO 3 and
BaCl2 solutions. As soon as the two substances are mixed, a white precipitate of AgCl is immediately formed.
Whether an exothermic reaction will occur spontaneously or not is beyond the scope of energetics. This
issue is best discussed under a separate topic, reaction kinetics.
Consider the conversion of graphite to diamond.
C (graphite) C (diamond), ΔH = +5Kj/mol
Why does this conversion not take place? The answer is not ‘because the reaction is endothermic’. Since
the enthalpy change of reaction is a small positive number, from an energetic point of view, the conversion
might be feasible. The reason why graphite can not be converted to diamond is the high activation energy
associated with the conversion. Activation energy is a purely kinetic term, and it will be discussed under
reaction kinetics.
5.4 Reversible processes
Consider the reversible conversion of hydrogen and nitrogen into ammonia during the Haber process.
N2(g) + 3H2(g) Ý 2NH3(g)
∆H = -92.4Kj
The enthalpy value of -92.4Kj refers the forward reaction. This reaction is exothermic, that is, the
formation of ammonia in the Haber process is accompanied by release of heat energy. The enthalpy of the
reverse reaction has the same magnitude but opposite sign. The decomposition of ammonia to form
hydrogen and nitrogen is therefore an endothermic process, with an enthalpy change of +92.4Kj.
5.5
Zero enthalpy reactions
In some reactions, the enthalpy change is zero. In such reactions, the energy used to break bonds in the
reactants is exactly equal to the energy released when new bonds are formed. The reactants and the
products are therefore at the same energy level (Fig 5.6).
Fig 5.6 Energy level diagram of a zero enthalpy
reaction
Page
132
Standard conditions for thermochemical measurements
In order to make meaningful comparison of thermochemical results, it is essential that measurements be
taken under standard conditions, and that the conditions of the system are the same before and after the
reaction.
The standard conditions of temperature and pressure for thermochemical
measurements are 298K ( 250) and 1 atmosphere.
An enthalpy change measured under these conditions is said to be standard and it is given the symbol ∆Hθ,
where the superscript ‘θ’ means ‘under standard conditions’.
5.7 Enthalpy terms
There are different enthalpy terms, each relating to a particular type of chemical or physical change. It is of
extreme importance that these terms be defined concisely and exactly to avoid ambiguity. More often than
not, errors in calculations come from a wrong or roughly accurate definition of one or more enthalpy terms.
By the time you are through with this section you should be able to define the following terms as completely
and as precisely as possible.
1. Enthalpy change of reaction
6. Electron affinity
2. Enthalpy change of formation
7. Ionization enthalpy (ionization energy)
3. Enthalpy change of combustion
8. Enthalpy change of hydration
4. Bond enthalpy (bond energy)
9. Enthalpy change of solution
5. Enthalpy change of atomization
10. Enthalpy change of neutralization
5.7. 1 Enthalpy change of reaction
Consider the balanced equation:
CuO(s) + Al(s)  Cu(s) + Al2O3(s)
... ∆Hθr
This is a balanced equation for the formation of the given products from the given reactants at 1 atm and
298K. The energy change for this reaction under these conditions is known as the standard enthalpy
change of reaction, ∆Hθr.
Page
133
The standard enthalpy change of a reaction (∆Hr) is the heat absorbed or
evolved(given out) when reactants combine to form products according to the balanced
equation, all measurements being taken under standard conditions of 1 atm and 298K.
For example
3O2(g) + 4NH3(g) 6H2O(l) + 2N2(g)
∆Hθr = -1 258Kj
This definition emphasizes that the reaction should be balanced. From the definition, we also learn that the
enthalpy change of reaction can be positive (energy absorbed) or negative (energy evolved), that is, the
reaction could be endothermic or exothermic.
5.7.2 Enthalpy change of formation
The standard enthalpy change of formation (∆Hθf ) of a substance is the heat
evolved or absorbed when 1 mole of the substance is formed from its elements in their
normal states under standard conditions of temperature (298K) and pressure (1 atm).
For example, the standard enthalpy change of formation of MgO refers to the energy
change for the reaction:
1
Mg(s) + O2 (g)  MgO(s)
2
∆H = -602Kjmol-1
Important points to note in this definition:
1. ∆Hθf refers to the formation of 1 mole of a substance. If the balanced equation does not produce 1 mole,
then it should be rewritten in such a way that one mole of product is formed. Consider the balanced
equation for the formation of MgO from its elements:
Mg(s) + O2 (g)  2MgO(s)
Written in this way, the equation does not represent the enthalpy change of formation of MgO. It
represents the enthalpy change of reaction for the written equation. We divide the equation by 2 so that 1
mol of product is formed.The units of ∆Hθf are always given per mole of substance formed.
2. ∆Hθf can be positive or negative. It all depends on the reaction being studied. Defining ∆H θf as ‘...the
energy absorbed...’ or ‘...the energy released ...’ is wrong as it gives the impression that ∆H fθ is always
endothermic (energy absorbed) or is always exothermic (energy released). The correct way to define ∆Hθf
should be ‘It is the energy change ...’ or ‘ It is the energy released or absorbed...’
3. State symbols are an important part of the definition, because all substances must be in their normal
states under standard conditions. Whenever an equation is written to illustrate ∆H θf , the correct state
symbols should be shown.
4. ∆Hθf refers to the formation of a substance from its elements. The reactant side of the equation should
therefore always show elements. The presence of compounds or radicals such as the Cl atom on the left
hand side of the equation renders the equation wrong if it is meant to illustrate ∆Hθf .
Page
134
5. It is clear from the given definition that the standard enthalpy formation of an element is zero, that is,
there is no heat change when an element is formed from itself.
Significance of enthalpies of formation

A negative enthalpy change of formation (exothermic) indicates that a substance is stable relative
to its elements. For instance, the enthalpy change of formation for MgO is -602Kjmol-1, indicating
that MgO is stable relative to elemental oxygen and magnesium metal. It will not easily decompose
to form oxygen and magnesium. Rather, Mg and O2 can combine to form MgO, since this results in
the formation of a more stable product.
Consider the interesting case of hydrogen peroxide, H2O2. Its enthalpy change of formation is
-133 Kjmol-1. Is the statement, ‘Hydrogen peroxide stable’ correct? The answer is no. It is well
known that hydrogen peroxide is unstable with respect to decomposition to form water and oxygen.
For this reason, it is always stored in dark bottles (to minimize exposure to light) and in a cool place
(to slow the decomposition process). The negative value of -133Kjmol-1 simply shows us that
hydrogen peroxide is stable relative to its elements, which are oxygen and hydrogen. It will not
easily decompose to form these gases, however, it can be unstable in other ways, for example, it
decomposes fairly easily to form water and oxygen.

5.7.3
Some substances have a positive enthalpy change of formation. Such substances are unstable
relative to their elements. An example is chlorine dioxide, whose enthalpy of formation is
+103Kjmol-1. Chlorine dioxide easily decomposes to chlorine and oxygen when warmed, because
the elements formed are relatively more stable.
Hess’ Law
There are certain cases where the enthalpy change of a process can not be found directly. Indirect methods
can then be used, which depend on Hess’ Law of constant heat summation.
Hess’ Law states that:
The energy change of a reaction is the same irrespective of the reaction route taken, provided
that all measurements are made under the same conditions.
Hess’ law can be used in conjuction with enthalpies of formation or combustion to determine enthalpy
changes that can not be found using direct methods.
Consider a reaction in which substances A and B react to form product C.
Page
135
A+B  C
This is the direct route. Now suppose that the reaction can be carried out indirectly by first producing an
intermediate X, then converting the intermediate to the desired product C.
The alternative route has two steps, each with its own enthalpy change, ∆H2 and ∆H3. Hess Law says that
the total enthalpy for the alternative route (∆H3 + ∆H2) is equal to the enthalpy of the direct route, ∆H1:
Enthalpy change of route 1 = enthalpy change of route 2
∆H1
= ∆H2 + ∆H3
The alternative route can be used to find the enthalpy of reaction, ∆H 1 if it is
impossible or difficult to find the enthalpy directly.
Hess’ Law is a direct consequence of the conservation of energy; energy can
neither be created nor be destroyed by changing the route of a reaction.
The example below shows how Hess’ Law can be used to find the enthalpy of
a reaction indirectly using the enthalpies of formation of the substances
involved in the reaction.
Fig 5.7 Energy cycle
for a hypothetical
reaction
Example
Methylhydrazine and dinitrogen tetraoxide burn spontaneously upon mixing. The reaction produces
large quantities of heat energy. This mixture can therefore be used as rocket fuel. How much energy is
released by the reaction?
5 N2O4 (l) + 4CH3NHNH2 (l)  4CO2 (g) + 12H2O (l) + 9N2 (g)
Data
∆Hθf (CH3NHNH2 (l))
∆Hθf (N2O4)
= +53kjmol-1
=
-20kjmol-1
∆Hθf (CO2 (g))
=
-393 kjmol-1
∆Hθf (H2O
= - 286kjmol-1
(l))
∆Hr =? ... I
Page
136
If we know the enthalpies of formation of each substance in the equation, then we can construct an
alternative route involving these enthalpies. If we add them up, the sum will be equal to the enthalpy of
reaction I. Fig 5.4 shows the energy cycle that can be used to calculate the enthalpy of reaction I using Hess’
Law. The energy cycle shows the direct reaction route and the indirect route (involving enthalpies of formation).
Fig 5.8
Energy cycle for calculating enthalpy of reaction using Hess’
Law.
To find the required enthalpy change of reaction, ∆H r, we add enthalpies of formation ∆H 1 to ∆H5. In this
case, the summation is anticlockwise as shown by the curved arrow. Remember that whatever the direction
of the arrow, it should always start at the reactants and end at the products. Drawing an arrow from the
starting point (reactants) to the ending point (products) may be useful in determining which is route 1 and
which route 2 is. It is also helpful in identifying which enthalpies ought to have their signs reversed. Moving
along the route arrow (in this case, the arrow labeled route 2), if we find that we are going against the
direction of a reaction arrow, we reverse the sign of ∆H. For example, at the position labeled A on route 2,
we are going against the reaction arrows labeled ∆H1 and ∆H2, so the sign of these enthalpy terms must be
reversed. The direction of route 2 is the same as the direction for the reactions labeled ∆H3, ∆H4 and ∆H5.
The signs of these enthalpy terms remain positive, that is, they should not be
reversed.
It is advisable to write a general equation in terms of ∆H before substituting any figures:
Enthalpy of Route 1 (direct route) = Enthalpy of route 2 (indirect route)
that is, ∆Hr = -5∆H1 + (- 4∆H2) + 4∆H3 + 12∆H4 + 0 …II




∆H1 and ∆H2: signs have been reversed to - as already explained.
∆H5 has been replaced by 0. The enthalpy change of formation of an element is zero.
Each enthalpy of formation is multiplied by the appropriate coefficient to agree with the definition
of enthalpy of formation, for example, the enthalpy change of formation of N2O4 refers to formation
of one mole of the compound. In the cycle, five moles are formed, so we multiply the given value
(∆H1) by 5.
NB: If a substance contains carbon (for example, CO 2 in Fig 5.4), the enthalpy of its formation
refers to formation of the compound from graphite, which is the more stable allotrope of carbon.
Substituting the given data into equation II:
∆Hr = [-5(-20) -4(53) + 4(-393) + 12(-286)] Kjmol-1 = -5 116 Kjmol-1
The reaction therefore produces a very large amount of energy. This explains its use in rocket fuel.
Example 2
The standard enthalpy changes of two reactions are given by the following equations
2Fe(s) + 3/2 O2(g)  Fe2O3(s)
C(s) + ½ O2(g)  CO(g)
∆Hθ = -822Kjmol-1
∆Hθ = -110Kjmol-1
What is the standard enthalpy change for the reaction?
Fe2O3(s) + 3C(s)  2Fe(s) + 3CO(g)
Solution
∆Hr = -∆H1 + ∆H2 (Fig 5.8.1)
Page
137
= -(-822) + 3(-110)
= +492 Kjmol-1
Fig 5.8.1
Fe2O 3 (s) + 3C(s)
Δ Hr
2Fe(s) + 3CO(g)
Δ H1 = -822
Δ H2 = 3(-110)
[2Fe(s) + 1/2O 2(g)] + [ 3C(s) + 3/2O 2(g)]
Measuring standard enthalpy of formation
Some enthalpies of formation can be measured directly using a bomb calorimeter. The enthalpies of
formation of carbon dioxide, magnesium oxide and many other oxides have been measured in this way
because they can conveniently be formed from the combustion of their respective elements. However, there
are many substances for which direct determination of enthalpy change of formation is impossible because
the compound can not be formed directly from its elements. An example is hydrogen peroxide. It can not
be synthesized directly from the combination of hydrogen and oxygen.
One indirect method involves using enthalpies of combustion.
5.7.4
Standard enthalpy change of combustion, ∆Hθc
The standard enthalpy change of combustion (∆Hθc) of a substance is the amount
of heat evolved when 1 mole of the substance burns completely in oxygen, the measurement of heat released being adjusted to standard conditions of temperature and
pressure.
For example, the enthalpy change of combustion of hydrogen refers to the equation:
H2 (g) + 1/2O2 (g)  H2O (l)
Page
138
Important points to note in this definition:
1. ∆Hθc refers to the combustion of one mole of a substance (of course it applies only to combustible
substances). If more than one mole of a substance is shown in the balanced equation, the equation should
be adjusted appropriately to show 1 mole of a substance being burnt, for example, the enthalpy change of
combustion of magnesium refers to the equation
Mg(s) +1/2O2 (g)  MgO(s) not 2Mg(s) + O2(g)  2 MgO(s)
Note that in this case, the equation for the enthalpy change of combustion is the same as that for the
enthalpy change of formation.
2. It is important that oxygen be present in excess so that there is complete combustion. This is particularly
important when organic substances burn. In a limited amount of oxygen, incomplete combustion tales
place. A mixture of products is formed and the amount of heat evolved is less than what would be evolved
if there was complete combustion.
3. The definition specifies that ∆Hc is the energy released; it does not generalize it as ‘... energy change...’.
This generalization may be used for a process that can be exothermic or endothermic. Combustion is always
exothermic .
Determination of enthalpy of combustion
A bomb calorimeter can be used for this purpose (Fig 5.9)
Fig 5.9
Automated bomb calorimeter
The apparatus is designed to minimize loss of heat to the surroundings. A sample of known mass is burnt
completely in the bomb calorimeter and the increase in temperature of a known volume of water
surrounding the bomb is noted. The equation below can then be used to calculate the enthalpy change of
combustion:
Heat produced = mc∆θ
m is the mass of water ( 1g ≈ 1 cm³)
c is the specific heat capacity of water = 4.2 JK-1g-1
Page
139
∆θ is the increase in temperature of water, in Kelvins or in degrees celcius.
Sample Data:
Volume of water in calorimeter = 400cm³
Initial temperature of water
= 100C
Final temperature of water
= 200C
Mass of ethanol burnt
= 0.90g
We wish to calculate the enthalpy change of combustion of ethanol.
The heat evolved when ethanol burns is given by:
∆H = mc∆θ
= 400 x 4.2 x 10
= 16 800J
Note:
 We use mass of water in this part of the calculation, not mass of the substance being burnt.
 The units are in joules since the value of c is given in these units.
Moles of ethanol used = m/Mr = 0.90/46
= 0.01957
∴ 0.01957 moles of ethanol produce 16 800J of energy when burnt.
By definition, the enthalpy change of combustion is given in terms of one mole of a substance burnt.
That is, 1 mole of ethanol produces
16 800/0.01957 = 858 456.8217 J
Therefore the enthalpy change of combustion of ethanol is 858.5Kjmol-1
This value can save as a rough guide to the heat energy evolved per mole when ethanol is completely burnt.
Limitations of the experiment include:
1. Heat loss to the environment.
2. Absorption of heat by the calorimeter and the thermometer.
3. Incomplete combustion is inevitable unless a large excess of pure oxygen is used
The value is therefore less than the actual value, which has been accurately determined to be
-1 371 Kjmol-1. Refined calorimeters are available, which are very accurate (and very expensive). These
apparatus are also calibrated to compensate for heat losses.
Enthalpy of combustion and molecular structure
Page
140
The enthalpies of combustion of the first four alkanes are given in Fig 5.10.
Methane
Ethane
Propane
Butane
∆Hθc /Kjmol-1
890
1 560
2 220
2 877
There is a clear trend in these values which is associated with molecular structure:
Fig 5.10
Enthalpy changes of combustion of the first four alkanes
There is an increase of about 670 KJmol-1 from one hydrocarbon to the next. This shows that each additional
CH2 group is contributing about 670 KJ mol-1. We conclude that the energies associated with a certain type
of bond is constant. The more bonds we have in the hydrocarbon, the more energy is released. This is not
surprising. Hydrocarbons store chemical energy in their bonds. The more bonds there are, the greater the
energy released per mole of the hydrocarbon. We therefore expect the higher Mr hydrocarbons to be better
fuels. However, the selection of fuels is not as easy as that. Consider the equations for the combustion of the
hydrocarbons:
CxHy + (x + y/4) O2  xCO2 +(y/2) O2
1. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
Page
141
2. C2H6 (g) + 7/2O2 (g)  2CO2 (g) + 3H2O (l)
3. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
4. C4H10 (g) + 13/2O2 (g)  4CO2 (g) + 5H2O (l)
It can be seen that even though more energy is released for longer chain alkanes, the demand for oxygen
also increases. For instance, 1 mole of methane will need only 1 mole of oxygen for complete combustion,
but 1 mole of butane will need 6.5 moles. The increase in demand for oxygen implies that there is more
incomplete combustion, that is, less heat energy would be produced than expected. Not only that; there will
also be wastage of the fuel since some of it will escape unburnt in the form of soot(carbon), carbon monoxide
and unburnt hydrocarbon molecules. Pollution therefore increases with increasing Mr of chain alkanes. Fig
5.5 shows the expected trend in heat of combustion when the alkanes are burnt, and the trend that is actually
obtained if combustion takes place in a limited supply of air.
Fig 5.11
The effect of number of carbon atoms on the heat
yield when the alkanes are burnt in a fixed volume of air.
A compromise must therefore be reached, and the fuel chosen must be able to burn almost completely in
air. Short chain alkanes such as methane burn efficiently with little pollution. However these fuels tend to
be gaseous or volatile liquids and pose a storage problem and fire risk. Longer chain fuels are liquid. They
are more convenient to handle, but as already noted, they demand a larger amount of oxygen, which might
not be present since air contains only 21% oxygen. The internal combustion engine is perhaps one of the
most inefficient devices invented by man. The petrol engine burns C8 hydrocarbons, but to a large extend
the combustion is incomplete because of the limited supply of air in the engine. It has been estimated that
only about 20% of the energy in the fuel is actually converted to the energy that drives the vehicle.
Significance of enthalpies of combustion
Page
142
The burning of fuels produce heat energy. This is an exothermic process in which the products are lower in
energy, and therefore more stable than the reactants.
Bond formation produces heat energy (it is exothermic), and bond breaking uses up heat energy (it is
endothermic). During combustion, more heat is produced by forming new bonds in products than is used
in breaking bonds in reactants. This implies that stronger bonds are formed in the products (CO 2, produced
when a hydrocarbon burns, has two double bonds, which are very strong). Stronger bonds means the
products are more stable than the reactants.
Using enthalpies of combustion to find enthalpy
of formation of a compound
The use of Hess’ Law in conjuction with enthalpy changes of formation to find the enthalpy change of a
reaction indirectly has already been discussed. Enthalpies of combustion can also be used in this way to
find the enthalpy of formation of a substance. This is particularly useful if the substance can not be formed
directly from its elements, for example, ethanol can not be formed directly by combining carbon, hydrogen
and oxygen.
Example 1
Calculate the enthalpy change of formation of ethanol given the following combustion data:
∆Hθc/ Kjmol-1
C (graphite)
-393.5
H2 (g)
- 285.8
C2H5OH (l)
- 1371.0
Solution
Let the direct route be the formation of ethanol from its elements. From this route we create another route
involving the combustion of each substance in the equation, that is if it can be burnt.
∆H1, ∆H2 and ∆H3 are the standard enthalpies of combustion of the substances involved in the formation
of ethanol.
Page
143
By Hess’ Law
Enthalpy of route 1 = enthalpy of route 2
∆Hθf = 2∆H1 + 3∆H2 -∆H3 … (i)
= [2(-393.5) + 3(-285.8) - (-1371)] Kjmol-1
= -273.4 KJmol-1
Notice the coefficients in equation 1. Enthalpy of combustion is defined per mole of substance burnt,
therefore if x moles are burnt, the total energy of combustion becomes x∆H. Also note the use of the +,signs; this is frequently a source of confusion among students. To avoid a mix up in signs, it is strongly
advised that you start by a writing a general algebraic expression such as equation (i) above, before you
substitute any figures.
Example 2
The standard enthalpy change of formation of CO2 and and H2O are -394 Kjmol-1 and -286 Kjmol-1
respectively. If the standard enthalpy change of combustion of propyne, C 3H4, is -1 938 Kjmol-1, what
is its standard enthalpy change of formation?
9189/1/O/N/2010
Solution
C3H4(g) + 4O 2(g)
ΔH1
3CO 2(g) + 2H2O(l)
x
ΔH2
ΔH3
[C(s) + 2H2(g)] + [3C(s) + 3O 2(g)] + [2H2(g) + O2(g)]
Let the standard enthalpy change of formation of propyne be x. By Hess’ Law
∆H1 = -x + ∆H2 + ∆H3
-1938 = -x + 3(-394) + 2(-286)
x = 184
The standard enthalpy change of formation of propyne is +184 Kjmol-1.
Page
144
5.7.5
Lattice Energy, ∆HL.E
We have discussed the relative stability of compounds with reference to enthalpies of formation.
The relative stability of ionic compounds is best explained in terms of their lattice energies.
Lattice Energy is the amount of heat energy evolved when 1.0 mole of a solid ionic
crystal is formed from its respective gaseous ions under standard conditions of 1.0
atm and 298K.
For example, the lattice energy of MgO is represented by the equation
Mg2+ (g) + O 2-(g)  MgO(s)
Important:
1. Lattice energies are always negative (exothermic). Energy is released whenever an ionic compound is
formed from its ions.
2. Lattice energy refers only to ionic compounds.
3. Lattice energy is defined per mole of crystal formed.
4. State symbols are important in the definition of lattice energy. Omitting then renders the
equation for lattice energy wrong.
Determination of lattice Energy
Direct determination of lattice energy is not possible. An indirect method which combines several enthalpy
terms is used. Suppose that we wish to form MgO. One method would be to form it from its elements in
their normal states at 298K and 1.0 atm:
Mg(s) + 1/2O2 (g)  MgO(s)
∆Hθ = standard enthalpy change
of formation of MgO.
An indirect method involves the following processes which ultimately converts the elements to their
respective gaseous ions. The energy released when these ions combine to form one mole of an ionic crystal
is referred to as lattice energy.
I. The elements are converted to gaseous atoms. The energy needed (absorbed) for this conversion is the
enthalpy change of atomization.
5.7.6
Standard Enthalpy Change of Atomization
The Standard Enthalpy Change of Atomization of an element refers to the
energy absorbed in forming 1.0 mole of gaseous atoms from that element in its
natural state under standard conditions of 1.0 atm and 298K.
Page
145
For example, the standard enthalpy of atomization of chlorine refers to the process
1/2Cl2(g)  Cl (g)
Note that the given equation shows the formation of 1 mole of chlorine atoms, not one atom. Enthalpies of
atomization are always endothermic because energy is required to pull atoms from each other and disperse
them into a gaseous state (recall that bond breaking is endothermic. During atomization, there is bond
breaking only and no bond formation). The term atomization also applies to metals, for example, the
standard enthalpy change of atomization of sodium metal refers to the energy released when the metal is
converted to 1 mole of gaseous sodium atoms, under standard conditions.
Na(s)  Na (g)
The energy change is also referred to as the enthalpy of sublimation of the metal because there is a direct
change of state from solid to gas. The process absorbs energy, which is used to overcome the electrostatic
attractions between positive metal atoms and a sea of delocalized electrons in the metal lattice.
II. The second step in the indirect formation of an ionic compound involves changing the gaseous metal
atoms to unipositive gaseous atoms. The energy needed for this conversion is the first ionization energy
of that metal.
5.7.7
Ionization Energy, ∆HIE
The First Ionization Energy of an element refers to the energy required(absorbed)
to knock out 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of
gaseous +1 ions, under standard conditions of 1.0 atm and 298K.
For example, the first ionization energy of magnesium refers to the process
Mg (g)  Mg+ (g) + e
Note that the term ionization energy applies to both metals and non-metals. However, the formation of an
ionic compound involves ionization of metals, so the following discussion will put emphasis on conversion
of gaseous metal atoms to ions.


Page
146

Ionization energies are always positive. Electrons are naturally held to the atom by attraction to the
nucleus. Therefore energy is required to remove them.
Ionization Energies increase across a Period in response to increasing nuclear charge (number of
protons). A nucleus with a larger number of protons has a stronger attraction for the outer-shell
electrons. Consequently, a larger amount of energy would be required to remove an electron from
this shell.
Ionization Energies decrease down a group, even though nuclear charge increases. This is because
down a group, additional shells are opened. This increases the size of the atoms and the distance of
the outer-shell electrons from the nucleus. Consequently, the valence shell electrons feel a weaker
attraction from the nucleus. Increase in atomic radii down a group has a greater effect than the
increase in proton number.
III. If the ionic compound to be formed consists of +2 ions, for example, magnesium oxide, then the +1
gaseous metal ions must be changed to gaseous +2 ions. The energy needed for this conversion is the second
ionization energy of that metal.
The Second Ionization Energy of an element refers to the energy required to knock
out 1.0 mole of electrons from 1.0 mole of gaseous +1 ions to form 1.0 mole of gaseous
+2 ions, under standard conditions of 1.0 atm and 298K.
For example, the second ionization energy of magnesium refers to the process
Mg+ (g)  Mg2+ (g) + e
The second ionization energy of an element is always higher than the first because the electron being
removed is now attracted to the atom by both the nucleus and the positive charge on the +1 ion. More energy
is therefore required to pull out the second electron.
IV. If the ionic compound to be formed consists of +3 ions, for example, aluminium oxide, then it is
necessary to convert the +2 metal ions to +3 ions. The energy absorbed in this conversion is known as the
third ionization energy for that metal.
The Third Ionization Energy of an element refers to the energy required to knock
out 1.0 mole of electrons from 1.0 mole of gaseous +2 ions to form 1.0 mole of gaseous
+3 ions, under standard conditions.
For example, the third ionization energy of aluminium refers to the process
Al2+(g)  Al3+(g) + e
The third ionization energy of an element is higher than the first and second ionization energies. This time,
the electron to be removed is under strong attraction by the +2 charge on the ion from which it is being
removed.
V. The next step in the formation of an ionic compound would be changing the gaseous non-metal
atoms(formed by the process of atomization) to gaseous -1 ions. The energy released by this conversion is
the first electron affinity of that element.
5.7.8
Electron Affinity
Page
147
The First Electron Affinity of an element refers to the energy released when
1.0 mole of gaseous atoms accept 1.0 mole of electrons to form 1.0 mole of -1
gaseous ions, under standard conditions.
For example, the first electron affinity of oxygen is represented by the equation
O(g) + e-  O -(g)
The first electron affinity affinity is exothermic. This shows that the process is favourable in terms of
energy. The incoming electron is stabilized by being attracted to the protons in the nucleus.
VI. If the ionic compound contains -2 ions, for example, O2- in MgO, then the -1 ions have to be converted
to -2 ions. This process absorbs energy and is known as the second electron affinity for that element.
The second electron affinity of an element refers to the energy absorbed to force
1.0 mole of electrons into the outer shell of 1.0 mole of gaseous -1 ions to form 1.0
mole of gaseous - 2 ions.
For example, the second electron affinity of oxygen refers to the process
O- (g) + e-  O2-(g)
The second electron affinity is endothermic because the incoming electron is being forced to enter the outer
shell of a negatively charged particle. There is repulsion between the atom and the incoming electron, and
so energy must be supplied to force the process.
VII. If the ionic compound contains an anion of -3 charge, for example, the nitride ion in magnesium
nitride, Mg3N2, then the -2 ions must be converted to the -3 state. This endothermic process is known as
the third electron affinity for that element.
The third electron affinity of an element refers to the energy required to force
1.0 mole of electrons into the outer shell of 1.0 mole of gaseous -2 ions to form 1.0
mole of gaseous - 3 ions.
For example, the third electron affinity of nitrogen refers to the process
N-2 (g) + e-  N3- (g)
The third electron affinity of an element is more positive (more endothermic) than the second electron
affinity. This is because the third electron has to enter the shell of an atom (ion) which already bears a large
negative charge (-2). The incoming electron experiences great repulsion from the charge on the ion so a
large amount of energy is required to force process to take place.
Page
148
VIII. The last step in the formation of the ionic compound is combining the gaseous positive ions from the
metal and the negative gaseous ions from the non-metal to form an ionic crystal. The energy released by
this process is the Lattice Energy of that ionic compound, which has already been defined.
Lattice energy, atomization, ionization and electron affinity are combined together with the enthalpy of
formation of the ionic compound, in an energy cycle known as the Born-Haber cycle. Using Hess’ Law, any
enthalpy term in the Born-Haber cycle can then be determined. This method can be used to determine
lattice enthalpies since they can not be determined directly.
The Born-Haber cycle for the formation of NaCl
We wish to find the lattice energy of NaCl. Let this be the direct route in Fig 5.12. This diagram has been
drawn so that the direction of the arrows is informative. An arrow pointing upwards denotes an endothermic process, and an arrow pointing downwards an exothermic process.
Fig 5.12
Born-Haber cycle for the formation of NaCl
∆H1 = Standard enthalpy change of formation of NaCl (exothermic) = -411Kj/mol
∆H2 = Standard enthalpy change of atomization of Na metal (endothermic) = +109Kj/mol
∆H3 = First ionization enthalpy of Na (endothermic) = +494Kj/mol
∆H4 = Standard enthalpy change of atomization of chlorine (endothermic) = +121Kj/mol
∆H5 = First electron affinity of chlorine (exothermic) = -367Kj/mol
∆HL.E = lattice enthalpy of NaCl (exothermic) = ?
By Hess’ law: Energy of direct route (∆HL.E) = Energy of indirect route
Page
149
∆HL.E = -∆H5 - ∆H4 - ∆H3 - ∆H2 + ∆H1
Substituting:
∆HL.E = -(-367)-(+121)-(+494)-(+109)+(-411)
= -768 Kj/mol
Note that lattice energies are always negative.
Next, we have to construct Born-Haber cycles for ionic compounds in which charges on the cation or anion
(or both) are greater than 1. However, before doing that, let us study another enthalpy term which always
appears in a Born-Haber cycle (even the one for NaCl discussed above) but is often overlooked. This
enthalpy term is known as bond enthalpy (that is, bond energy).
5.7.9 Bond Enthalpy
Bond enthalpy is the average energy needed to break one mole of a specific
covalent bond in the gaseous phase, to form separate gaseous atoms.
For example, the bond energy of chlorine refers to the energy required to break one
mole of the Cl-Cl bond.
Cl-Cl (g)  2Cl- (g)
The Born-Haber cycle drawn for NaCl (Fig 5.12) involved the conversion
½Cl-Cl(g)  Cl(g) ... (i)
This has been defined as the atomization of chlorine. Now, notice that this process involves the breaking of
a covalent bond, so it can also be expressed in terms of bond energy.
Enthalpy change of atomization of chlorine = ½(Cl-Cl) bond energy
The co-efficient ½ is a direct consequence of the definition of bond energy. Bond enthalpies are quoted per
mole of covalent bond broken. Now, in equation (i) above, only half a mole of the bond is being broken.
The atomization of a covalent bond and bond energy differ mainly in that atomization refers to the
formation (product side) of 1 mole of gaseous atoms, as in (i) above, but bond energy emphasizes the
breaking of one mole of a covalent bond(reactant side).
From the data booklet, the bond enthalpy of chlorine is +242Kj/mol. The enthalpy change of atomization
of chlorine is therefore ½(242) = +121Kj/mol. In your exams you may notice that the Data Booklets that
are provided do not contain the enthalpies of atomization of covalent bonds such as the
Cl-Cl bond. The reader should be able to determine these enthalpies of atomization from relevant bond
energies, usually given in the Data Booklet.
Page
150
Factors that affect the magnitude of bond enthalpies
Bond enthalpies are a direct measure of the strength of a covalent bond. You will notice that bond enthalpies
are always positive (endothermic), that is, energy is required to separate the two atoms of a bond. The pair
or pairs of electrons shared between the two atoms of the bond experience attraction from the nuclei of both
atoms of that bond. It is this electrostatic attraction that holds the two atoms of the bond together. The
stronger the electrostatic attraction, the stronger the bond and the greater the energy required to break it
(that is, the more endothermic the bond enthalpy). One very important factor that affects bond enthalpy,
and hence, bond strength, is atomic size.
Consider the bond enthalpies of the hydrogen halides (HX) given in Table 5.1.
The trend is clear; the H-X bond becomes weaker (less
-1
Benergy is requiredo to break it) as the
n radius
d
of the halogen
atom increases. When one or both atoms of a covalent bond
H-Cl in size, overlap 431
increase
of atomic orbitals during bond
formation becomes poor. As a result, the bond pair of
H-Br
366
electrons is a large distance from the nuclei of both atoms of
the
bond. Electrostatic attraction
H-I
299 between the bond pair and
the nuclei therefore becomes weak.
Table 5.1
Q
A
Energies of the H-X bond
Explain why the atomic radii of Cl, Br and I increases in that order.
Going down the Group, a new shell is opened form one element to the next. Increase in number of
shells results in an increase in atomic size.
Another important factor that affects bond energy is nature of the bond. Multiple bonds (double and triple)
are shorter and stronger than single bonds.
5.7.10 More Born-Haber cycles
The following Born-Haber cycles involve formation of ions whose magnitude is greater than 1. Once more,
the direction of arrows has been meant to inform the reader whether the process involved is exothermic or
endothermic.
The Born-Haber cycle for the formation of MgO (Fig 5.13)
Data: ∆H1 = ∆Hθf (MgO) = -602Kjmol-1
∆H2 = ∆Hatomization (Mg) = +150Kjmol-1
∆H3 = ∆HI.E1 (Mg) = +736Kjmol-1
∆H6 = e.a1 (O) = -142Kjmol-1
By Hess Law
Enthalpy of route 1 = enthalpy of route 2
∆H7 = e.a2 (O) = +844Kjmol-1
L.E = -∆H7 - ∆H6 - ∆H5 - ∆H4 - ∆H3 - ∆H2 + ∆H1
151
∆H4 = ∆HI.E2 (Mg) = +1 450Kjmol-1
Page
Calculation
The aim is to determine the Lattice Energy of
MgO
More information
The O=O bond energy is 496Kj/mol.
= - (+844)-(-142)-(+248)-(+1450)-(+736)
- (+150) + (-602)
= -3 888Kjmol-1
2-
2+
O (g) + Mg (g)
2+
Mg (g) + O (g)
H4
H6
H5
H7
2+
Mg (g) + O (g)
Mg (g) + 1/2O2(g)
direct
route
H3
Mg(g) + 1/2O2(g)
lattice energy
H2
Mg(s) + 1/2O2(g)
Fig 5.13


H1
Born - Haber cycle for the formation of MgO
∆H5 is the enthalpy change of atomization of oxygen, which is equal to ½ the bond energy of the
O=O bond.
The terms ∆H2 to ∆H7 are against the arrow for route 2. The signs of these terms are reversed
accordingly. ∆H1 is in the same direction as the arrow for route 2, so the sign for ∆H 1 remains +.
Born – Haber cycle for Al2O3
Fig 5.14 is a Born-Haber cycle for the formation of Al2O3.
Data : ∆H1 = ∆Hθf(Al2O3) = -1669Kjmol-1
∆H2 = ∆Hatomization(Al) =
∆H3 = ∆HI.E1(Al) =
314Kjmol-1
577Kjmol-1
∆H4 = ∆HI.E2(Al) = 1820Kjmol-1
Page
152
∆H5 =∆HI.E3(Al) = 2740
∆H7 = e.a1 (O) =
-142Kjmol-1
∆H8 = e.a2 (O) =
+844Kjmol-1
More information
The O=O bond energy is 496Kj/mol.
Calculation
Enthalpy of route 1 = enthalpy of route 2
L.E = -3∆H8 - 3∆H7 - ∆H6 - 2∆H5 - 2∆H4 -2∆H3
-2∆H2 + ∆H1 ... (i)
= - 3(+844) - 3(-142) - (+744) - 2(+2 740)
- 2(+1 820) - 2(+577) - 2(314) + (-1 669)
= -15 421Kjmol-1
Fig 5.14
Born - Haber cycle for the formation of Al2O3
Notice the use of co-efficients in equation (i). This is a direct consequence of the way the enthalpy terms
are defined. For example, ∆H2 represents the enthalpy of atomization of Al metal, which is the energy
required to produce 1.0 mole of gaseous atoms of Al. Now, in Fig 5.14, two moles of Al atoms are produced
during the atomization process, so the value of ∆H θatomization must be multiplied by 2 to obtain the total
energy absorbed in that step.
Also notice how ∆H6, the total enthalpy change of atomization of oxygen has been obtained.
(∆Hatomization = 3/2 x Bond Energy = 3/2 x 496 = +744Kj)
Page
153
A Born-Haber cycle can be used to find the value of any other enthalpy term involved in it, if the rest of the
enthalpy terms are known.
Q
A
Explain why lattice enthalpies are always negative.
Bond formation is exothermic. During the formation of an ionic crystal from its respective
gaseous ions, there are no bonds broken (no energy is absorbed). There is only bond formation in
which strong ionic bonds are formed and this is accompanied by evolution of large amounts of
heat energy.
Significance of lattice energy
The magnitude of lattice energy is a measure of the strength of an ionic bond. The more negative the value
of lattice energy, the lower in energy the ionic compound and the more stable it is. We thus have a way of
comparing the relative stabilities of ionic compounds.
What factors affect the magnitude of lattice energy?
The lattice enthalpies of four ionic compounds are given in Table 5.2.
NB Lattice Enthalpies are often given in Data
Booklets without the negative sign. The reader
should always remember that Lattice
Enthalpies are negative.
The following factors affect the magnitude of
Lattice Energy
Table 5.2

Lattice energies of some
ionic compounds

Size of ions (compare NaCl, NaBr and
NaI in Table 5.2, in which size of the
halide ion changes)
Size of charge on ions (compare NaCl and MgCl2 in the table above, in which charge on the metal
ion changes)
Lattice energy varies inversely as the size of an ion, that is, the larger the ion, the smaller the lattice energy
and vice-versa. We thus expect, for instance, the lattice energy of NaCl to be smaller (less exothermic/less
negative) than that of LiCl because the sodium ion is larger than the lithium ion. Compare the lattice
enthalpies of NaCl, NaBr and NaI in Table 5.2 above. The values become smaller in that order due to an
increase in the radius of the halide ion.
Page
154
When comparing lattice energies, always remember that the more negative value represents the larger
lattice energy, that is, the more negative the value, the more exothermic is the formation of the compound
from its ions. To prevent confusion we may refer to the magnitude of the lattice energy, disregarding the
sign as in table 5.2 above.
Why do ionic radii affect the magnitude of lattice energy?
1. Distance of nucleus from outer shell
Compare The Lattice Enthalpies of LiCl and NaCl, which are -846 and -771Kjmol-1 respectively. LiCl has the
larger Lattice Enthalpy because the Li+ ion is smaller than the Na+ ion in NaCl.
Being larger, the nucleus of the Na+ ion is a greater distance from the outermost shell of the ion. This nucleus
therefore has a weaker attraction of the anion, Cl-. The Li+ ion is smaller and its nucleus is closer to the
electrons in the outer-shell of the Cl- ion. The attraction between cation and anion is therefore stronger in
LiCl. When LiCl is formed, very strong ionic attractions are formed, accompanied by the release of large
amounts of heat energy. The ionic attractions in NaCl are weaker, and less energy is released when this
compound is formed from its ions.
2. The effect of charge density
We may think of charge density as the ‘concentration’ of a charge, that is, how much charge is distributed
in a given volume. The Li+ and the Na+ ion have the same size of charge. However, being larger, the sodium
ion has the smaller charge density, that is, its charge is dispersed over a larger volume. Such an ion is not
very effective at attracting an anion. The ionic bond in NaCl is therefore weaker than in LiCl. As already
mentioned, a relatively weak ionic bond is accompanied by the release of a relatively small amount of energy
(smaller value of lattice energy).
How size of charge affects Lattice Energy
Table 5.2 shows that MgCl2 has a larger Lattice Energy than NaCl. The larger charge on the Mg2+ ion results
in a stronger electrostatic attraction with the Cl- ion. That is, the ionic bonds in MgCl2 are stronger and more
heat energy is produced when the bonds are formed.
The concept of charge density is also relevant in explaining the differences in lattice enthalpies of NaCl and
MgCl2. The Mg2+ ion has a larger charge, and yet its ionic radius is smaller. It therefore has the larger charge
density, that is, it has a large charge spread in a small volume. Such an ion has a relatively strong attraction
for an anion.
Quantifying Lattice Energy
Lattice Energy α
Product of charges on cation and anion
sum of radii of cation and anion
5.7.11 More on Bond Energies
That is
In words

Lattice Enthalpy is directly proportional to size of charge. The
larger the charge on one or both of
the ions, the greater the lattice
energy.
Page
155
 Lattice Enthalpy is inversely proq- strength
Whereof qionic
+ = charge
We have seen thatq+the
bonds on
in cation
ionic compounds can be measured in terms of Lattice
LE 
portional to ionic radius. The larq- = charge on anion
r ++ r Energy of the compound.
ger the radius of one or both ions,
+ = radius of
In covalent substances, we use bondrenthalpies
tocation
compare the strength of covalent
bonds
and the
relative
the smaller
the lattice
energy.
r
=
radius
of
anion
stabilities of covalent substances. Bond energy is defined as the amount of energy required to break one
5.7.11
More on bond energies
We have already defined bond energy as the amount of energy required to break one mole of a gaseous
bond to form separate gaseous atoms. For example, the bond energy of the C-H bond refers to the process
C-H(g)  C(s) + H(g)
Bond energies are always positive, that is, energy is required to break bonds. The larger the bond energy
(the more endothermic it is), the stronger the bond. A substance with strong covalent bonds is stable and
so resists undergoing chemical changes.
Average bond energies
The bond energies given in Data Booklets are averages. This means that a particular bond energy is
obtained by averaging the bond enthalpies for that bond in different compounds. Consider the O-H bond.
Its bond energy differs depending on where it is found, for example, the O-H bonds in water and ethanol
have slightly different bond energies. The O-H bond energy given in Data Booklets is therefore an average.
Determination of the average C-H bond energy in methane
Methane is atomized into its constituent gaseous atoms. This involves breaking four C-H bonds:
H
H
C
H
4C - H(g) ...(i)
H
We may find the energy change of process (i) indirectly by using the standard enthalpy change of
formation of methane and the standard enthalpy change of atomization of carbon and hydrogen. These
enthalpies are linked by an energy cycle in Fig 5.15.
∆Hr is the energy required to break all four C-H
bonds in methane
By Hess’ Law,
∆Hr = - ∆Hθ1 + ∆Hθ2 + 4∆Hθ3
... (ii)
Where ∆Hθ1 = standard enthalpy change of formation of methane
Fig 5.15 An energy cycle for the
determination of the C-H
bond energy
Substituting and evaluating equation 1:
= -75 Kjmol-1
∆Hθ2 = standard enthalpy change of atomization of carbon
= +715Kjmol-1
∆Hθ3 = standard enthalpy change of atomization of hydrogen = +218Kjmol-1
∆Hθr = - (-75) + 715 + 4(218)
=
+ 1 662kJ
Page
156
Therefore 1 662 Kj of energy are required to break four C-H bonds in methane. The average C-H bond
energy is therefore
1 662
4
= +415.5Kjmol-1
That is, breaking 1 mole of gaseous C-H bonds in methane would require 415.5 Kj of energy.
The strength of the C-H bond in other compounds can be found in a similar way. It is found that in most
cases, the values agree, showing that a definite quantity of energy (the bond energy) is associated with a
particular type of bond. However, in some cases there is a significant difference in the value of the bond
energy because of the particular environment of the C-H bond. So what is the C-H bond energy,
disregarding the compound in which it is found? The answer has been found by averaging the C-H bond
energies in different compounds. The value recorded in data booklets is 412 KJmol -1.
The C-C bond energy
We may use ethane to compute a value for the C-C bond energy.
H
ethane
H
H
C
C
H
H
H
First we calculate the total energy required to break all the bonds in ethane (six C-H bonds and one C-C
bond).
∆Hr, the energy needed to atomize ethane by
breaking all bonds, is found by Hess’ Law:
∆Hr = -∆H2 + 2∆H3+ 6∆H4 ... (iii)
Data:
Fig 5.16
Energy cycle for determination
of the C-C bond energy in ethane
∆H2 = enthalpy change of formation of
ethane = -85KJ/mol
∆H3 = enthalpy change of atomization of
carbon = +715KJ/mol
∆H4 = enthalpy change of atomization of
hydrogen = + 218KJ/mol
Substituting these values in equation (iii)
∆Hr = - (-85) + 2(715) + 6(218) = +2 823Kj
This is the enthalpy change of atomization of ethane, that is, the energy needed to break six C-H bonds and
one C-C bond.
E(C-C) + 6E(C-H) = 2 823,
where E(C-C) means ‘energy of the C-C bond’ and E(C-H) means ‘energy of the C-H bond’.
The energy of the C-H bond is 412Kjmol-1, so
E(C-C) + 6(412) = 2 823
Page
157
E(C-C) = 2 823 – 2472
= +351Kjmol-1
We would expect all C-C bonds to have this energy, irrespective of the molecule in which the bond is found.
In most cases the values obtained are similar. In a few cases, the special environment in which the C-C bond
is found leads to a significantly different value. The C-C bond energy recorded in data booklets is an average
of the C-C bond energies in a wide range of compounds. This value is 348Kj/mol.
Uses of Bond Energies
Bond energies are useful in the following ways:



Comparing the strength of bonds
Understanding reaction mechanisms, that is, how reactions take place.
Estimating enthalpy changes of reactions
Comparing the strength of bonds using Bone Energies
Table 5.3 compares some Bond Energies.
As seen in this table, some bonds are relatively stronger than
others. A substance which contains strong bonds is relatively
inert, that is, it won’t react easily. This is the case with the
nitrogen molecule which has a bond energy of +944Kj/mol.
Compare this with the Cl-Cl bond energy, which is only +242
Kj/mol. This partly explains why chlorine is more reactive than
nitrogen. It is easier to break the Cl-Cl bond energy during
chemical reactions than it is to break the N to N bond in the
nitrogen molecule.
However, there are other factors apart from bond energies,
which affect the reactivity and stability of substances.
Table 5.3 A comparison of
some bond energies
Factors affecting the relative strength of a bond


Multiple bonds are stronger than single bonds. Compare the energies of the N-N and NɼN bonds
in Table 5.3.
Bond strength is also affected by bond length. In general, shorter bonds are stronger than longer
ones. Bond length is itself related to the size of the atoms that make up that bond. The bond
strengths of the halogens show a trend:
Page
158
Halogen
bond
length/nm
Cl-Cl
0.199
Br-Br
0.228
I-I
0.267
Bond length increases down the group. Going down the
group from one element to the next, an additional shell is
opened. This increases the size of the halogen atom, leading
to poor overlap of orbitals during bond formation.This poor
overlap of orbitals results in a longer bond, because the
atomic orbitals can not approach each other as closely as
they would if the orbitals were small.
The bond energies of the halogens are expected to decrease down the group, due to increasing bond
length. Indeed, this is observed:
Halogen
bond energy/
Kj mol-1
Cl-Cl
242
Br-Br
193
I-I
151
The bonds become weaker down the group. But does this
make the halogens more reactive? This question must be
answered with care. It all depends on the type of reaction we
are discussing. In a large number of their reactions, the
halogens act as oxidizing agents. In the process, they are
reduced, that is they gain electrons. Whether the halogen is
relatively more or less reactive therefore depends on its
The smaller the halogen atom, the more willing it is to gain an electron, and the more reactive it is. In this
sense, chlorine is the most reactive of the three halogens given above. The chlorine atoms in Cl 2 are quite
small, and they have a few shells intervening between the nucleus and the outer shell. An electron is
therefore easily attracted and stabilized in the outer shell because it will be close to the nucleus to which it
feels a strong attraction.
Using bond energies to estimate the enthalpy change of a reaction
The standard enthalpy change of a reaction, ∆Hθr has already been defined as the energy absorbed or released when a chemical reaction takes place according to the written balanced equation. Bond energies
can be used to determine the value of ∆Hr.
This method depends on the fact that bond breaking is endothermic (absorbs energy) and bond breaking is
exothermic (gives out heat energy). For instance, the O to O double bond has an energy of 497 Kj/mol. This
means that 497 Kj of energy are absorbed when one mole of the bond breaks, and that 497Kj of energy are
released when 1 mole of the bond forms.
A chemical reaction therefore involves two opposing processes, one that absorbs heat energy (bond
breaking) and another that releases heat energy (bond formation). The enthalpy change of a reaction is then
found by
∆Hr = (Total energy used to break bonds in reactants) + (Total energy released when
bonds form in products)
Example 1
Estimate the enthalpy change, ∆Hr, for the reaction below.
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
Solution
Calculating the total energy absorbed:
Page
159
This is the energy needed to break four moles of C-H bonds and two moles of O=O bonds (reactant side).
H
H C
Energy absorbed = 4E(C-H) + 2(O=O)
H + 2(O = O)
Data (from Data Booklet)
C-H bond energy = 412 Kjmol-1
H
O=O bond energy = 496 Kjmol-1
(4C-H bonds and 2O=O bond)
∴ Energy absorbed = (4 x 412) + 2(496) = +2 640 Kj
Calculating the total energy released:
This is the energy released when 2 moles of C=O bonds and four moles of O-H bonds are formed:
O= C =O + 2 O
Total energy released = 2E(C=O) + 4E (O-H)
= 2(743) + 4(463) (using the Data Booklet)
H H
= - 3 338 Kj
2C=O bonds and 4O-H bonds
Note that this value is negative since it represents energy released.
Enthalpy of reaction = Enthalpy of bond breaking (positive) + Enthalpy of bond formation (negative)
= 2 640 + (-3 338)
= -698 Kj
For practice
A mixture of hydrazine and fluorine is a potential fuel for rockets. How much energy would be released
when the two substances react?
N2H4 (g) + 2F2 (g)  N2 (g) + 4HF (g)
Data
N-H
O=O
Page
160
NɼN
H-F
N-N
Bond Energy/Kjmol-1
388
496
944
562
163
Extra information
Hydrazine contains 1 N-N bond and 4 N-H bonds.
Answer
-962Kj
We now look at enthalpies associated with processes that take place in aqueous solution, namely,
dissolution of ionic compounds and acid- base reactions.
5.7.12
Enthalpy change of solution
The standard enthalpy change of solution, ∆Hθsol, is the energy released or
absorbed when 1.0 mole of a substance dissolves in a large volume of water to
form an infinitely dilute solution under standard conditions.
For example, the standard enthalpy change of solution for NaCl refers to the
process:
NaCl(s)  Na+ (aq) + Cl-(aq), ∆Hθsol = +3.9 Kj mol-1




Standard enthalpies of solution are quoted per mole of substance that dissolves in a solvent. In
most cases, the solvent refers to water.
The term standard enthalpy of solution refers to both ionic and covalent substances.
∆Hθsol can be endothermic or exothermic; it all depends on which substance is involved.
Enthalpies of solution refer to the formation of infinitely dilute solutions.
An solution is said to be ‘infinitely dilute’ if there is a sufficiently large excess of water so that
adding any more doesn't cause any further heat to be absorbed or evolved.
Dissolution of an ionic solid in water
When sodium hydroxide dissolves in water, heat energy is evolved and the solution becomes warmer. When
ammonium chloride dissolves in water, heat energy is absorbed and the solution becomes cooler. Why are
some dissolution processes endothermic and others exothermic?
To understand enthalpies of solution of ionic compounds, we divide the dissolution process into two stages:


1 mole of the solid ionic crystal is dissociated into its respective gaseous ions. This process is the
reverse of lattice energy. The process is therefore endothermic (lattice energy is exothermic).
Energy is obviously needed to force oppositely charged particles apart.
The gaseous ions are then dissolved in a large volume of water to make an infinitely dilute
solution. The energy evolved by this step is known as the enthalpy change of hydration, ∆Hθhyd.
Page
161
Standard enthalpy change of hydration is the energy evolved when 1.0 mole of a gaseous ion
(cation or anion) dissolves in a large volume of water to form an infinitely dilute solution, under standard
conditions.
For example, the enthalpy change of hydration of the Na+ ion refers to the process
Na+ (g) + H2O  Na+ (aq)
∆Hθhyd. = -390KJ/mol.


∆Hθhyd is always exothermic because it results in the formation electrostatic attractions between
the ions and the water molecules. These attractions are actually relatively weak bonds. The
formation of bonds is always accompanied by release of energy.
Different ions produce different amounts of heat energy when they dissolve in water.
Table 5.4 below compares the enthalpies of hydration of some ions.
Cations
Ion
-∆Hθhydration /
Kjmol-1
Anions
Ion
-∆Hθhydration /
Kjmol-1
H+
1 090
F-
506
Li+
519
Cl-
364
Na+
406
Br-
336
K+
322
Rb+
301
Mg2+
1 917
Ca2+
1 650
Al3+
4 690
Table 5.4
Hydration Enthalpies of selected
ions. All values are negative and
are in Kjmol-1.
Factors that affect the magnitude of
hydration energy
It is clear from the Table 5.3 that the magnitude of
hydration energy depends on
 the size of charge on the ion
 the size of the ion( ionic radius)
 number of protons in the nucleus.
The first two factors above contribute to the charge
density of the ion.
Size of charge
The larger the charge on the ion, the more
negative the hydration energy. Compare the
enthalpies of hydration of Na+, Mg2+ and Al³+,
which become more negative in that order.
When an ion has a large charge, it can form strong
electrostatic attractions with water. Strong bonds
are formed, accompanied by the release of large
amounts of heat energy.
Ionic radius
Enthalpy change of hydration is inversely proportional to ionic radius. The larger the ion, the smaller the
Enthalpy of hydration. Compare the enthalpies of hydration of the ions of Groups (I) and (II). Going down
each group, additional shells are opened and the ions become larger. The nucleus is then a large distance
from the surface of the ion. Attraction of water molecules by the nucleus of the ion is therefore smaller, that
is, enthalpy change of hydration decreases (becomes less negative).
Number of protons in the nucleus of the ion (nuclear charge)
Page
162
Going along a period, for example from Na to Mg and Al, the number of protons in the nucleus increases.
The nucleus’ attraction for solvent molecules therefore increases, that is, the hydration energy of the ions
increases. The increase in hydration energy from Na+ to Al3+ can also be explained in terms of increase in
charge density as the ionic radii decreases due to increasing proton number.
Charge density
An ion with a relatively large charge and a relatively small ionic radius (for example, Al3+) has a large charge
density. Such an ion has a very powerful attraction for water molecules, resulting in the release of large
amounts of heat energy. The hydration energies of Na+, Mg2+ and Al3+ increases in that order due the
increase in charge density of the ions
The Hydration Enthalpies of Na+, Mg2+ and Al3+ (Period 3) increases in that order due to




increase in ionic charge
increase in nuclear charge (proton number)
decrease in ionic radii
increase in charge density
The Hydration Enthalpies of Groups (I) and (II) ions decrease down the group due to


Increase in ionic radii
decrease in charge density
Finding enthalpy of solution by Hess’ Law
As an example, consider the dissolution of NaCl in water. The energy change associated with this process is
the enthalpy change of solution, ∆Hsln. This value can be determined from an energy cycle (Fig 5.17).
By Hess’ Law:
Enthalpy of Route 1 = Enthalpy of Route
∆HθSln = - ∆H1+ ∆H2+ ∆H3
= - LE (NaCl) + ∆Hθhyd (Na+) + ∆Hθhyd (Cl-) … (i)
The Lattice Energy of NaCl is - 771 Kjmol-1.
∴ ∆HθSln = - (-776) + (- 390) + (-381)
= +5 Kjmol-1
The dissolution of NaCl in water is therefore an endothermic process.
Page
163
Fig 5.17 Energy cycle for the determination
of the enthalpy change of solution of an ionic
compound
5.7.13
Standard Enthalpy Change of Neutralization, ∆Hneut
This is the energy released when an acid and a base completely neutralize each other in aqueous
solution to form 1.0 mole of water, under standard conditions.
For example, the Standard Enthalpy Change of Neutralization for the reaction between sulphuric
acid and sodium hydroxide pertains to the equation
½ H2SO4 (aq) + NaOH (aq)  ½ NaCl (aq) + H2O (l)



The assumption is that the reaction takes place in a volume of water large enough to form an
infinitely dilute solution.
∆Hθneut is exothermic. This is because strong bonds are formed (in water) resulting in the release of
energy.
The units are Kj per mol (of water produced).
Table 5.5 below shows enthalpies of neutralization for some acid, base pairs. All values are negative and
have units of Kjmol-1.
Acid
Base
-H neut
HCl
KOH
57.1
HCl
NaOH
57.2
HNO3
HNO3
HCl
CH3COOH
HCN
HF
NaOH
1/2Ba(OH)2
57.3
58.1
NH3
52.2
NaOH
55.2
NH3
5.3
NaOH
68.6
Table 5.5 Enthalpies of neutralization

Page
164


The neutralization of a strong acid by a strong base
always produces about 57Kj per mole of water formed.
This is expected since the same overall reaction takes
place :
H+ (aq) + OH-(aq)  H2O(l) (after cancelling out
spectator ions)
This explains why the definition is made in terms of
amount of water formed. Even for dibasic acids and
alkalis such as sulphuric acid and barium hydroxide,
the value is the same because we define the enthalpy
change of neutralization per mole of water. If two or
more moles of water are shown in the balanced
chemical equation, then we should divide throughout
by the appropriate number so that we remain with one
mole of water on the product side.
If either the acid or the base is weak, the enthalpy of neutralization would be less exothermic than
-57Kjmol-1. Let us use HF and NaOH as an example. HF is a weak acid. Its ionization in water to
produce the H+ ion is not so energetically favorable because the H-F bond is quite strong. The
neutralization of NaOH by HF is therefore more endothermic that -57 Kjmol-1.
The Enthalpy Change of Neutralization is even smaller if both the acid and the alkali are weak, for
example, the enthalpy change of neutralization of NH3 (weak base) by HCN (weak acid) is only
-5.3 Kjmol-1.
Experimental determination of enthalpy of neutralization
Calorimetric measurements can be used to find the enthalpies of neutralization of an acid by a base. The
important thing in this determination is to make sure that loss of heat to the environment is negligible and
that the solution is stirred well to evenly distribute heat, otherwise pockets of heat energy are created (water
is a poor conductor of heat) and the thermometer will record a wrong temperature. Another issue is loss of
heat energy to the calorimeter and the thermometer. Advanced calorimeters are calibrated to cater for loss
of heat to the environment and to the apparatus. This gives very accurate values of enthalpies of
neutralization.
Q
Equal volumes of 2.0 moldm-3 NaOH and 2.0 moldm-3 HCl solutions are mixed in a plastic
cup and the temperature of the mixture changes from 200C to 250C . Explain how this
information can be used to determine a value for the enthalpy change of neutralization per
mole of the acid.
The recorded increase in temperature when the two solutions are mixed is smaller
than the actual value due to loss of heat to the environment. Suggest how a more accurate
value of the temperature change can be obtained without modifying the experimental
set up.
A
Let volume of NaOH
= y cm3 = volume of HCl
Total volume after mixing = 2y
Heat , Q, produced by reaction = mc∆θ,
where m = mass of solution = total volume in cm3 = 2y
c
= specific heat capacity of water = 4.2 Jg-1K-1
∆θ = increase in temperature = 5
Q = 2y x 4.2 x 5 = 42y joules
2 moles of acid produce 42y joules. 1 mole of the acid therefore produces 21y joules
The value of temperature change can be corrected graphically. Mark on an appropriately
scaled graph paper the initial temperature and the highest temperature attained. Now allow
the solution to cool, recording the drop in temperature at regular time intervals. On the graph
paper draw the ‘cooling curve’ and extraplate it backwards to the y (temperature ) axis, as
illustrated below.
extrapolation of
cooling curve
Page
165
z
x
y
x = initial temperature
cooling curve y = maximum temperature recorded
z = new maximum after extrapolation
The corrected temperature change is (z-x) which is slightly larger than the uncorrected value
(y - x)
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise 5: Miscelleneous problems
text box.]
1.
Use the following data to calculate the standard enthalpy change of reaction (i) below
4NH3(g) + 5O2(g) Ý 4NO(g) + 6H2O(g) ... (i)
Data:
Compound
∆Hθf / Kjmol-1
NH3(g)
-46.1
NO(g)
+90.
H2O(g)
-241.8
9701/11/O/N/2009
2. Heats of combustion of ethane, acetylene and hydrogen are -1 393Kjmol-1, -1 310Kjmol-1 and
-285 Kjmol-1 respectively.
What is the enthalpy change for the hydrogenation of acetylene?
C2H2 + 2H2  C2H6
9189/3/O/N/2008
3. (a)
(b)
State Hess’ Law
(i)
Write an equation to define the lattice energy of calcium oxide.
(ii)
Construct a Born-Haber cycle for the formation of calcium oxide.
(iii)
Use the cycle to calculate the lattice energy from the following information, and any
relevant data from the Data Booklet.
∆H atomization of calcium = +178Kjmol-1
∆H formation of calcium oxide = -635Kjmol-1
Page
166
∆H electron affinity (1st and 2nd) for oxygen = +657Kjmol-1
(iv)
Explain how and why the lattice energy of calcium oxide is different from that of
strontium oxide.
(c) An ice pack is used as a simple treatment for minor strains and sprains. It consists of a thick
plastic bag containing ammonium nitrate powder, and a second thinner plastic bag full of water.
If the bag is given a sharp blow, the thinner bag bursts and the ammonium nitrate dissolves
endothermically. The drop in temperature depends on the mass of the ammonium nitrate used
and the volume of water.
Given that ∆Hθ solution for ammonium nitrate is +26Kjmol-1 and the specific heat capacity of
the resultant solution is 4.2 JgK-1, calculate the mass of ammonium nitrate that would decrease
the temperature of 500cm3 of water by 250C.
9189/1/O/N/2011
4 (a)
Page
167
(b)
(c)
Write an equation to represent the lattice energy of sodium oxide, Na 2O.
The Born-Haber cycle shown may be used to calculate the lattice energy of sodium
oxide.
(i)
Identify the species A and B in the cycle, including the appropriate state symbols.
(ii)
Identify the enthalpy changes labelled by the numbers 1 to 4 in the cycle.
Use your cycle, the following data, and further data from the Data Booklet to calculate a value for
the lattice energy of sodium oxide.
Data:
enthalpy change of atomisation for Na(s)
+107 kJ mol–1
first electron affinity of oxygen
–141 kJ mol–1
second electron affinity of oxygen
+798 kJ mol–1
enthalpy change of formation of Na2O(s)
–414 kJ mol–1
enthalpy change of atomization for oxygen = half the bond energy for O2.
(d) (i) How would you expect the magnitude of lattice energy of magnesium oxide to
compare with that of sodium oxide? Explain your reasoning.
(ii)
State a use of magnesium oxide, and explain how the use relates to your answer in
part (d) (i).
9701/4/O/N/2002
5. Use the following data, together with relevant data from the Data Booklet, to construct a Born-Haber
cycle and calculate a value for the lattice energy of zinc chloride.
standard enthalpy change of formation of ZnCl2
standard enthalpy change of atomisation of Zn(s)
electron affinity per mole of chlorine atoms
–415 kJ mol–1
+131 kJ mol–1
–349 kJ mol–1
9701/4/M/J/2007
7. From an appropriately drawn and labelled Born-Haber cycle of CaF2, write down an expression to
find the second electron affinity of oxygen. Calculate this value using the following information as
well as any other relevant data from the Data Booklet.
∆Hθ atomization of Ca = +193 Kjmol-1
First electron affinity of oxygen = -142 Kjmol-1
∆Hθ formation of calcium fluoride = -1 214 Kjmol-1
Lattice energy of calcium fluoride = -2 602 Kjmol-1
Page
168
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
A similar experiment can be simulated in the lab using a plastic cup (plastic is a good insulator) as the
reaction vessel, but the value obtained is only an estimate because loss of heat to the environment and to
Glossary Sheet
the apparatus can not be avoided.
Examiners frequently demand precise definitions of enthalpy terms. Even if they do not, calculations in energetics
frequently depend on how certain enthalpy terms are defined. If your definitions and understanding of these terms
is hazy, even the simplest problems in Chemical Energetics will prove to be a real challenge. For this reason, we
have brought together the definitions of all the enthalpy terms discussed in this chapter. Memorize them, and more
importantly understand them.
1. The Standard Enthalpy Change of Reaction (∆Hr) is the heat absorbed or evolved when reactants
combine to form products according to the balanced equation, all measurements being taken under standard
conditions of 1 atm and 298K.
For example
3O2 (g) + 4NH3 (g) 6H2O (l) + 2N2 (g)
∆Hθr = -1 258Kj
2. The Standard Enthalpy Change of Formation (∆Hθf) of a substance is the heat evolved or absorbed
when 1 mole of the substance is formed from its elements in their normal states under standard conditions of
temperature (298K) and pressure (1 atm).
For example, the standard enthalpy change of formation of MgO refers to the energy change for the reaction:
1
Mg(s) + O2 (g)  MgO(s)
2
∆H = -602Kjmol-1
3. The Standard Enthalpy Change of Combustion (∆HCθ) of a substance is the amount of heat evolved when
1 mole of the substance burns completely in oxygen, the measurement of heat released being adjusted to standard
conditions of temperature and pressure.
For example, the enthalpy change of combustion of hydrogen refers to the equation:
H2 (g) + 1/2O2 (g)  H2O (l)
4. Lattice Energy is the amount of energy evolved when 1.0 mole of a solid ionic crystal is formed from its
respective gaseous ions under standard conditions of 1.0atm and 298K.
For example, the lattice energy of MgO is represented by the equation
Mg2+(g) + O 2-(g)  MgO(s)
5. The Standard Enthalpy Change of Atomization of an element refers to the energy absorbed in
forming 1.0 mole of gaseous atoms from that element in its natural state under standard conditions of 1.0 atm and
298K.
For example, the standard enthalpy of atomization of chlorine refers to the process
Page
169
1/2Cl2(g)  Cl (g)
6. The First Ionization Energy of an element refers to the energy required to knock out 1 mole of electrons
from 1 mole of gaseous atoms to form 1 mole of gaseous +1 ions, under standard conditions of 1.0 atm and 298K.
For example, the first ionization energy of magnesium refers to the process
Mg (g)  Mg+ (g) + e7. The Second Ionization Energy of an element refers to the energy required to knock out 1.0 mole of
electrons from 1.0 mole of gaseous +1 ions to form 1.0 mole of gaseous +2 ions, under standard conditions of 1.0
atm and 298K.
For example, the second ionization energy of magnesium refers to the process
Mg+ (g)  Mg2+ (g) + e8. The Third Ionization Energy of an element refers to the energy required to knock out 1.0 mole of electrons
from 1.0 mole of gaseous +2 ions to form 1.0 mole of gaseous +3 ions, under standard conditions.
For example, the third ionization energy of aluminium refers to the process
Al2+(g)  Al3+(g) + e9. The First Electron Affinity of an element refers to the energy released when 1.0 mole of gaseous atoms
accept 1.0 mole of electrons to form one mole of gaseous -1 ions, under standard conditions.
For example, the first electron affinity of oxygen is represented by the equation
O(g) + e-  O -(g)
10. The second electron affinity of an element refers to the energy absorbed to force 1.0 mole of electrons
into the outer shell of 1.0 mole of gaseous -1 ions to form 1.0 mole of gaseous - 2 ions.
For example, the second electron affinity of oxygen refers to the process
O –(g) + e-  O 2-(g)
11. The third electron affinity of an element refers to the energy required to force 1.0 mole of electrons into
the outer shell of 1.0 mole of gaseous -2 ions to form 1.0 mole of gaseous - 3 ions.
For example, the third electron affinity of nitrogen refers to the process
N-2(g) + e-  N3-(g)
12. Bond enthalpy is the average energy needed to break one mole of a specific covalent bond in the gaseous
phase, to form separate gaseous atoms.
For example, the bond energy of chlorine refers to the energy required to break one mole of the Cl-Cl bond.
Cl-Cl (g)  2Cl-(g)
13. The standard enthalpy change of solution, ∆Hθsol, is the energy released or absorbed when 1.0 mole of
a substance dissolves in a large volume of water to form an infinitely dilute solution under standard conditions.
For example, the standard enthalpy change of solution for NaCl refers to the process:
Page
170
NaCl(s)  Na+ (aq) + Cl- (aq), ∆Hθsol = +3.9 KJ /mol
14. Standard enthalpy change of hydration is the energy evolved when 1.0 mole of a gaseous ion (cation or
anion) dissolves in a large volume of water to form an infinitely dilute solution, under standard conditions.
For example, the enthalpy change of hydration of the Na+ ion refers to the process
Na+ (g)  Na+ (aq)
∆Hθhyd. = -3.90Kj/mol.
15. Standard Enthalpy Change of Neutralization is the energy released when an acid and a base completely
neutralize each other in aqueous solution to form 1.0 mole of water, under standard conditions.
For example, the Standard Enthalpy Change of Neutralization for the reaction between sulphuric acid and
sodium hydroxide pertains to the equation
Page
171
½ H2SO4 (aq) + NaOH (aq)  ½ NaCl (aq) + H2O (l)
CHAPTER
6
Electrochemistry
In Chapter 2 we discussed the composition of an atom in terms of sub-atomic particles. It was
mentioned that electrons have a direct effect on the chemical properties of substances. Chemical reactions
result in a rearrangement of atoms and electrons in the outer-shells of these atoms. This rearrangement
of electrons determines, to a large extend, the stability of the products of a reaction relative to the reactants.
There are certain reactions which involve a total transfer of electrons between reactants. Such reactions
are known as redox reactions (that is, reduction-oxidation reactions).
Electrochemistry is concerned with the study of redox processes. By the time you are through with this
chapter, you should be able to








6.1
decide whether a given reaction is redox or not, on the basis of oxidation numbers
determine the number of electrons that are transferred during redox reactions
identify the oxidizing and the reducing agent in a redox reaction
predict the energetic feasibility of a redox reaction in terms of Electrode Potentials
construct electronically and stoichiometrically balanced redox equations
describe the key concepts involved in electrochemical cells, and predict the voltage out-put of a
simple cell
explain the principle of electrolysis, and predict the products of a given electrolytic process
Calculate quantities involved during electrolytic processes, for example, mass of metal deposited
and volume of gas produced.
Oxidation and Reduction
Page
172
Two processes, known as reduction and oxidation, are involved in a redox reaction. These terms may be
defined in terms of transfer of oxygen /hydrogen atoms, or transfer of electrons (Table 6.1). In
electrochemistry, emphasis is on the transfer (loss or gain) of electrons. A large number of redox
reactions do not involve the transfer of oxygen or hydrogen atoms. However, for most organic redox
reactions, gain or loss of hydrogen or oxygen is more apparent than transfer of electrons.
Oxidation is
Useful Mnemonic
Reduction is
Loss of electrons
Gain of electrons
Gain of oxygen
Loss of oxygen
Loss of hydrogen
Gain of hydrogen
An increase in oxidation
number
A decrease in oxidation
number
Table 6.1
6.2
Oxidation Is Loss (of
electrons or hydrogen)
Reduction Is Gain (of
electrons or hydrogen)
Definitions of reduction and oxidation
The concept of oxidation numbers
Oxidation numbers help in the determination of number of electrons transferred in redox reactions, and in
determining whether a reaction is redox or not.
The oxidation number (oxidation state) of an atom is the charge it would have if it existed as a stable ion.
The oxidation number therefore describes the relative state of oxidation or reduction of the atom of an
element. For example, in nitrogen monoxide, nitrogen has an oxidation state of -2. It is in a relatively
reduced state, because it is possible for nitrogen to attain higher oxidation states, for example, the +5 state
in HNO3.
Panel 1
Guidelines for assigning oxidation numbers
1. An element or neutral atom is assigned an oxidation
number of zero.
except in hydrides where it is -1. The oxidation state of
2. The oxidation number for a monatomic ion is the
charge on that ion, e.g. for
Na+,
the
except in the peroxides (where it is -1) and in OF2
(where it is +2). The oxidation state of halogens in a
written.)
large number of their compounds is -1, for example, the
the different elements present is equal to the charge
on the ion, e.g. in the
SO42-
ion , the sum of the
oxidation state of S and four O atoms is equal to -2.
4. In neutral molecules, the algebraic sum of the oxidation
oxidation state of Cl in NaCl is -1. However, there are
many exceptions to this generalization.
7. The oxidation state of metals in their compounds is
always positive. In their stable compounds, this
oxidation number is equal to group number, e.g the
numbers of the different elements present is equal to
oxidation sate of Al in Al2O3 is +3, since Al is in Group
zero.
(III). The transition metals exhibit variable oxidation
5. In any compound made up of two elements, the more
electronegative atom will have a negative oxidation
173
oxygen in all of its compounds is always equal to -2,
oxidation state of sodium is +1 (the + sign should be
3. In a polyatomic ion, the sum of oxidation numbers of
Page
6. The oxidation state of hydrogen is always equal to +1,
number and the less electronegative will have a
positive oxidation number.
states in their compounds.
Examples
Solutions
Determine the oxidation number
1. -2 (Guideline 6)
of the indicated element in each
2. Let oxidation state of N be x.
5.
x -2
x + -2(4) = -2
The oxidation states of H and O
of the following compounds.
are +1 and -2 respectively.
+1 x -2
2. N in HNO3
in MnO42- is +6.
7.
Using Guideline 4,
3. S in H2S
x =6
The oxidation state of Mn in
H N O3
1. O in H2O
Mn O42-
K2 Cr2O7
+1 x -2
1 + x -2(3) = 0
+1(2) + 2x -2(7) = 0
x=5
2x = 12
4. Cl in ClO-
The oxidation state of N in
x=6
5. Mn in MnO42-
HNO3 is +5.
3. Let the oxidation state of S
6. Mn in MnO4-
The oxidation state of Cr in
K2 Cr2O7 is +6.
be x.
7. Cr in K2Cr2O7 ( potassium
dichromate)
H2 S
The rest of the problems
+1 x
are left as an exercise to
+1(2) + x = 0
8. Cr in K2CrO4 (potassium
chromate)
the reader.
x = -2
The oxidation state of S in
9. V in VO2+ and in VO2+
H2S is -2
4.
Cl Ox -2
x - 2 = -1(Guideline 3)
x=1
The oxidation state of
Cl in ClO- is +1.
Identifying redox reactions using oxidation numbers
Page
174
One limitation of defining redox reactions in terms of gain or loss of hydrogen or oxygen is that a large
number of redox reactions do not involve hydrogen or oxygen at all. If we define a redox reaction in terms
of electron transfer we have an easy way of identifying redox reaction. Since reduction is gain of electrons,
it must be accompanied by a decrease in oxidation number of the atom which is being reduced (the atom
gains negative charge). Oxidation is loss of electrons. It is therefore accompanied by an increase in the
oxidation number of the atom which is being oxidized since the atom loses negative charge.
Q
Which of the following conversions are redox?
(i) CaCO3  CaO + CO2
(ii) K2Cr2O7  K2CrO4
(iii) VO2+ + Zn + 2H+  V3+ + Zn2+ + H2O
A
(i) and (ii) are not redox reactions. All atoms have the same oxidation state on both sides of the
equation. (iii) is redox because the oxidation state of Vanadium changes from +4 in VO 2+ to +3
in V3+.
NB. It is usually not necessary to calculate the oxidation state of every atom in the equation. For example,
in reaction (iii), check using vanadium only. Being a transition element, it is the one whose oxidation state
is likely to change.
Let us analyze reaction 3 further and determine
 The substance which is reduced
 The substance which is oxidized

the reducing agent and the oxidizing agent
 the number of electrons transferred in the reaction
VO2+ is reduced because V experiences a decrease in oxidation number. Zn is oxidized because its oxidation
state increases (from 0 to +2). It loses two electrons, so its oxidation number increases by 2. It gives these
electrons to VO2+. In other words, Zn causes VO2+ to be reduced (to gain electrons). Zn is therefore the
reducing agent, but in carrying out its function it is oxidized. We may also say VO2+ takes electrons from
Zn, that is, it causes Zn to undergo oxidation. VO 2+ is therefore the oxidizing agent, and in the process, it is
reduced.

An oxidizing agent causes another substance to be oxidized (to lose electrons). A reducing agent
causes another substance to be reduced (to gain electrons).

An oxidizing agent is the species which is reduced in the reaction. The reducing agent is the
species which is oxidized in the reaction.

Reduction and oxidation always happen together. If we identify one substance which is oxidized
in a reaction, there must be another substance which is being reduced.
We also need to know the number of electrons that are transferred in the reaction.
VO2+(aq) + Zn(s) +2H+(aq)  V³+(aq) + Zn2+(aq) + H2O(l)
Zn loses two electrons and becomes Zn2+, but VO2+ needs only one electron to be reduced from the +4
state to the + 3 state. Two moles of VO 2+ are therefore needed per mole of Zn. Thus, though the equation
above looks balanced, it in fact is not, because it does not account for the other electron from Zn. Where
does it go? The correct (balanced) equation is shown below:
Page
175
2VO2+ (aq) + Zn(s) +4H+ (aq)  2 V³+ (aq) + Zn2+ (aq) + 2H2O (l) ... (i)
The reaction involves the transfer of two moles of electrons from one mole of Zn to two moles of VO2+.
It should also be noted that there is a relationship between the stoichiometry of the reaction and the
number of electrons transferred. The following equation generalizes this relationship.
In a redox reaction,
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐫𝐞𝐝𝐮𝐜𝐞𝐝 𝐬𝐩𝐞𝐜𝐢𝐞𝐬
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐨𝐱𝐢𝐝𝐢𝐳𝐞𝐝 𝐬𝐩𝐞𝐜𝐢𝐞𝐬
=
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧𝐬 𝐥𝐨𝐬𝐭 𝐛𝐲 𝐨𝐱𝐢𝐝𝐢𝐳𝐞𝐝 𝐬𝐩𝐞𝐜𝐢𝐞𝐬
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧𝐬 𝐠𝐚𝐢𝐧𝐞𝐝 𝐛𝐲 𝐫𝐞𝐝𝐮𝐜𝐞𝐝 𝐬𝐩𝐞𝐜𝐢𝐞𝐬
... (I)
Example
In an experiment, 3.0 moles of Zn were required to reduce 2.0 moles of an aqueous ion (X) of
Vanadium to VÆÄ (aq). What is the oxidation state of V in the ion X?
Solution
The reduced species is X and the oxidized species is Zn
Moles of electrons lost by oxidized species = 2 (Zn is always oxidized from 0 to +2 state)
Let the oxidation state of V in ion X be p. Since X is reduced to +2 state, the number of electrons
gained by X = p-2 (the oxidation number of V changes from a higher value, p, to a lower number, +2).
Using formula (I) above:
2
3
2
= p−2
Solving for p yields:
p= 5, which is the oxidation number of V in species X.
Further information
Species X, in which vanadium
has an oxidation number of +5
could be the vanadate ion, VO3(stable in alkaline medium) or
the vanadic ion VO2+, which is
stable in acidic medium.
Balancing redox equations
Half equations
Page
176
It has already been mentioned that in a redox reaction, there is simultaneous oxidation and reduction. It is
therefore possible to resolve the redox reaction into two equations; one showing what is happening to the
oxidized species, and the other showing what is happening to the reduced species.
These equations are known as the oxidation half equation and the reduction half equation respectively.
They are known as half equations because in practice they can not happen on their own. They only occur in
a full reaction which can be obtained by adding together the two half equations.
Example 1
Construct balanced half equations for the reduction of VO2+ by Zn to V3+ in aqueous solution. Hence
construct a balanced equation for the reaction.
Solution
In VO2+ , vanadium has an oxidation number of +5. For this species to become V 3+ , it must gain 5-3 = 2
electrons.
VO2+ + 2e  V3+
Next, balance oxygen atoms by adding two water molecules to the right hand side (RHS) of the equation.
VO2+ + 2e  V3+ + 2H2O
Next balance hydrogen atoms by adding 4H+ ions to the left hand side (LHS) of the equation
4H+ + VO2+ + 2e  V3+ + 2H2O ... (i)
In general, if O atoms are short on the product side, balance by adding the correct number of H2O molecules
on that side. This necessitates the addition of the correct number of H + ions to the reactant side.
Reaction (i) is the reduction half-equation (it involves gain of electrons). This reduction requires an acidic
medium (H+ ions). The role of the hydrogen ions is to combine with O atoms in VO 2+ to form water.
The oxidation half equation is straight forward. Zn loses two electrons and becomes Zn2+.
Zn  Zn2++ 2e ... (ii)
Bring the reduction and oxidation half equations together and add them vertically. This gives the net
equation for the reaction between Zn and VO2+ in the presence of an acid. Notice that the number of
electrons in the two half-equations must be equal, so they cancel out. This shows that all the electrons
produced by the reducing agent are taken up by the oxidizing agent.
4H+ + VO2+ + 2e-  V3+ + 2H2O
Zn 
Net equation
Zn2++
2e-
... (i) reduction half-equation
... (ii) oxidation half-equation
4H+ + VO2+ + Zn  V3+ + Zn2++ 2H2O ... (iii)
NB A redox reaction must be both stoichiometrically and electronically balanced. That is, it must be balanced both
in numbers of atoms present on the two sides of the reaction and in the number of electrons involved in the
reduction and the oxidation half-equations.
Example 2
Page
177
Construct a balanced equation for the oxidation of iron (II) chloride to iron (III) chloride by
potassium permanganate in aqueous solution. The permanganate is reduced to manganese (II)
chloride.
Solution
The chloride ion is a spectator ion in this reaction and can be ignored for convenience.
The reduction half-equation involves the reduction of MnO4- to Mn+2. The oxidation number of Mn in
MnO4- is +7. To form Mn+2, MnO4- must therefore gain 7-2 = 5 electrons.
MnO4- + 5e  Mn+2
Balance O by adding 4H2O to the RHS then balance H by adding 8H+ to the LHS.
8H+ + MnO4-+ 5e  Mn2+ + 4H2O ... ( i) reduction half equation
The oxidation half equation is the conversion of Fe2+ to Fe3+ . In this process, Fe2+ loses one electron:
Fe2+  Fe3+ + e ... (ii)
oxidation half equation
Fe2+ loses one electron and yet MnO4- needs five to be reduced to Mn2+. We therefore multiply equation
(ii) by 5 to balance the number of electrons. Equation (i) and (ii) are now added up vertically to get the
overall equation:
8H+ + MnO4-+ 5e  Mn2+ + 4H2O
5Fe2+  5Fe3+ + 5e
Net equation 8H+ + 5Fe2+ + MnO4-  Mn2+ + 5Fe3+ + 4H2O
The equation shows that an acidified solution of potassium permanganate oxidizes iron (II) chloride to
iron (III) chloride. The purple color of potassium permanganate disappears and the mixture turns
yellowish brown due to the formation of Fe (III ) chloride.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
6.1 in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Page
178
Balance the following redox reactions
(a)
Cr2O72- + Fe2+  Cr3+ + Fe3+
(b)
VO2+ + Sn2+  V3+ + Sn4+
(c)
Fe3+ + I-  Fe2+ + I2
(d)
Cr2O72- + SO2  Cr3+ + H+ + SO42-
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text Electrode
box.]
6.3
potentials
Consider a metal electrode M in contact with a solution of its ions (For example, a copper rod immersed in
copper (II) sulphate solution).
Two opposing processes take place:

Oxidation of the metal to its ions
M(s)  Mn+ (aq) + ne-
reduction of the metal ions back to the metal.
Mn+ (aq) + ne-  M(s)
In other words, the following equilibrium is established
Fig 6.1 A half cell containing
an electrode M in contact with
a solution of its ions.
Mn+ (aq) + ne- Ý M(s)
There is a therefore a flow of electrons between the electrode and its ions,
the metal loses electrons, which are transferred to the ions in solution.
The process reaches dynamic equilibrium in which the rate of the forward reaction is equal to the rate of
the reverse reaction. There is therefore no net change to the metal or to the solution.
The set up in Fig 6.1 is a half cell because only one electrode is involved. It is possible to connect this cell to
another half cell and provide a complete circuit for the flow of electrons between the two cells. A voltmeter
can then be connected to the external circuit to measure the potential difference between the electrodes in
the two half cells.
We may need to compare the potentials due to different electrodes. To do this, we connect a half cell of the
electrode M to a standard reference cell. By using different electrodes, it is possible to compare the potential
differences of the half-cell relative to the reference cell. The standard reference cell used is a half cell known
as the standard hydrogen electrode, S.H.E (Fig 6.2); so called because the ‘electrode’ is hydrogen
gas in contact with its ions, H+. The potential difference recorded between the electrode M and the SHE is
known as the electrode potential (or reduction potential) of electrode M. In this set up, it is necessary to
make sure that standard conditions are employed. The electrode potentials obtained under these
conditions are said to be standard. This makes the comparison of electrode potentials of different electrodes
meaningful. The standard conditions used are
Temperature: 298K (250C)
Pressure: 1.0 atm
All concentrations are 1.0 moldm-3
In the SHE, hydrogen gas dissociates into its ions
H2 (g)  2H+(aq) + 2eThe Pt metal provides an inert surface for this reaction, and also acts as a catalyst, since hydrogen on its
own does not easily dissociate into hydrogen ions. Pt is particularly selected on the basis of its inertness and
catalytic ability. It does not ionize in the acidic medium in which it is immersed; otherwise it would generate
its own voltage.
The hydrogen ions in solution accept electrons provided by
hydrogen gas. The ions are thus reduced to H2 (g).
Page
179
2H+ (aq) + 2e H2 (g)
An equilibrium is therefore established on the platinum
electrode
2H+ (aq) + 2e Ý H2 (g)
This equilibrium involves movement of electrons between the Pt electrode and the solution of H + ions. At
equilibrium there is a potential difference between the electrode and the solution.
By convention, the SHE is assigned a voltage of zero. The value
recorded by the voltmeter when the SHE is connected to an
electrode M is therefore attributed to the electrode. This value is
known as the electrode potential of the electrode M.
The standard electrode potential, Eθ, of an electrode is the
potential difference between a half cell made up of that
electrode and its ions, and a standard hydrogen electrode,
under standard conditions.
Standard Electrode Potentials are measured in Volts (V).
Fig 6. 2
The standard hydrogen
electrode
Fig 6.3
Fig 6.3 illustrates how the electrode potential of Zinc can be
determined by linking a Zn(s)/Zn2+(aq) half cell to the SHE. The
arrows indicate the direction of flow of electrons.
There are two half cells involved in this set-up. One cell is
Zn(s)⃒Zn2+(aq), read as ‘Zinc electrode in contact with its ions’
. The other half cell is the SHE, Pt(s)⃒H2(g),H+(aq), read as
‘hydrogen gas in contact with its ions over a platinum electrode’.
Set-up for the determination of the electrode potential of Zn
metal
In the Zn(s)⃒Zn2+(aq) half cell zinc is oxidized to Zn2+ ions
Zn(s) ÝZn2+(aq) + 2e- ...(i)
Page
180
In the SHE, a similar process takes place, in which hydrogen is oxidized H+ ions.
H2 (g) Ý 2H+ + 2e- ... (ii)
Notice that each atom in H2 (g) provides an electron during dissociation, so a total of two electrons are
produced per molecule of hydrogen gas. Both half cells therefore produce electrons, and which ever of the
two cells is more negative in terms of these electrons will supply electrons in the external circuit.

In this case, zinc metal provides electrons in the external circuit. This is because zinc metal has a
greater tendency to lose electrons (to be oxidized) compared with hydrogen gas. The Zinc electrode
is therefore relatively negative, whilst the Pt electrode on the SHE side is relatively positive. We say
there is a potential difference between the two electrodes. It is this potential difference (Voltage)
that forces a flow of electrons (current) in the external circuit. Notice the use of the word relative.
Neither electrode (Zn or Pt) has a full positive or negative charge. Instead, one electrode (Zn) ionizes more readily so it becomes richer in electrons than the Pt electrode.

The voltmeter measures the potential difference between the Zn and the Pt electrode. The direction
in which the needle of the voltmeter moves shows that electrons flow from Zn to Pt. In this case,
the needle of the voltmeter moves to the negative side of the zero mark. The electrode
potential of zinc is therefore negative.

By convention (that is, not by measurement), the SHE electrode is assigned a voltage of zero, so the
voltage reading is the standard electrode potential of Zinc. This method of assigning a zero value to
the SHE half-cell is valid because the aim is to compare the electrode potentials of different electrodes. The comparison is legitimate as long as the voltage of the SHE is taken as a constant (the
constant zero is used for its convinience)

Notice the presence of a salt bridge in Fig 6.3. Its function is to complete the circuit between the
two half cells so that current can flow. The salt bridge contains an aqueous salt such as KNO 3. Flow
of current in the salt bridge is in the form of movement of ions, not electrons. Another
important function of the salt bridge is to maintain electroneutrality. This will be explained later.

In Fig 6.3, the voltmeter records a negative value of -0.76V. The negative sign indicates that the
needle of the voltmeter sweeps to the negative side of zero. It shows that electrons flow from zinc
to the Pt electrode. The value -0.76 is the standard electrode potential of Zinc and it pertains to
forward reaction of the equilibrium
Zn2+ (aq) + 2e- Ý Zn(s)
Electrode potentials are also known as reduction potentials because they refer (by convention) to the
reduction half equations.
An important note on the +, - notation
The sign of the electrode potential (+ or -) indicates the direction of flow of electrons. A negative sign shows that
electrons flow from the electrode M to the SHE, that is, the electrode M has the greater tendency to release electrons
than hydrogen gas. A positive value shows that electrons flow from the SHE to the electrode M, that is, hydrogen gas
has the greater tendency to release electrons than the electrode M.
An example of a positive electrode potential is that of Ag. This value is +0.80V, and it refers to the forward
reaction of the equilibrium
Page
181
Ag+(aq) + e- Ý Ag(s)
The electrode here is Ag metal immersed in its ions, Ag +.The positive value shows that hydrogen (in SHE)
ionizes more readily than silver metal. The Pt surface in SHE therefore becomes richer in electrons
(supplied by the dissociation of H2) than the Ag electrode. Electrons therefore flow from the SHE to the
Ag(s)⃒Ag+(aq) half cell. The needle of the voltmeter deflects to the positive side of zero, that is, a positive
value is recorded.
Table 6.2 gives the electrode potentials for some elements. The table includes non-metallic elements.
The measurement of the electrode potentials of such electrodes will be discussed later.
The significance of electrode potentials
First the reader should recall that electrode potentials are given in terms of reduction (gain of
electrons). For example, the electrode potential of zinc metal, -0.76, refers to the forward reduction
reaction in the equilibrium
Zn2+(aq) + 2e- Ý Zn(s) ... (i)
Reading the equilibrium in reverse gives
Page
182
Electrode
Equilibrium
Standard
Electrode
Potential/
V
Mg
Mg2+(aq) + 2e- Ý
Mg(s)
-2.38
Al
Al3+ + 3e- Ý Al(s)
-1.66
Zn(s
Zn2+(aq) + 2e- Ý Zn(s)
- 0.76
Pb(s)
Pb2+(aq) + 2e- Ý Mg(s)
-0.13
H2
H+(aq) + e- Ý ½ H2(g)
0.00
Cu
Cu2+(aq) + 2e- Ý Cu(s)
+0.34
Ag
Ag+(aq) + e- Ý Ag(s)
+0.80
I2
I2(g) + 2e- Ý 2I-(aq)
+0.54
Cl2
Cl2(g) + 2e- Ý 2Cl-(aq)
+1.36
Table 6.2 Standard Electrode Potentials,
measured at 298K and for 1.0 moldm-3 solutions.
For gases such as chlorine, the reaction takes
place on the surface of an inert metal,
particularly platinum.
Zn(s) Ý Zn2+ (aq) + 2e- ...(ii)
for which the potential is +0.76V. The positive
value after reversing equation (i) indicates that Zn
metal readily loses electrons, compared to hydrogen
gas. Thus, as already explained, electrons will flow
from Zinc to SHE.
But why did we reverse the equilibrium?
The electrode here is a metal, and metals usually
undergo oxidation in their reactions. We have
therefore reversed the equilibrium to show the oxidation of zinc. The sign of the electrode potential
after reversing the reduction half equation helps in
assessing the ease with which a metal can be
oxidized. If the reduction potential is negative before
reversing, as is the case with most metals, it becomes
positive after reversing. The positive value shows
that under standard conditions, the metal can easily
be oxidized to its ions, that is, it is quite reactive. The
more positive the potential after reversing, the more
reactive the metal.
For metals, it is recommended that the reader develops
the habit of reading the redox potentials in reverse. This is
because the reverse reactions show oxidation, which is the
reaction likely to be undergone by metals. A positive
potential after the reversal shows that the metal is
relatively easily oxidized to its ions.
By referring to Table 6.2, you can see that both magnesium and aluminium are expected to be very reactive
metals. Magnesium is indeed very reactive, but aluminium shows an inertness which is due to kinetic factors
to be explained later.
It possible to arrange metals in order of reactivity by comparing their reduction potentials. The more
negative the electrode potential before reversing, or the more positive the value after reversing,
the more reactive the metal. Some metals have a positive reduction potential, for example, copper whose
electrode potential is +0.34V. That means the reverse oxidation reaction
Cu(s) Ý Cu2+(aq) + 2ehas a negative reduction potential of -0.34V. The negative sign shows that the oxidation of copper metal to
Cu2+ is not energetically favourable. In other words, copper is a relatively unreactive metal. In Table
6.1, silver has the most positive electrode potential, so it is the least reactive (most stable) of the metals in
this table. All metals with positive electrode potentials are below hydrogen in the reactivity series. When
connected to a SHE, electrons flow from the SHE to the metal electrode, showing that hydrogen is oxidized
instead of the metal.
Q
A
Arrange the metals in Table 6.2 above in order of decreasing reactivity (that is, from the
most reactive to the least reactive)
Mg, Al, Zn, Pb, Cu, Ag
6.4 The electrochemical series
This is an arrangement of the metals in order of their standard electrode potentials, starting with the most
negative (that is, the most positive if the electrode potentials are considered in reverse). Part of the
electrochemical series is shown in Fig 6.4 (Hydrogen is included in this list as a reference).
A metal which is high in the reactivity series (large and negative electrode potential), for example, Mg shows
the following properties


it readily gives up electrons to form ions, for example, Mg(s) Mg2+ + 2e-. These electrons are
gained by another substance in a redox reaction. This other substance is therefore reduced. In other
words, a metal with a large and negative electrode potential (high up the electrochemical series) is
a good reducing agent.
it is very reactive
On the other hand, a metal which is at the bottom of the electrochemical series
Page
183

does not readily give up electrons. Instead, its ions readily gain electrons (for example, from a
metal higher in the series) to form the neutral metal, e.g
Ag+(aq) + e-  Ag(s)
Note that the electrode potential for this reaction is positive (= +0.80V. See Table 6.2). This shows
that the conversion of the ions to the metal is energetically feasible. However, the reverse reaction
in which Ag is oxidized to its ions is negative. This shows that the oxidation of Ag metal to its ions
is energetically unfavourable. In other words, Ag metal is relatively inert.
Metal
most
reactive
least
reactive
6.5
Reaction
Standard
Electrode
Potential/ V
Mg
Mg2+(aq) + 2e- Ý Mg(s)
-2.38
Al
Al3+ + 3e- Ý Al(s)
-1.66
Zn
Zn2+(aq) + 2e- Ý Zn(s)
- 0.76
Pb
Pb2+(aq) + 2e- Ý Mg(s)
-0.13
H2
H+(aq) + e- Ý ½ H2(g)
0.00
Cu
Cu2+(aq) + 2e- Ý Cu(s)
+0.34
Ag
Ag+(aq) + e- Ý Ag(s)
+0.80
Fig 6.4 Shortened form of the electrochemical series
of metals
Most
negative
Eθ
Most
positive
Eθ
Standard Electrode potentials for non metals
The standard electrode potential of chlorine is +1.36V and it refers to forward reaction of the equilibrium
Cl2 (g) + 2e- Ý 2Cl-(aq) ... (i)
Fig 6.5 shows the set up for the determination of the electrode potential of a gas such as Cl 2. The design of
the cell is necessitated by the fact that a gas can not be dipped into a solution of its ions as is the case with
metal electrodes. Instead, it gets into contact with its ions, in this case, Cl - ions, over a platinum metal
surface.
Page
184
Understanding the electrode potentials of the halogens
The electrode potentials of metals can be understood by reversing the relevant equilibrium so that the
reaction shows oxidation of the metal. For the Group (VII) elements, it is not necessary to do this reversal,
because the halogens usually undergo reduction, in which they gain electrons to form the halide ions.
The halogens all have positive electrode potentials which shows that the reduction of the gas to its ions is
energetically favourable. The halogens are therefore good at accepting electrons. In so doing, they act as
oxidizing agents. The more positive the value of the reduction potential, the easier it is for the halogen to
be reduced (to accept electrons, for example, from a metal).
Compare the Standard electrode potentials of Cl2, Br2 and I2 given in Fig 6.6. The decrease in the reduction
potentials of the halogens down the group explains why they become less reactive as the group is
descended.
Fig 6.5
The set up for the determination of the electrode
potential of chlorine gas

Chlorine, with the most positive electrode potential, has the greatest ability to accept electrons. In
other words, it easily causes other substances, for examples, metals, to lose electrons. Chlorine is
therefore a powerful oxidizing agent.

Iodine, with the smallest electrode potential, has the least ability to accept electrons. It is
therefore the weakest oxidizing agent of the three halogens.

The halogens therefore become weaker oxidizing agents down they group, as they become less
willing to accept electrons. This trend can be explained in terms of atomic radii. Going down the
group, atoms of the halogen become larger due to the opening of new shells. Large atoms are not
so good at accepting electrons in their outer shell. This is because the outer shell is a large distance
from the nucleus, so the incoming electron feels only a weak attraction from the nucleus. This weak
attraction implies that the electron is not effectively stabilized, so its incorporation into the outer
shell of the atom is not so favourable.
most
Eθ/V
reactive
Page
185
Cl2
least
reactive
Cl2(g) + 2e- Ý 2Cl-(aq)
+1.36
+1.07
Br2(g) + 2e- Ý 2Br-(aq)
Br2
Fig 6.6 The Standard Electrode Potentials of
- Ý 2I-(aq)become less positive
I2 I2(g) the
+0.54
+ 2ehalogens
down the group. The reactivity of the
elements decrease in that order.
most
positive Eθ
least positive
Eθ
Electrode potentials for a redox couple that involves ions
The Data Booklet gives the standard electrode potential of the Fe3+(aq)/Fe2+(aq) couple as +0.77V. This
corresponds to the equilibrium
Fe3+(aq) + e- Ý Fe2+(aq)
Fe3+, having the larger oxidation number, is the one that is gaining electrons, that is, it is being reduced to
Fe2+ (reduction is accompanied by a decrease in oxidation number). The value of the electrode potential for
this redox couple is found by setting up a half cell shown in Fig 6.7 and connecting it to the SHE.
Fig 6.7
6.6
A half cell for the determination of the
Fe3+/Fe2+ electrode potential
Using Standard Electrode Potentials to predict redox reactions
Eθ values can be used to determine the feasibility of a redox reaction. If a reaction has a net cell potential
which is positive, it is energetically feasible and can take place under standard conditions. In terms of
thermodynamics, the reaction has a negative enthalpy of reaction, that is, the products are lower in energy
and more stable than the reactants.
A negative cell potential corresponds with a positive enthalpy of reaction. Such a reaction is not energetically favourable.
Example 1
Predict if an aqueous solution of Fe (II) can convert aqueous potassium permanganate to Mn 2+.
Page
186
Solution
If a reaction occurs between these two substances, Fe (II) will be the reducing agent and will be oxidized to
Fe (III). MnO4- would be the oxidizing agent and would be reduced to Mn 2+ (Mn in MnO4- is in a high
oxidation state (+7) so it is likely to be converted to a lower oxidation state.
Fe (II) is in a lower oxidation state so it is likely to be converted to a higher oxidation state).
A balanced equation can then be constructed by combining the reduction half equation and the oxidation
half equation. The relevant reduction potentials are shown below.
Eθ/V
MnO4-(aq) + 8H+ (aq) + 5e- Ý Mn2+ + 4H2O
Fe3+ + e- Ý Fe2+
+1.52
... (i)
+0.77 ... (ii)
These two half equations have been taken as they are in the Data Booklet, that is, in terms of reduction.
The electrode potential of MnO4-/Mn2+ is more positive, which means that MnO4- is the one which is
more likely to undergo reduction, that is, it is likely to behave as an oxidizing agent. Next we reverse
reaction (ii), together with the sign of its electrode potential. We do this for two reasons. First, we intend
to bring Fe2+ to the same reactant side as MnO4- so that the two substances can react. Secondly, we need
to show the oxidation of Fe2+ to Fe3+. Since MnO4- undergoes reduction, Fe2+ must undergo oxidation. We
now have a reduction half equation and an oxidation half equation.
Notice that equation (ii) must be multiplied by 5 to balance the number of electrons with those shown in
equation (i). The net equation is obtained by adding the two equations vertically. The electrode potentials
are also added vertically to obtain the cell potential, Eθ.
Eθ/V
MnO4-(aq) + 8H+ (aq) + 5e- Ý Mn2+ + 4H2O
+1.52
(i)
Ý
-0.77
(ii)
5Fe2+
5Fe3+
+
5e-
Cell reaction MnO4-(aq) + 8H+ (aq) + 5Fe2+ Ý Mn2+ + 5Fe3+ + 4H2O
+0.75 (iii)
Since the net cell potential of the reaction is positive, we conclude that the reaction is indeed feasible under
standard conditions. MnO4- ions oxidize Fe2+ ions to Fe3+. Meanwhile, MnO4- is reduced to Mn2+, resulting
in the decolourization of the purple solution.
The following working order is useful in the determination of the feasibility and outcome of a redox reaction.

Page
187


The first step is to select the correct half equations from the Data Booklet. First, write down these equilibria as
they are in the Data Booklet, that is, in terms of reduction.
Next, consider which of the two equations has the more positive electrode potential. This gives you the species
which is likely to undergo reduction (This species is on the right hand side of the equation). This equation is
therefore your reduction half equation.
Now reverse the second equation, that is, the one with the less positive (or, more negative) electrode potential.
This gives you the oxidation half equation, and also brings reactants to the same side. Do not forget that when
you reverse an equation, you reverse the sign of its electrode potential as well.


Balance the number of electrons and add the two half equations vertically. The electrode potentials are also added
vertically, but take care that you have reversed the less positive electrode potential before adding. Also check that
you have the correct reactants on the reactant side (For example, MnO4-(aq) + 5Fe2+ (aq) , not MnO4-(aq) +
5Fe3+).
The value you obtain after adding the two electrode potentials is the cell potential. If it is positive, then the
reaction is feasible under standard conditions. If the value is negative, then the reaction is not energetically
feasible under standard conditions. However, the reaction might still take place if reaction conditions are altered,
for example by heating or using concentrated solutions, rather than 1.o moldm-3 solutions.
Example 2
Predict if an aqueous solution of Fe (II) can convert aqueous potassium dichromate to Cr 3+(aq).
Solution
Reduction potentials as extracted from the Data Booklet:
Eθ/V
Cr2O72-(aq)
+
14H+
(aq) +
6e-
Ý
2Cr3+
+ 7H2O (l)
Fe3+ (aq) + e- Ý Fe2+(aq)
+1.33
+0.77
Reverse the (second) equation with the less positive electrode potential and the sign of the electrode potential,
balance electrons and then add the two equations vertically:
Cr2O72-(aq) + 14H+ (aq) + 6e- Ý 2Cr3+ + 7H2O (l)
6Fe2+ (aq) Ý 6Fe3+ (aq) + 6eNet reaction Cr2O72-(aq) + 14H+ (aq) + 6Fe2+ (aq) Ý 2Cr3+ + 7H2O (l) + 6Fe3+ (aq)
+1.33
-0.77
+0.56V
Since the net cell potential is positive, the reaction does take place. Potassium dichromate, just like KMnO 4,
oxidizes light green Fe2+ (aq) to yellowish brown Fe3+ (aq). Meanwhile, the orange dichromate is reduced to
dark green Cr3+ (aq).
Page
188
Oxidizing power of the oxoanions
MnO4-, Cr2O72-, ClO3- and VO3- are example of oxoanions. They are all rich in oxygen and contain an element
in a high oxidation state, which can therefore be reduced to lower oxidation states by suitable reducing
agents. The oxoanions are therefore good oxidizing agents; for example, both KMnO 4 and k2Cr2O7 can
oxidize Fe2+ to Fe3+ in aqueous solution. As oxidizing agents, the oxoanions usually require an acidic
medium. The role of the acid is to remove the oxygen atoms from the oxidizing agent, converting them to
water. In other words, the oxidizing agent loses oxygen, that is, it is reduced. Reduction of the oxoanion
therefore involves both the loss of oxygen and the gain of electrons by an element in the oxoanion. For
example, when Cr2O72- is reduced to Cr3+, each Cr atom in Cr2O72- gains 3 electrons, since the oxidation
number of Cr changes from +6 to +3. 1 mole of Cr2O72- therefore gains 6moles of electrons.
You can arrange oxidizing agents in order of their strength by comparing their reduction potentials. A
powerful oxidizing agent has a large and positive reduction potential. A less positive (more negative)
reduction potential shows that the substance is a weaker oxidizing agent. We can therefore conclude that
the permanganate ion, MnO4- , whose reduction potential is + 1.52, is a stronger oxidizing agent than the
dichromate ion, Cr2O72-, whose reduction potential is +1.33V.
The issue of spectator ions
Consider the reaction that occurs between dichromate and iron (II) ions.
Cr2O72-(aq) + 14H+ (aq) + 6Fe2+ (aq) Ý 2Cr3+ + 7H2O (l) + 6Fe3+ (aq)
Notice that this is an ionic equation in which spectator ions have been ignored. The full equation may be
written depending on which spectator ions are present.
Q
Construct a balanced full equation for the aqueous reaction between K2Cr2O7 and iron (II)
sulphate in the presence of sulphuric acid.
Redox titrations
A redox titration involves a redox reaction which takes place in aqueous solution. This procedure can be
used to determine the amount (moles, mass or concentration) of a reducing or oxidizing agent involved in
a redox reaction. Some redox titrations are self indicating, that is, they involve clear colour changes at the
end point. An indicator is not necessary in such titrations.
Example
During a titration, 25 cm3 of 0.1 moldm-3 KMnO4- were required to exactly neutralize 20.00 cm3 of
an aqueous solution containing Fe2+.
(a) Which reagent was placed in the burette?
(b) Determine the concentration of Fe2+ in the solution
(c) Explain why it is possible to determine the end point of the reaction without using an indicator.
Solution
(a) KMnO4. The reagent whose concentration is known is always placed in the burette.
Page
189
(b)
MnO4- (aq) + 8H+ (aq) + 5Fe2+ Ý Mn2+ + 5Fe3+ + 4H2O ... (i)
moles of MnO4- that reacted = C x V = 0.1 x
25
1000
= 0.0025
MnO4- : Fe2+ = 1 : 5 (equation (i))
∴ moles of Fe2+ that reacted = 0.0025 x 5 = 0.0125
Concentration of Fe2+ = n/V = 0.0125/0.020 = 0.625 moldm-3
(c) The reaction involves decolourization of potassium permanganate. The end point is reached when one
last drop of KMnO4 from the burette causes the colour of the titration mixture to change to pale pink
(colour of very dilute KMnO4)
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
6.2in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Predict
whether the following pairs of substances can react in a redox reaction. Where a reaction takes
place,
 construct a balanced equation and calculate the cell potential.
 identify the oxidizing and the reducing agent
1. KMnO4 and HCl (that is, MnO4- and Cl-)
2. KMnO4 and HBr-
3. K2Cr2O7 and HCl
4. VO2+ and Zn
5. Cu2+ and Mg
6. Cu2+ and Ag
7. Cl2 and Fe2+
8. Br2 and Fe2+
9. I2 and Fe2+
10. Na2S2O3 and I2
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
6.7
The redox series
This is an arrangement of electrode potentials, starting with the most positive and ending with the most
negative. A short form of the redox series is shown in Table 6.3.
Electrode reaction
Page
190
F2 + 2e- Ý 2F-
Eθ/V
+2.87
S2O82- + 2e- Ý 2SO42-
+2.01
H2O2 + 2H+ + 2e- Ý 2H2O
+1.77
MnO4- + 8H+ + 5e- Ý Mn2+ + 4H2O
+1.52
PbO2 + 4H+ + 2e- Ý Pb2+ + 2H2O
+1.47
Cl2 + 2e- Ý 2Cl-
+1.36
Cr2O72- + 14H+ + 6e- Ý 2Cr3+ + 7H2O
+1.33
Br2 + 2e- Ý 2Br-
+1.07
Ag+ + e- Ý Ag
+0.80
Fe3+ + e- Ý Fe2+
+0.77
I2 + 2e- Ý 2I-
+0.54
O2 + 2H2O + 4e- Ý 4OH-
+o.40
Cu2+ + 2e- Ý Cu
+0.34
Sn4+ + 2e- Ý Sn2+
+0.15
2H+ + 2e- Ý H2
+0.00
Pb2+ + 2e- Ý Pb
-0.13
Sn2+ + 2e- Ý Sn
-0.14
Fe2+ + 2e- Ý Fe
-0.44
Zn2+ + 2e- Ý Zn
-0.76
Mg2+ + 2e- Ý Mg
-2.38
Ca2+ + 2e- Ý Ca
-2.87
K+ + e- Ý K
-2.92
Table 6.3
The redox series
Interpretation of the redox series
A large positive electrode potential shows that the species on the left hand side of the equation is a good
oxidizing agent. For example, the electrode reaction
F2 + 2e Ý 2F-
Page
191
has an electrode potential of +2.87V. The large and positive value of Eθ shows that the forward reaction is
very energetically feasible. Thus, a large number of substances are able to convert fluorine to its ions. In
these reactions, F2 is reduced and it therefore acts as an oxidizing agent.
Less positive values of Eθ shows that the species on the left hand side is a weaker oxidizing agent. For
example, the electrode potentials of the halogens become less positive down the group, showing that their
oxidizing power decreases down the group.
A large negative electrode potential shows that the species on the right hand side is a good reducing
agent. This is the case for the groups (I) and (II) metals, for example,
Mg2+ + 2e- Ý Mg, Eθ/V = -2.38
If you read the electrode reaction in reverse, you have the oxidation reaction
Mg Ý Mg2+ + 2e-, Eθ/V = +2.38
The large and positive electrode potential shows that indeed, the oxidation of Mg is very likely. Mg reacts
with a large number of oxidizing agents, to form Mg 2+ compounds. Whilst Mg metal is being oxidized, it
acts as a reducing agent.
Predicting redox reactions using the redox series
Consider the two electrode equations below
Eθ/V
Cl2 + 2e- Ý 2ClBr2 +
2e-
Ý
2Br-
+1.36
+ 1.07
Both electrode potentials are positive, showing that the species on the left hand side are good oxidizing
agents. However, the first electrode potential is more positive and the second is more negative. This means
that chlorine is a better oxidizing agent than Br2. It also means that Br- is a better reducing agent than Cl-.
We would therefore expect a reaction to take place if we mixed Cl 2 and Br- ions. Chlorine would act as an
oxidizing agent and Br- would act as a reducing agent. In the reaction, chlorine would be reduced to Cl - but
Br- would be oxidized to Br2.
Eθ/V
Cl2 + 2e- Ý 2Cl2Br-
Ý Br2 +
+1.36
2e-
- 1.07
Cl2 + 2Br- Br2 + 2Cl-
+0.29
Note that the reduction equation involving bromine has been reversed to show the oxidation of Br- . The net
reaction has a positive Eθcell value, showing that the reaction is feasible. This is an example of a redox
displacement reaction. A halogen higher up in the redox series (higher up in the periodic table) will displace
the one below it from its solution. In the reaction above, chlorine displaces Br2 from Br- (aq). Br- ions are
thus removed from solution, and are replaced by Cl- ions.
Similarly, we would expect Br2 to convert I- (aq) to iodine. Meanwhile, Br2 is reduced to Br-(aq)
[Type a quote from the document or the summary of an interesting point. You can position the text
box6.3
anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Ex
text box.]
1. A part of the redox series involving transition metals is given below
Electrode
reaction
Fe3+ + e- Ý Fe2+
Eθ/V
+ 2e Ý Cu
+0.34
Ý
Cu+
+0.15
Fe2+ + 2e- Ý Fe
-0.44
Cu2+
+
e-
+0.77
(a) Cu+ and Cu+ (b) CuSO4(aq) and Fe (c) Cu+ and Fe
Page
192
Cu2+
Use the information in the table to predict if a reaction is possible
under standard conditions between the following pairs of substances.
Where a reaction occurs, construct a balanced equation and identify
the oxidizing and reducing agent.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Limitations of electrode potentials
(i) Effect of concentration
Standard electrode potentials refer to standard conditions of 298K, 1 atm and 1.0 moldm-3
solutions. Any predictions made using these conditions might differ from what is actually observed,
depending on the conditions employed. For example, if a reaction is predicted to be feasible because it has
a positive Eθcell value, in practice the reaction might not take place if very dilute solutions are used. Consider
the reaction between Cr2O72- and Fe2+. The two half equations and the net reaction are shown below. The
Eθcell shows that the reaction is feasible under standard conditions.
Eθ/V
Cr2O72-(aq) + 14H+ (aq) + 6e- Ý 2Cr3+ + 7H2O (l)
+1.33
6Fe2+ (aq) Ý 6Fe3+ (aq) + 6e-
-0.77
Cr2O72-(aq) + 14H+ (aq) + 6Fe2+ (aq) Ý 2Cr3+ + 7H2O (l) + 6Fe3+ (aq)
+0.56
Suppose that a very dilute solution of Cr2O72- is used. The decrease in the concentration of Cr2O72- results in
the electrode potential of the first reaction being reduced from +1.33V. Suppose that the new value of the
electrode potential is 1.00V.
The Eθcell for the reaction then becomes 1.00 + (-0.77) = +0.23V
Note that the Ecell is now less positive than +0.56V, which would be the E θcell under standard conditions. A
less positive cell potential implies that the reaction becomes less feasible. In fact, the reaction will not take
place if the dilution reduces the cell potential to a negative value.
Now consider what would happen if the concentration of Fe2+ was reduced from 1.0 moldm-3. Reducing the
concentration of Fe2+ would make the electrode potential of the reversed reaction Fe2+ (aq) Ý Fe3+ (aq) +
e- more negative than -0.77V, for example, it may become -0.80V. In other words, the oxidation of Fe2+ to
Fe3+ becomes less likely . The cell potential for the reaction between Fe 2+ and Cr2O72- becomes 1.33+ (-0.80)
= +0.53V. Once more, the cell potential becomes less positive (that is, more negative) and the reaction
becomes less likely.
In some cases, electrode potentials tell us that a reaction will not take place, but you should remember that
the reaction might in fact take place if non standard conditions are used. Consider the reaction between
MnO2 and chloride ions. The relevant half equations are shown below.
Eθ/V
MnO2 + 4H+ + 2e- Ý Mn2+ + 2H2O
½ Cl2 + e- Ý Cl-
+1.23
+1.36
If MnO2 is to react with Cl- ions, then the second equation must be reversed to bring the reactants to the
same side.
Eθ/V
Page
193
MnO2 +
4H+ +
2e-
Ý
Mn2+
+ 2H2O
2Cl- Ý Cl2 + 2e-
+1.23
-1.36
MnO2 + 4H+ + 2Cl- Ý Mn2+ + 2H2O + Cl2 -0.13V
Since the cell potential is negative, we do not expect MnO 2 to oxidize chloride ions under standard
conditions. Now, consider what happens when concentrated HCl is used. The concentration of chloride ions
is then greater than 1.0 moldm-3 and this makes the electrode potential of the reaction
2Cl- Ý Cl2 + 2emore positive than -1.36V, that is, the reaction becomes more feasible. Suppose that the new electrode
potential changes from -1.36V to -1.30V. Similarly, an increase in the concentration of hydrogen ions would
make the electrode potential of the first reaction more positive than +1.23V, for example, it may become
+1.40V. The cell potential for the reaction between MnO 2 and Cl- then becomes 1.40-1.30
= +0.10V. The positive value shows that under these conditions, MnO 2 converts Cl- ions to chlorine. In fact,
this is the method used to synthesize chlorine in the lab.
(ii) Electrode potentials and reaction rates
One limitation of electrode potentials is that they fail to predict the speed of a redox reaction. Consider the
reaction between peroxodisulphate (S2O82-) ions and iodide ions.
The relevant electrode potentials are given below.
Eθ/V
S2O82-+ 2e– Ý 2SO42–
I2 +
2e–
Ý
2I–
+2.01
+0.54
Reversing the second electrode potential and combining the two half equations gives
S2O82- + I2 Ý 2SO42– + 2I-
Eθcell = +1.47 V
The large and positive Eθcell value shows that the reaction is energetically feasible. Peroxodisulphate ions
are expected to oxidize iodide ions to iodine. Meanwhile, the peroxodisulphate ions are reduced to sulphate
ions. In practice, this reaction is so slow that for practical considerations we may say it does not take place
at all. The reaction has a very high activation energy due to the fact that the reacting ions have the same
charge, so they naturally repel.
Electrode potentials often predict that a reaction is feasible, but they do not tell us if in practice the
reaction can actually takes place.
6.8
Electrochemical cells
Page
194
An electrochemical cell is a device that contains chemicals which react through reduction and oxidation to
produce a voltage. This voltage can be used to induce a flow of electrons (current) in the external circuit. A
simple electrochemical cell can easily be set up by connecting two half cells by means of a salt bridge and
an external circuit. A wet cell consists of solid electrodes immersed in a liquid. Fig 6.8 shows a wet cell
made up of a Zn(s)/Zn2+ (aq) half cell and a Cu(s)/Cu2+ half cell. This cell, known as the Danielli Cell, uses
Cu and Zn electrodes.
The relevant reduction potentials are
Eθ/V
Cu2+ (aq) + 2e- Ý Cu(s)
Zn2+
(aq) +
2e-
Ý Zn(s)
+0.34
(i)
-0.76 (ii)
Since copper has the more positive electrode potential we expect Cu2+ ions to be reduced to Cu metal.
Reaction (i) is therefore the reduction half equation. Reversing reaction (ii) gives us the oxidation half
equation
Zn(s) Ý Zn2+ (aq) + 2e

+0.76V
Zinc is therefore the one which loses electrons, that is, it undergoes oxidation.
The metal with the more negative reduction potential (before reversing) undergoes oxidation.
This is because after reversing the equation, the oxidation reaction becomes more positive, that
is, more feasible.

Zinc is therefore the source of electrons in this cell. The electrons travel in the external circuit in
the direction shown by the arrows in Fig 6.8. The electrons are conducted into the Cu2+ solution,
leading to reduction of the Cu2+ ions to copper atoms.
Cu2+ (aq) + 2e- Ý Cu(s) Eθ/V = +0.34.

The zinc electrode thus dissolves as it releases Zn2+ ions into solution. Meanwhile, the copper
electrode gains mass as atoms of copper formed from Cu2+ stick on it.The overall cell reaction and
the cell potential can be worked out.
Eθ/V
reduction half equation
Cu2+ (aq) + 2e- Ý Cu(s)
+0.34
oxidation half equation
Zn(s) Ý Zn2+ (aq) + 2e-
+0.76 (ii)
Page
195
Cu2+ (aq) + Zn(s)  Cu(s) + Zn2+ (aq)
Fig 6.8

+1.10
(i)
(iii)
A Danielli cell
The net equation (iii) shows the reduction of Cu2+ to Cu metal by zinc. Meanwhile, the zinc metal
dissolves as it is oxidized to Zn2+. The cell potential of the electrochemical cell is +1.1oV. This
voltage pushes electrons into the external circuit. This flow of electrons (from Zn to Cu), known
as a current can be put to use, for example, to light an electric lamp.

All electrochemical cells work using the same principle. There are two half cells. In one half cell,
an electrode undergoes oxidation and in the process provides electrons, which flow to the other
half cell. In the other half cell, the electrons from the oxidized electrode cause a reduction reaction
to take place. A net redox reaction occurs, involving a reducing agent and an oxidizing agent. In
the Cu/Zn cell above, the reducing agent is Zn metal and the oxidizing agent is Cu 2+.
Electrochemical cells and the reactivity series
You probably recall that a metal higher up in the reactivity series has a large and negative reduction
potential. Such a metal is easily oxidized, that is, it is a good reducing agent. It will easily reduce the ions of
a metal lower than it in the electrochemical series. If you study reaction (iii) you will notice that Zinc, which
is higher than Cu in the electrochemical series, reduces Cu2+ ions to Cu metal. This is an example of a redox
displacement reaction. Cu2+ ions are displaced from solution (as cu metal) by Zn. At the end of the reaction,
Zn2+ ions have taken the place of Cu2+ ions in solution.
The reader should realize that such a redox displacement reaction can easily be carried out in the lab without
setting up an electrochemical cell. For example, when a Zn rod is dipped into a solution of CuSO4, a reaction
occurs between zinc metal and Cu2+ ions. The following observations are made:
 the zinc rod dissolves and is soon coated pink
 the blue colour of CuSO4 fades
Zinc metal reduces Cu2+ ions to copper atoms, which stick onto the Zn rod, forming a pink coat. Meanwhile,
Zn metal is oxidized to Zn2+. There is a transfer of electrons from the zinc metal to the Cu 2+ ions. In an
electrochemical cell, the reducing agent (Zn) is separated from the oxidizing agent (Cu2+) by half cells. This
allows electrons from the reducing agent to flow in an external circuit to the oxidizing agent. These electrons
can be employed to do useful work, for example, lighting a bulb, before they reach the oxidizing agent.
Q
Predict if a redox displacement is possible when the following substances are
mixed. Where a reaction occurs
(i)
Construct a balanced equation
(ii)
Calculate the Eθcell
(iii)
State the oxidizing agent and the reducing agent.
(a) Cu2+(aq) and Mg(s)
(b) Cu2+(aq) and Ag(s)
Page
196
(c) Zn2+(aq) and Mg(s)
A
A metal higher up in the electrochemical series will displace the ions of a metal which is
lower from its solution.
(a) Mg, being higher than copper in the electrochemical series, will displace Cu 2+ ions
from solution. In other words, Mg is more reactive than Cu, and Mg 2+ ions are more
stable than Mg metal, so the conversion Mg Mg2+ is favourable. On the other hand,
Cu metal is more stable than Cu2+. In the presence of Mg metal, Cu2+ ions are
converted (reduced) to Cu metal.
Cu2+ (aq) + Mg(s)  Cu2+(s) + Mg2+ (aq)
Eθcell =+1.10V
The positive value of the cell potential shows that the reaction is energetically feasible
under standard conditions.
Questions (b) and (c) are left as an exercise to the reader.
Anode and cathode in an electrochemical cell
In Fig 6.8, zinc metal undergoes oxidation, so it becomes the source of electrons. Copper metal, being less
willing to release electrons, becomes relatively positively charged. Electrons therefore flow from the zinc
electrode to the relatively positive copper electrode. Which then is the anode and cathode in this
electrochemical cell? There is a strong temptation to say that zinc is the cathode (since it is relatively
negatively charged compared with Cu) and Cu is the anode. This is incorrect.
Zinc is the anode and copper is the cathode. The confusion for some students comes from the way they
define the terms ‘cathode’ and ‘anode’. It is frequently taught in juniour chemistry that an anode is positively
charged and a cathode is negatively charged. Although this definition appears to work well in juniour
chemistry, it presents students with unnecessary difficulties at ‘A’ level. Here is the proper definition of the
terms ‘cathode’ and ‘anode’.
An anode is an electrode where oxidation takes place. In electrochemical cells, the anode
is the negative electrode.
A cathode is an electrode where reduction takes place. In an electrochemical cell, the
cathode is the positive electrode.
In an electrochemical cell, the metal with the more negative reduction potential (as taken from the Data
Booklet) undergoes oxidation. It is the anode. The electrode with the more positive reduction potential
provides a surface where reduction takes place. It is the cathode.
Page
197
More about the salt bridge
It was mentioned in an earlier discussion that the salt bridge has two functions; completing the circuit
between two half cells and maintaining electroneutrality (preventing a build up of charges which would
cause the electrochemical cell to stop working).
The salt bridge contains an aqueous solution of a salt, for example, KNO 3. Consider what happens when the
Zn/Cu cell (Fig 6.8) is working. At the anode, Zinc metal dissolves, increasing the concentration of Zn 2+
ions in solution. A build up of positive ions will eventually prevent the zinc from dissolving any further. That
means the zinc stops supplying electrons, and the cell stops working. The salt bridge supplies negative ions
(NO3-) to neutralize the excess positive charge from Zn2+ ions. NO3- ions therefore move from the salt bridge
to the Zn2+ solution. At the cathode, Cu2+ ions are removed from solution as they form copper metal.
Removal of positive charges makes the solution relatively negatively charged due to the presence of an
excess of SO42- ions. The build up of negative charge reduces the tendency of electrons, which are also
negatively charged, to enter the solution in this half cell. This stops the reduction reaction at the cathode.
The salt bridge provides positive ions (K+) that counter the excess negative charge. K+ ions and NO3- ions
therefore move in opposite directions in the salt bridge, and this movement transports electric charge, thus
completing the circuit between the two half cells.
The shorthand notation for describing electrochemical cells
The Danielli cell is made up of two half cells. The Zn/Zn2+ and the Cu/Cu2+ half cells.
The full cell can be shown using a shorthand notation.
Zn(s) ⃒ Zn2+(aq) ⃒⃒ Cu2+(aq) ⃒ Cu(s)




The electrodes, in this case Zn and Cu, are written on the far right and far left of the notation.
A single vertical line (⃒ ) represents the junction between an electrode and its ions.
Double lines, broken or full, ( ⃒⃒ ) represent the salt bridge.
The electrode with the more negative electrode potential, that is, the anode, is written at the left
hand - side cell. The one with the more positive electrode potential (cathode) is shown on the right
hand side.
Consider the electrochemical cell whose cell notation is
Al(s) ⃒ Al3+(aq) ⃒⃒ Pb2+(aq) ⃒ Pb(s)
The anode is Al whilst the cathode is Pb. The electrode on the left loses and supplies electrons in the external
circuit, that is, it is oxidized
Al(s)  Al3+ (aq) + 3e- Eθ = +1.66V ... (i)
(after reversing the reduction potential)
Reduction occurs at the cathode
Pb2+(aq) + 2e  Pb(s)
Eθ = -0.13V ...(ii)
Multiplying (i) by 2 and equation (ii) by 3 to balance the electrons, and then adding the two equations
together gives the net reaction
2Al(s) + 3Pb2+(aq)  2Al3+(aq) + 3Pb(s) ...(iii)
For this reaction, Eθcell = +1.66 + (-0.13) = +1.53V
The relationship Eθ cell = EθR - EθL
The cell potential of an electrochemical cell can be obtained by using the formula
Page
198
Eθ cell = EθR - EθL,
where EθR is the reduction potential of the right hand side cell and EθL is the reduction potential of the left
hand side cell. For example, the cell potential for reaction (iii) above is -0.13 -(-1.66) = +1.53
This is a useful relationship, but it may cause unnecessary difficulties if the following points are overlooked.

This relationship is only valid in assessing feasibility of the reaction if the anode (more negative
reduction potential, as given in the Data Booklet) is on the left hand side and the cathode is on the
right hand side. This is the case with the cell
Al(s) ⃒ Al3+(aq) ⃒⃒ Pb2+(aq) ⃒ Pb(s)


When using this formula, both reduction potentials are taken as they are in the Data Booklet, that
is, no reduction potential should be reversed. For the cell above, the reduction potential for the left
cell (Al/Al3+) is -1.66V and that of the right cell (Pb/Pb2+) is -0.34.
This relationship is often used to predict the direction of electron flow, not necessarily feasibility.
Consider the following problem from a past paper
Chlorine gas and iron (II) ions react together in aqueous solution as follows.
Cl2 + 2Fe2+  2Cl– + 2Fe3+
The following diagram shows the apparatus needed to measure the E cell for the above
reaction.
Use the Data Booklet to calculate the Eθcell for this reaction, and hence decide which
direction (left to right, or right to left) electrons would flow through the voltmeter V
when switch S is closed.
[UCLES 9701/J2008 paper 4 .q1]
Page
199
The relevant reduction potentials from the Data Booklet are
Eθ/V
Cl2 + 2e- Ý 2Cl-
+1.36
Fe3+ + e- Ý Fe2+
+0.77
If we use the convention ‘left cell is the more negative and right cell is the more positive’, then
Eθ cell = EθR - EθL = 1.36 - 0.77 = +o.59V
Whilst this value indicates that the reaction is feasible, this is not the answer examiners expected. The
question clearly required students to interpret the diagram as drawn, that is, the left cell is Cl 2/Cl- (more
positive) and the right one is Pt/Fe2+, Fe3+.
The Eθcell is therefore 0.77-1.36 = -0.59V.
The negative cell potential in this case does not mean the reaction is not feasible. It is simply a result of
which cell is on the right and which one is on the left. The first value (+0.59V) indicates that the reaction is
feasible and that electrons flow from the anode (oxidation) to the cathode (reduction). The negative value 0.59V indicates that the convention ‘left cell is the more negative and right cell is the more positive’ has
not been used since in the diagram the anode (more negative) is the right cell. Electrons therefore flow
from the right to the left.
Do not forget
 that the anode has the more negative(less positive) cell potential. Oxidation takes place here, releasing electrons.
 Consequently, electrons will always flow from the anode to the cathode. The actual direction in
which electrons flow in an electrochemical cell depends on which side , right or left, the anode is.
Q
An electrochemical cell can be set up in which the overall reaction is
Mg(s) + Cl2 (g)  Mg2+ (aq) + 2Cl-(aq) ... (I)
(a)
By means of a Data Booklet, write down two equations that can be used to construct
equation (I).
(b)
Page
200
A
Hence
(i)
Construct a fully labelled diagram of the electrochemical cell represented by
equation (I)
(ii)
Show the short-hand notation of the cell and calculate the cell potential of
equation (I)
(a)
(b)(i)
Cl2(g) + 2e- Ý 2Cl- (aq)
Eθ/V
+1.36
Mg2+ + 2e- Ý Mg(s)
-2.38
One half cell (oxidation) should contain magnesium metal dipped in an aqueous
solution of Mg2+(for example, MgSO4(aq) ). The other half cell (reduction) should have
chlorine gas passing over a platinum electrode in a solution containing Cl - ions. The
two half cells are connected by an external circuit and a salt bridge.
(ii)
Mg(s) ⃒ Mg2+ (aq) ‖ Cl-(aq), Cl2(g)⃒ Pt(s)
Eθ cell = EθR - EθL = 1.36 -(-2.38) = +3.74V
[Type a quote from the document or the summary of an interesting point. You can position the text
Exercise
box anywhere
6.4in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
1. Construct a fully labelled diagram for the electrochemical cell represented by the short-hand
notation
Mg(s) ⃒ Mg2+(aq) ⃒⃒ Ag+(aq) ⃒Ag(s)
(a) Indicate by means of arrows on the diagram the direction of electron flow.
(b) Construct a net equation for the cell reaction and by means of a Data Booklet calculate the
cell potential
2.
An electrochemical cell is set up to determine the voltage of an electrochemical cell that involves
chlorine gas and magnesium metal ( Fig 6.4.1 ).
(a) Labell the diagram as completely as possible.
(b)
Show the short-hand notation for this cell.
(b)
Construct a net equation for the cell reaction and calculate the
cell potential
Fig 6.4.1
Page
201
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
6.8.1
Practical electrochemical cells
text box.]
The Danielli cell is quite useful in demonstrating the principles behind an electrochemical cell. Its major
disadvantage is its cumbersome nature which limits its portability.
Commercial electrochemical cells, known as batteries, have been designed to be a convenient source of
energy. These include the dry cell, the lead-acid battery and the fuel cell.
1 The dry cell
The most common dry cell is the Zinc-Carbon cell. The cell is said to be dry because it does not contain
liquids. Fig 6.9 shows a typical dry cell.
Fig 6.9 A Zinc - Carbon
dry cell
At the anode, oxidation of zinc metal takes place
Zn(s)  Zn2+ (aq) + 2e-
Eθ/V = +0.76V ... (i)
At the cathode, NH4+ ions are reduced to ammonia and hydrogen.
NH4+ (aq) + 2e-  2NH3 (aq) + H2 (g)
Eθ/V = +0.74 ... (ii)
The net reaction is therefore
Zn(s) + NH4+ (aq)  Zn2+ (aq) + 2NH3 (aq) + H2 (g)





Zn acts as a reducing agent. It reduces NH4+ ions to NH3.
NH4+ has a nitrogen atom in a high oxidation state (+4). NH 4+ can therefore act as an oxidizing
agent. In this reaction it oxidizes Zn metal to Zn2+.
As the cell works, the zinc casing becomes thinner as zinc metal is dissolved to Zn2+. This explains
why dry cells may eventually start to leak.
Reaction (ii) produces hydrogen which prevents further conversion of NH4+ ions to NH3 (Le
Chatelier’s principle). This inhibition of reaction (ii) causes the dry cell to stop working (the cell is
said to be polarized). To prevent this polarization, MnO2 is included in the cell. It oxidizes the
hydrogen to water. This explains why dry cells often ooze a liquid after they have been working for
a long time.
The dry cell produces a maximum voltage of 1.5V. Dry cells can be combined to give batteries with
the desired size of voltage, for example, connecting 3 cells in series gives a maximum voltage of
4.5V. Dry cells are not rechargeable. This is because reaction (ii) can not be reversed, since
hydrogen is lost from this reaction as water. The cells soon stop working, that is, they become flat.
Dry cells find widespread use because they are light and compact. They are used, for example, in
transistor radios and torches.
Page
202

Eθcell = 1.50V
Effect of size of the dry cell
Size of a dry cell does not affect the voltage out put, provided the chemical composition is the same. This is
because the reactions involved are the same, and these reactions take place under the same conditions.
However, smaller cells contain a smaller mass of reactants, so they become flat more quickly.
2. The lead-acid battery
This is a well known example of a secondary (rechargeable) battery. The anode (negative electrode) is a
plate of lead metal and the cathode is lead (IV) oxide, PbO2, present as a coat on a lead metal plate. Both
the anode and the cathode are immersed in an electrolyte of moderately concentrated sulphuric acid (Fig
6.10)
During discharge
At the cathode (positive terminal) PbO2 is reduced to Pb2+ by electrodes coming from the anode. The Pb2+
ions immediately combine with SO42- ions from H2SO4 to from white and insoluble PbSO4.
At the anode (negative terminal), the lead plates are oxidized to Pb 2+. Once more, the Pb2+ ions combine
with SO42- ions to form PbSO4. The two half equations are shown below
Eθ/V
cathode reduction half reaction
anode oxidation half reaction
PbO2(s) + 4H+ + 2e- Ý Pb2+ + 2H2O
Pb(s) Ý Pb2+ + 2e
PbO2(s) + 4H+ + Pb(s) Ý 2Pb2+ + 2H2O
+1.87 ...(i)
+0.13 ...(ii)
+2.00... (iii)
Page
203
Note that the reduction potential for equation (i) is more positive than the one quoted in the Data Booklet
(+1.47V). This is because the acid (H+) used in the battery is fairly concentrated and this drives the
equilibrium to the right, that is, the reaction becomes more favourable in the forward direction.

Under standard conditions, one cell in the lead-acid battery would produce a voltage of +1.34V. In
practice, each cell produces a maximum of 2V. This is because the conditions used are not standard.
Fairly concentrated sulphuric acid is used, and this makes the reduction potential for reaction (i)
more positive than +1.47V.

In practice, the lead-acid battery usually contains six cells, each producing a maximum voltage of
2.00V. The maximum voltage that the battery can produce is therefore 12.00V.

The net cell reaction (reaction (iii)) shows PbO 2 acting as an oxidizing agent and Pb metal as a
reducing agent. This reaction is expected, and does not take place only in the lead-acid battery. It
can be carried out in the lab by mixing PbO2, Pb and H2SO4. Compounds of lead are more stable in
their +2 oxidation states. Lead (IV) compounds are unstable relative to the +2 state. Thus PbO 2 is
easily reduced to Pb2+. Meanwhile, it acts as an oxidizing agent. In this case, it oxidizes Pb (which,
like most metals, is a reducing agent) to Pb2+. The Pb2+ ions immediately combine with sulphate
ions from H2SO4 to form a white precipitate of PbSO4.
Fig 6.10 The lead - acid accumulator (‘car’ battery)
During charging
The major advantage of the lead-acid battery is that it can be recharged. This is done by passing an electric
current from an external source into the battery. In vehicles, the charging process is automated. As the
engine runs, it drives a dynamo or alternator, which in turn generates an electric current. This current is
then passed into the battery. The flow of electrons in to the battery essentially reverses the anode and
cathode half reactions. Half of the Pb 2+ produced in reaction (iii) are reduced to Pb metal, which is
deposited on the negative (anode) lead metal plate. The rest of the Pb 2+ are oxidized to PbO2(s) which forms
a deposit on the positive (cathode) lead metal plate.
Eθ/V
Pb2+ + 2e Ý Pb(s)
Page
204
Pb2+
+ 2H2O Ý PbO2(s) +
-0.13
4H+
+
2e-
-1.87
These reactions have negative potentials and so will not take place under normal conditions. The processes
are actually driven by electrical energy from an external source.
The lead-acid battery: key facts at a
glance
Anode (negative terminal): Zinc plate
Cathode (positive terminal): PbO2(s)
coated on lead plates
Electrolyte: H2SO4
Charging
Pb2+ is converted to PbO2 (cathode) and to Pb
metal (anode)
Cell Notation
Pb(s)⃒ Pb2+(aq) ⃒⃒ Pb2+(aq) ⃒PbO2(s)
The anode and cathode are shown in bold.
Discharge reaction:
Cathode reaction (reduction) : PbO2 reduced to
Pb2+
Anode reaction (oxidation) : Pb oxidized to Pb2+
Cell potential: +2.00V. Total for 6 cells
= 12.00V
Advantage: rechargeable.
Disadvantages :
(i) Heavy due to the presence of lead
and sulphuric acid.
(ii) Needs careful disposal because lead
compounds are poisonous.
3. The fuel cell
A fuel cell generates electrical energy from the net redox reaction between oxygen and hydrogen
gases
2H₂ (g) + O₂ (g) → 2H₂O (l)
Eθ = +1.23V ... (i)
The hydrogen and oxygen gases are fed into the cell through different outlets from an external source (Fig
6.11). In this reaction, hydrogen is used as a fuel. Notice that this reaction is comparable to the exothermic
combustion of hydrogen in oxygen. The key fact is that in an electrochemical cell, the aim is not to produce
heat energy but electrical energy.
Use of an electrolyte
An alkali or an acid can be used as an electrolyte. Fig 6.11 shows an alkaline fuel cell.
graphite
Page
205
graphite
Fig 6.11
An alkaline hydrogen fuel cell
The half equations
Equation (i), like all redox reactions, can be resolved into a reduction half equation and an oxidation half
equation. At the anode, catalytic oxidation of hydrogen gas occurs to produce H+ ions.
H₂ (g) ⇌ 2H⁺ + 2e⁻ (ii)
The anode (negative electrode) is made of porous graphite which is coated with a catalyst of nickel and
nickel (II) oxide. The graphite electrode provides an inert surface on which hydrogen is oxidized,
catalyzed by nickel and nickel (II) oxide.
Apart from acting as an electrolyte, the alkali supplies OH⁻ ions which remove the H⁺ ions produced in
reaction (ii).
2OH⁻ (aq) + 2H⁺ (aq) → 2H₂O ... (iii)
This speeds up the anode reaction by pushing equilibrium (ii) to the right, that is, more hydrogen gas
molecules are oxidized to replace the OH⁻ ions that are lost from the equilibrium as water (Le Chatelier’s
principle).
By combining equations (ii) and (iii) we obtain the half equation for the oxidation of hydrogen gas at the
anode in the presence of an alkali (OH⁻ ions). Note that this oxidation half equation actually involves two
types of reactions.
H₂ (g) ⇌ 2H⁺ + 2e⁻
2OH⁻ (aq) + 2H⁺ (aq) → H₂O
Anode half equation:
... (ii) [oxidation]
... (iii)
[neutralization]
H₂ (g) + 2OH⁻ (aq) → 2H₂O (l) + 2e Eθ = +0.83V ... (iv)
Note that the oxidation of hydrogen gas in the absence of OH⁻ ions (reaction ii) would give an Eθ value of
zero. The presence of OH⁻ ions therefore makes the Eθ value more positive, and the oxidation half reaction
more feasible.
The anode reaction produces electrons which then travel in the external circuit towards the cathode. This
flow of electrons can be used to operate a device (the load).
The reduction half equation (cathode)
An O atom is reduced by gaining two electrons which come from the anode via an external circuit.
O + 2e → O2⁻
Since there are two oxygen atoms in a molecule of oxygen, a total of four electrons would be gained per
molecule of oxygen
O₂ + 4e⁻ ⇌ 2O2⁻
... (v)
The oxide ions are too unstable to have an independent existence. They rapidly react with water, to produce
OH⁻ ions. This reaction can be understood in this way: the negatively charged oxide ion (a strong base) is
attracted to a relatively positively charged hydrogen atom in water (water behaves as an acid). It then
extracts this hydrogen atom, resulting in the formation of two OH⁻ ions.
Page
206
O2⁻ + H₂O → 2OH⁻ ... (vi)
δ+
δδ+
2OH-
By combining equations (v) and (vi), we obtain the reduction (cathode) half equation.
O₂ + 4e⁻ ⇌ 2O2⁻
2O2⁻ + 2H₂O Ý 4OH⁻
Reduction half equation:
O₂ (g) + 4e⁻ + 2H₂O (l) → 4OH⁻ (aq) Eθ = +0.40V ... (vii)
The cathode is porous graphite coated with nickel which catalyzes the reduction of oxygen.
The OH⁻ ions generated at the cathode continuously diffuse through the cell to the anode, where they aid
the oxidation of hydrogen .
We now combine equation (iv) (anode half reaction) and equation (vii) (cathode half reaction) to obtain
the cell reaction that takes place in the fuel cell (Equation (iv) is multiplied by 2 to balance the number of
electrons).
Eθ/V
+0.83
anode
2H₂ (g) + 4OH⁻ (aq) Ý 4H₂O (l) + 4e
cathode
O₂ (g) + 4e⁻ + 2H₂O (l) Ý 4OH⁻ (aq)
+0.40
2H₂ (g) + O₂ (g) Ý 2H₂O (l)
+1.23
Net:
... (iv)
... (vii)
Notice that the same amount of OH- ions that are used at the negative electrode are produced at the positive
electrode. The concentration of electrolyte is therefore expected to remain constant. However,
concentration of NaOH is reduced due to the formation of water. To reduce the dilution problem, water is
drained out of the cell as soon as it is formed.
Use of an acid electrolyte
It is possible to use an acid as an electrolyte. At the anode, hydrogen is oxidized
H₂ (g) ⇌ 2H⁺ (aq) + 2e⁻ ...
Eθ/V = +0. 00V
... (viii)
At the cathode, oxygen is reduced
O₂ (g) + 4H⁺ (g) +4e⁻ ⇌ 2H2O (l)
Eθ/V = +1.23V ... (ix)
Notice that in acidic solution we have selected those half equations which involve hydrogen ions.Similarly,
for an alkaline electrolyte, we used those half equations which involve hydroxyl ions. Combining
equations (viii) and (ix), we obtain the overall equation for the fuel cell
Page
207
Net equation:
2H₂ (g) ⇌ 4H⁺ (aq) + 4e⁻
O₂ (g) + 4H⁺ (g) +4e⁻ ⇌ 2H₂O (l)
Eθ/V
0. 00V
+1.23V
2H₂ (g) + O₂ (g) →2H₂O (l)
+1.23V
We obtain the same equation as before. The Eθ cell is not changed by changing the electrolyte. Overally,
hydrogen ions consumed at the cathode are generated at the anode. Concentration of electrolyte therefore
remains constant. In view of this, it is safe to assume that alkali or acid acts as a catalyst. It speeds up the
reaction between oxygen and hydrogen, it is regenerated at the end of the reaction, it does not change nature
of the product, and it does not change the Eθcell of the reaction.
Question and answer section : The fuel cell
Q. Why is a fuel cell known by this name?
A. The overall reaction in the fuel cell is 2H₂ (g) + O₂ (g) →2H₂O. From this reaction, hydrogen acts like a fuel
because it is oxidized by oxygen to produce (electrical) energy.
Q. A fuel cell may use NaOH or KOH solutions as the electrolyte. What is the advantage of KOH over NaOH?
A. In both cases, there is a carbonate problem in which OH⁻ ions neutralize acidic CO₂ from the atmosphere to
form a carbonate.
OH⁻ (aq) + CO₂ (g) → H⁺ (aq) + CO₃2-(aq)
This results in the precipitation of carbonates, for example, Na2CO3 if NaOH is used. These carbonates may
poison the nickel catalyst and clog the cell. KOH is preferred as the electrolyte because potassium carbonate has
a better solubility than sodium carbonate.
Q. The voltage obtained by using an acid and an alkaline electrolyte is the same. Why then is an alkali preferred as
the electrolyte?
A. An alkaline electrolyte has more advantages than an acid catalyst.



It allows for a wide choice of catalysts.
Alkaline systems work well at room temperature.
The cell and electrodes can be built from low cost carbon and plastics.
Q. What are the possible advantages and disadvantages of the hydrogen fuel cell?
Page
208
A. The advantages include

It can supply a continuous current, which is suitable in devices which require a steady and reliable source of
current, for example, in a spaceship.

It is one of the greenest sources of energy to be developed. There is no pollution associated with it, since the
sole product of the reaction is water.
The main disadvantages include

The development of a fuel cell is rather expensive. This has limited the widespread use of fuel cells.

The storage of hydrogen still presents a problem because of its explosive nature. Effective methods of storing
hydrogen before it is fed into the fuel cell have been found, but more research is still needed to improve them
and make their use widespread.

The fuel cell is large and cumbersome. Once more, this prevents it from being a versatile source of electrical
energy.

The source of hydrogen gas is still a challenge since it is not found in air. Producing hydrogen by the
electrolysis of water is not a solution to the energy problem since the process consumes huge quantities of
electrical energy. A method has been developed for producing hydrogen gas from methane, but the demands
of the process makes its application limited.
6.9
Electrolysis
Electrolysis is a process whereby an electric current is used to decompose a compound into one or more
elements. An electrolytic cell is essentially the reverse of an electrochemical cell, as illustrated in Table
6.4. Electrolytic cells differ in design and complexity, but they have the same basic components.



An external source of current, for example a battery
Electrodes. These provide surfaces on which reduction and oxidation reactions take place. The
positive electrode, connected to the positive terminal of the power source, is known as the anode.
This is where oxidation takes place, just as in electrochemical cells. The negative electrode is
connected to the negative terminal of the power source and is known as the cathode. This is where
reduction takes place. The anode is so named because it attracts anions (negative ions). Similarly,
the cathode attracts cations (positive ions).
An electrolyte. This is a liquid conductor which allows current to flow between the anode and
the cathode.
Electrolytic cell
Electrochemical cell
Takes in electric current from an external
source
Produces a current in an external circuit
Current from an external source drives
chemical reactions in the cell
Chemical reactions in the cell produce an
electric current.
The anode is the positive electrode and
the cathode is the negative electrode
The anode is the negative electrode and the
cathode is the positive electrode
Table 6.4
Comparison of electrolytic and electrochemical cell
6.9.1 Electrolysis of water
Fig 6.12 shows a simple set up for the electrolysis of water. During this process, water is decomposed into
its elements, that is H2 and O2.


Page
209

The water is acidified with a little sulphuric acid to improve its conductivity. It is also possible to
improve conductivity of water by adding a salt such as sodium sulphate.
A suitable indicator, for example, litmus, can be added to the water to demonstrate the changes in
pH that take place at each electrode during the electrolysis. However, for the indicator to give
reliable results, an acid or alkali should not be added to the water. Instead, a salt such as sodium
sulphate can be added to improve the conductivity of water. The advantage of sodium sulphate is
that it has no ions which can be discharged during the electrolysis process.
Any suitable material, for example graphite, can be used as the anode and the cathode. This
material must be chemically inert and it must be a good conductor of electricity.
At the anode
Water is oxidized to oxygen. The relevant equation for the oxidation of acidified water can be extracted
from the Data Booklet.
2H2O  O2 + 4H+ + 4e-
Eθ cell = -1.23 ... (i)
Here we have selected an equation involving H+ ions, since the water has been acidified. The hydrogen ions
produced at the anode are attracted to the negatively charged cathode.
At the cathode
H+ ions are reduced to hydrogen gas.
2H+ + 2e-  H2
Eθ cell = 0.00V ... (ii)
The net equation for the electrolysis of water is obtained by combining equations (i) and (ii).
2H2O  O2 + 4H+ + 4e+ 4e-  2H2
4H+
2H2O  O2 (g) + 2H2 (g)
Eθ/V
-1.23V ... (i)
0.00V ... (ii)
-1.23
... (iii)
The negative value indicates that the reaction is not
likely to take place spontaneously. It is not energetically
feasible, and energy must be supplied to drive it. In this
case, the reaction is driven by electrical energy from an
external source.
It is also possible to carry out the electrolysis in an
alkaline environment. Once more, the role of the alkali
is to improve the conductivity of water. The relevant half
equations for an alkaline environment are
Cathode (reduction): 4H2O + 4e−  H2 + 4OHAnode (oxidation): 4OH-  O2+ 2 H2O+ 4eAt the cathode, electrons from an external source
reduce water to form OH- ions which then drift to the
anode where they are oxidized. The net equation is
obtained by adding the two equations together and
subtracting 2H2O from both sides
Fig 6.12 Apparatus used to demonstrate
the electrolysis of water
Cathode (reduction)
Anode (oxidation)
4H2O + 4e−  2H2 + 4OH4OH-  O2+ 2 H2O+ 4e-
Page
210
2H2O  O2 + 2H2
Q
A
Explain why during the electrolysis of water, hydrogen and oxygen are produced in the
ratio of 2: 1
The formula of water shows that each molecule contains two hydrogen atoms and one oxygen
atom. Consequently, hydrogen and oxygen are produced in the ratio of 2:1 when water undergoes
electrolysis.
The electrolysis of aqueous sodium sulphate is essentially the electrolysis of pure water. The role of the
sodium sulphate is simply to improve the conductivity of water. The ions in sodium sulphate are not
discharged. Since the solution is neutral, the electrode reactions involve OH - and H+ ions from the autodissociation of water.
H2O Ý 2H+ + OHThe H+ ions are attracted to the cathode whey they are reduced to hydrogen.
2H+ + 2e-  H2
... (iii)
The removal of H+ ions from the solution around the cathode results in an accumulation of OH - ions. If red
litmus indicator had been added to the water, it will change to blue, showing that pH around the cathode
increases.
At the anode, OH- ions are oxidized.
4OH– O2 + 2H2O + 4e- ... (iv)
The removal of OH- around the anode increases concentration of H + ions, so if litmus indicator had been
added to the solution, it would turn red around the anode.
The net equation for the electrolysis of sodium sulphate solution is obtained by combining equations (iii)
and (iv)(after balancing number of electreons)
4H+ + 4OH–  2H2 + O2 + 2H2O
The left hand side of the reaction involves 4OH- and 4H+ ions, which is essentially 4H2O, so the reaction
can be rewritten as
4H2O  2H2 + O2 + 2H2O
and subtracting 2H2O from both sides gives
2H2O  2H2 + O2
The same net equation is obtained, irrespective of which electrolyte is used.
The industrial importance of the electrolysis of water
Electrolysis of water is a method that is used in many countries to produce hydrogen and oxygen gases. The
importance of the process comes from the fact that hydrogen can not be obtained from the fractional
distillation of air (there is very little hydrogen in the atmosphere).
The uses of oxygen include



Page
211

In hospitals for patience in ICU (Intensive Care Unit)
Mixed with acetylene (ethyne) for use in oxyacetylene torches. These are very hot flames used for
welding.
Used in hospitals to destroy bacteria that cause gangrene. These bacteria thrive in anoxic (oxygen
deficient) environments and are killed by oxygen.
Rockets use liquid oxygen to burn fuel and generate lift.
The uses of hydrogen include


in the manufacture of ammonia by the Haber-process.
In the manufacture of margarine.


In the manufacture of hydrochloric acid.
In the reduction of some metal ores.
Many countries are considering finding other sources of hydrogen. This is because the electrolysis of
water on an industrial scale consumes large quantities of electricity, which makes the process expensive. In
some countries, hydrogen is now being obtained from methane.
6.9.2
Electrolysis of aqueous salt solutions
The products of the electrolysis of a salt solution depend on


the concentration of the solution
the electrode potential of the ions involved
Consider the electrolysis of NaCl (aq). The ions present in the solution are shown in Table 6.5.
The cations (H+ and Na+) are attracted to the cathode. The large and negative value of the electrode potential
for Na shows that it can not be discharged in water. In other words, Na metal is so reactive that if it were to
be produced in water, it would instantly react with water to form NaOH. So at the cathode,
hydrogen is preferentially discharged.
2H+ + e- Ý H2
If the electrode potential of a metal is negative, the metal can not be formed from its ions in aqueous
solution by electrolysis
Table
6.5
Cations
Electrode Potentials/V
Anions
Electrode Potentials/V
Na+
Na+ + e- Ý Na
-2.71
Cl-
Cl2 + 2e- Ý 2Cl-(aq)
+1.36
H+
2H+ + 2e- Ý H2
-0.00
OH-
O2 + 2H2O + 4e- Ý 4OH-
+0.40
NB: OH- and H+ ions come from the auto-dissociation of water.
The anions (Cl- and OH-) are attracted to the anode. The discharge (oxidation) reactions for these ions are
the reverse of the reduction potentials given in the Data Booklet.
 Cl2 +
4OH-  O2 + 2H2O + 4e-
Page
212
2Cl-(aq)
2e-
Eθ/V
-1.36 ... (i)
-0.40 ... (ii)
Both processes have negative electrode potentials, showing that they are not likely to take place, unless if
energy is supplied, for example, in the form of an electric current. However, the electrode potential for the
second reaction is more positive, so we would expect O2 to be discharged in preference to Cl2. This is what
actually happens if the NaCl solution is dilute. However, if a concentrated solution of NaCl is used, chlorine
will be discharged instead of oxygen. Increasing the concentration of Cl - ions makes the electrode potential
of reaction (i) more positive than -1.36V. At a certain concentration of Cl - ions, the electrode potential
becomes more positive than the one for the discharge of O2. In that case, chlorine will be
discharged in preference to O2.
In the electrolysis of a dilute solution of NaCl, oxygen is discharged at the anode in preference to
chlorine. This is expected from a comparison of the relevant electrode potentials. However, when the solution is concentrated, the deviation from standard conditions implies that chlorine can be discharged instead of oxygen. Note that to easily understand the discharge of anions, you need to read their reduction
potentials in reverse.
The production of chlorine by the electrolysis of
concentrated aqueous NaCl (Brine)
Fig 6.13 shows the electrolytic cell for the manufacture of chlorine by the electrolysis of brine.
At the cathode, hydrogen gas is produced due to the reduction of H + ions from water. The anode attracts
OH- and Cl- ions, but in this case, it is Cl- ions that are discharged, as already explained.


This method of producing Chlorine is known as the Chlor-alkali process, because it produces chlorine as well as an alkali. The other product is hydrogen.
The cell used is known as the diaphragm cell because it makes use of a porous diaphragm to separate the anode compartment from the cathode compartment. This is important in ensuring that
chlorine produced at the anode does not mix with NaOH which accumulates in the cathode
compartment.
Fig 6.13 Diaphragm cell for the electrolysis of brine
Page
213

At the anode, Cl- ions are discharged in preference to OH- ions.
2Cl-  Cl2 + 2eThis leaves behind OH- ions, which then drift through the diaphragm to the cathode
compartment. The diaphragm is made up of a porous substance to ensure that solution rich in
NaOH can pass through it.

The level of liquid is higher in the anode compartment, ensuring that solution can only flow in one
direction, that is, from anode to cathode. This ensures that NaOH in the cathode compartment does
not mix with chlorine produced at the anode.
 At the cathode, H+ ions are discharged in preference to Na+ ions. Na+ ions therefore accumulate in
the cathode compartment and they attract OH - ions from the anode compartment. The solution in
the cathode compartment therefore gradually becomes richer in NaOH. This solution is drained
out and the NaCl present in it is removed by crystallization.
The electrolysis of brine therefore produces three products; hydrogen, chlorine and NaOH. The net
equation for the electrolytic process is
2NaCl (aq) + 2H2O (l)  Cl2 (aq) + 2NaOH (aq) + H2 (g)
This equation is useful in determining the ratio in which the three products are produced. The ratio of Cl 2:
NaOH: H2 is 1: 2: 1. However, this reaction is an oversimplification and is misleading. It gives the wrong
impression that NaCl has a reaction with water. NaCl does not react with water. It simply dissolves in it.
The three products of the Chlor-alkali process have very important industrial uses, as shown Table 6.6.
Sodium hydroxide
Used in the manufacture of



soap and
detergents
rayon fibres
paper
Table 6.6
Chlorine
Used in the manufacture of



organic solvents
household bleaches
swimming pool
cleaning chemicals
Hydrogen
Used in the production of



ammonia
margarine
metals from their
ores
Uses of the products of the Chlor-alkali process
A fourth product, sodium chlorate, is produced when the contents of the anode and cathode compartments
are allowed to mix. This is done by removing the diaphragm and stirring the contents of the cell. This allows
chlorine and NaOH to mix and react according to the equation.
Cl2 (g) + 2NaOH (aq)  NaClO (aq) + NaCl (aq) + H2O (l)
The uses of sodium chlorate, NaClO, include the manufacture of water treatment chemicals.
Preferential discharge of aqueous cations during electrolysis
Page
214
Discharge of cations during electrolysis takes place at the cathode by reduction.
Mn+ (aq) + ne- Ý M(s) ... (i)
For metal cations, the order in which discharge takes place in aqueous solution can be deduced from the
electrochemical series. It is very difficult to discharge ions at the top of the series. In fact, if the reduction
potential for reaction (i) is negative, it is impossible to discharge the ions as a metal. A negative electrode
potential implies that the formation of the metal in aqueous solution is not energetically favourable. In
other words, if the metal were to be produced, it would immediately react with water to produce the ions
once more. Examples of metals which can not be discharged in aqueous solution include K, Na and Mg.
Metals that are below hydrogen in the electrochemical series, for example Ag and Cu have positive reduction
potentials. This shows that reduction of the metal ions is possible in aqueous solution, forming a metal.
Table 6.7 is an arrangement of metal cations according to the ease with which they can be discharged at the
cathode.
Q
A
During the electrolysis of an aqueous solution containing equal numbers of
moles of Cu2+, Ag+ and Mg2+ ions, state and explain the order in which the
ions would be discharged.
Ag+ ions are discharged first, since the Ag+/Ag electrode potential is the most positive. When
discharged, Ag metal would be the most stable. The next iron to be discharged is Cu 2+. Mg2+, with
a negative reduction potential, can not discharged in aqueous solution.
Preferential discharge of anions during electrolysis
Discharge of anions takes place at the anode by oxidation. During this process, the anions lose their
negative charge (electrons) . Consider the discharge of Cl- ions during electrolysis.
Cl2 + 2e- Ý 2Cl-
Eθ = +1.36V
This reduction equation must be read in reverse inorder to show oxidation of the anion to form chlorine
gas.
2Cl- Ý Cl2 + 2e-
Eθ/V = -1.36V ... (i)
Compare with the equation for the discharge of OH- ions as oxygen gas
4OH- Ý 2H2O + O2 + 4e- Eθ/V = -0.40 (after reversing) ... (ii)
Since the oxidation process for OH- has a more positive Eθ value, it implies that in the electrolysis of a dilute
solution containing OH- and Cl- ions, OH- ions would be preferentially. Table 6.7 is an arrangement of
anions according to their ease of discharge during electrolysis.
Page
215
Cation
Electrode potential
Eθ/V
Cl-
Cl2 + 2e- Ý 2Cl-
+1.36
Br-
Br2 + 2e- Ý 2Br-
+1.07
I-
I2 + 2e- Ý 2I-
+0.54
OH-
2H2O + O2 + 4e- Ý
4OH-
+0.40
Table 6.7
Preferential discharge of anions
Most difficult
to discharge
Easiest to
to discharge




The anion with the most positive reduction potential is the most difficult to discharge during electrolysis. When read in reverse, this process has the most negative potential, so it is the least likely
to take place.
The anion with the least positive reduction potential is the easiest to discharge.
Polyatomic ions such as SO 42- are the most difficult to discharge. In fact, they are usually not
discharged during electrolysis.
The predictions given above apply only under standard conditions, that is, the assumption made is
that the ions have a concentration of 1.0 moldm-3 and that the electrolysis is carried out at 298K.
Changing concentration affects the identity of the substance discharged. For example, during the
electrolysis of concentrated NaCl, Cl 2 and not O2 is produced at the anode.
Electrolytic purification of copper
Impure copper is purified by making it the anode in an electrolytic cell containing copper (II) sulphate
solution. The cathode is made of pure copper (Fig 6.14).
The copper anodes dissolve through oxidation, releasing Cu2+ ions, which then migrate to the cathode
Cu(s)  Cu2+ + 2eAt the cathode, Cu2+ ions are reduced, forming copper metal, which sticks onto the cathode.
Cu2+ + 2e-  Cu(s)
The anodes therefore get thinner whilst the cathodes gain mass and become thicker. At the end of the
process, the cathodes, containing pure copper, are removed from the electrolytic cell. The purification
process removes certain impurities, particularly zinc, gold and silver.We can use electrode potentials to
explain how these impurities are removed from the impure copper.
Page
216
Fig 6.14
Electrolytic purification of copper
Equation
Reduction
potential/ V
(i)
Zn2+(aq) + 2e- Ý Zn(s)
-0.76
(ii)
Cu2+(aq) + 2e- Ý Cu(s)
+0.34
(iii)
Ag+ (aq) + e- Ý Ag(s)
+0.80
Table 6.8 Preferential discharge of ions
involved in the purification of copper by
electrolysis.


The conversion of zinc metal to Zn2+ has a
positive reduction potential (reverse of
equation (i)), that is, at the anode, any zinc
in the copper easily turns to Zn2+ ions. On
the other hand , the reduction of Zn2+ ions
to zinc metal has a negative reduction potential, that is, the reduction of Zn2+ ions to
zinc metal is not likely in aqueous solution.
The ions therefore stay in solution.
The reduction potentials of silver and gold
are more positive than that of copper (they
are more stable). These two impurities do
not easily dissolve at the anode.
Instead, they remain in their solid metallic state and drop to the bottom of the cell as the impure copper
anode dissolves.
6.9.3 Electrolysis of molten salts
Page
217
The electrolysis of molten (liquid) salts is a good method for producing reactive metals such as sodium and
magnesium. These metals can not be produced by electrolysis of the aqueous salt solution, as already
explained. The major disadvantage of this method is that huge amounts of heat energy are required to melt
most salts and keep them liquid during the electrolytic process.
The heat energy is required to overcome the electrostatic attractions between anions and cations in the ionic
crystal. This causes breakdown of the lattice, and makes the ions mobile. The molten solution can therefore
conduct electricity because the ions carry electric charge when they move from one point to another.
The electrolysis of a pure molten salt is straightforward because there are no other ions to consider except
those that came from the ionic compound.
Fig 6.15 below shows a set up for the electrolysis of molten lead (II) bromide in the lab.
Fig 6.15 Electrolysis of molten lead (II) bromide
PbBr2 is decomposed by heat according to the equation
PbBr2 Ý Pb2+ + 2Br-
Pb2+ ions migrate to the cathode where they are reduced to lead metal.
Pb2+ + 2e-  Pb (l)
Meanwhile, Br- ions drift to the anode whey they are oxidized to Br 2.
2Br-  Br2 (g) + 2eExtraction of Al by the electrolysis of molten aluminium (III) oxide
The major ore of aluminium is bauxite, which is impure Al2O3. Fig 6.16 shows the electrolytic cell for the
extraction of Al from molten Al2O3 by the Hall-Héroult process.
Since the melting point of bauxite is very high (about 2 0000C), a fluxing agent is added. This is a substance
which lowers the melting point of another. In this case, cryolite, Na3AlF6, is used. Its presence lowers the
melting point of bauxite to about 9500C . This saves energy and makes it easy to keep the ore molten.
Cryolite also acts as an electrolyte to improve conductivity of the cell.
Fig 6.16 Electrolytic cell for the extraction of aluminium by the Hall - Héroult process.
Al2O3 is decomposed by heat according to the equation
Al2O3  2Al3+ + 3O2At the anode, oxide ions are oxidized
O2-  O2 + 4eAt the cathode, Al3+ ions are reduced to Al.
Page
218
Al3+ + 3e-  Al
The molten aluminium drains out of the cell from the bottom.
Anodization of Aluminium
The large and negative reduction potential for Al shows that it is a very reactive metal, that is, it is very
unstable relative to its ions, Al3+.
Al3+ + 3e- Ý Al(s)
Eθ = -1.66V
Reading in reverse gives
Al(s)  Al3+ + 3e-
Eθ = +1.66V
The large and positive value of this reaction indicates that the oxidation of aluminium to Al3+ is energetically
favourable.However, in practice, aluminium is an unreactive metal. In fact, some of its uses depend on this
property (for example, it is used to make cooking pots). The inertness of Al metal is due to kinetic factors.
In air, freshly prepared aluminium rapidly undergoes a superficial (surface) reaction with oxygen, forming
a thin, tough and impermeable layer of Al 2O3. This oxide layer is tenacious, that is, it sticks to the metal
surface so tightly that it is not easily removed. Being impermeable to air and water, the oxide layer protects
the metal underneath from further corrosion.
The aluminium oxide layer on aluminium objects can be made thicker by anodizing. This is achieved by
making the aluminium object the anode in the electrolysis of dilute sulphuric acid.
Oxygen is discharged at the anode
4OH-  O2 + 2H2O + 4e- ... (i)
The oxygen then reacts with the aluminium object to form Al 2O3
Al(s) + O2(g)  Al2O3(s)
Question and answer section: electrolysis of molten Al2O3
Q.
In the early nineteenth century, attempts were made to extract aluminium from Al2O3 using sodium
metal.
a) Explain the chemical principles underlying this method of extraction, supporting your answer with an
equation.
b) What are the major disadvantages of this method?
A. a)
Na metal is higher in the electrochemical cell than aluminium and is therefore a more powerful reducing
agent (than Al itself). Na is therefore able to reduce Al2O3 by removing oxygen atoms.
Al2O3 + 6Na  2Al + 3Na2O
b) Na metal is difficult to handle because of its very reactive nature, for example, it can easily burn in air.
The production of Na itself (by electrolysis) is an expensive process.
Q.
Explain the use of graphite anodes in the electrolytic cell for the extraction of aluminium.
A.
Graphite has a very high melting point, so it is not decomposed by the high temperature used in the
process. Graphite also has the advantage of being relatively cheap.
Page
219
Q.
Explain why the anodes must continuously be replaced during the extraction of aluminium.
A.
Oxygen produced at the anode oxidizes carbon in graphite to CO2. This reaction is promoted by the high
operating temperature of the cell. The anodes are therefore gradually consumed and must be replaced
continuously.
Q.
Construct a balanced net equation for the production of Al by electrolysis of Al2O3.
A.
Anode reaction
3[2O2-  O2 + 4e-]
Cathode reaction 4[Al3+ + 3e-  Al]
Net reaction
6O2- + 4Al3+  3O2 + 4Al
Combining the ions on the left side gives the net reaction as
2Al2O3  3O2 + 4Al
Q
The net reaction for the electrolysis of Al2O3 using the Hall- Héroult process is sometimes given as
2Al2O3 + 3C  4Al + 3CO2
What is the justification for writing the net equation like this? Explain why this equation is an
oversimplification of the electrolytic process.
A.
2Al2O3  3O2 + 4Al ... (i)
The oxygen produced at the anode oxidizes graphite to CO2
3O2 + 3C  3CO2 ... (ii)
Combining equations (i) and (ii) gives the net equation
2Al2O3 + 3C  4Al + 3CO2
This equation is an oversimplification because it gives the wrong impression that Al2O3 is reduced by
carbon to Al metal. The conversion of Al2O3 to Al in this process is not achieved by the use of a
reducing agent, but by the use of an electric current (as a reducing agent, carbon is too weak to reduce
Al2O3 ). Moreover, the equation wrongly shows a one step reduction, not an electrolytic process which
is followed by the reaction of graphite with oxygen.
Q.
Explain the need to keep current high and voltage low in the electrolytic production of Al by the HallHéroult process.
A
High current helps to keep temperature high so that the ore remains molten. Low voltage minimizes the
discharge of fluoride ions at the anode as the poisonous fluorine gas. The fluoride ions come from the
electrolyte, Na3AlF6.
Q
Page
220
A
State two uses of cryolite in the extraction of Al.
-Electrolyte
-Lowers melting point of bauxite
Q.
In table form, give three uses of aluminium in relation to its properties.
A.
Use


Resists corrosion
Good conductor of heat
Overhead power cables



Light
Resistant to corrosion
Good conductor of
Electricity (it is a better
conductor than Cu).
Aircraft bodies

Light
Making cooking pots
6.10
Property
Quantitative electrolysis
It is possible to calculate the amount of substance produced during electrolysis, for example, the amount of
metal deposited at the cathode and the volume of gas produced at the anode.
6.10.1
Electric charge and the Faraday constant
The quantity of electricity passed during electrolysis is measured in coulombs (C). This quantity, also known
as electric charge, Q, depends on the duration (time) of the process and also on the size of current.
that is , Q = I x t
Where I = current in amperes
t = time in seconds
Since 1 electron has a charge of -1.60 × 10–19 C, one mole of electrons has a charge of
–1.60 × 10–19 C x (6, 022 x 1022) = 96 352C
This figure, often rounded off to 96 500C/mol, is known as the Faraday constant, F.
The Faraday constant, F is the size of charge on 1 mole of electrons
Page
221
F ≈ 96 500 C
This value is obtained by multiplying the Avogadro number L and the charge e on an
electron, that is
F = Le = 96 500 C
6.10.2 Amount of metal deposited at the cathode during electrolysis
A metal is deposited by the reduction of its ions at the cathode
Mn+ (aq) + ne- Ý M(s)
Consider the discharge of copper during the electrolysis of CuSO 4 (aq).
Cu2+ (aq) + 2e- Ý Cu(s)
2 moles of electrons are required to discharge 1 mole of Cu. The number of electrons transferred during the
process depends on the charge on the ion. For a +1 ion such as Ag+, the number of electrons transferred would be 1 mole.
Ag+ (aq) + e- Ý Ag(s)
The number of moles of electrons required to discharge 1 mole of a metal from its ions is equal to
the charge on the ions.
For copper, the size of electric charge required to discharge 1 mole of metal
= 2 moles of electrons x charge on one mole of electrons
= 2F ( = 2 x 96 500C)
The charge required to discharge 1 mole of a metal from its ions is given by
Q = zF,
where z is the size of charge on the ion and F is the Faraday constant.
Since a charge of zF discharges 1 mole of metal, a charge Q would discharge
𝐐
𝐳𝐱𝐅
moles of metal.
The number of moles, n of metal discharged during electrolysis is given by
n=
𝐐
𝐳𝐱𝐅
where Q = It ... (i)
The mass, m of metal deposited is given by n x Ar, where n is as given in equation (i) above.
The mass m of metal deposited during electrolysis is given by
m=
𝐐
𝐳𝐱𝐅
x Ar
... (ii)
Page
222
You may use equations (i) and (ii) in exams without proof, but it is certainly helpful to know how the
equations are derived.
Further analysis of equation (ii)
Rearranging equation (ii) gives
m=
𝐀𝐫
𝐳𝐱𝐅
x Q ... (iii)
For a particular metal ion, Ar, z and F are constant, so equation (iii) can be expressed in terms of Q as
m= 𝐊xQ
... (iv)
where K is a constant equal to
𝐀𝐫
𝐳𝐱𝐅
Now equation (iv) is that of a straight line, showing direct variation between mass of metal deposited and
the size of charge that passes during the period of electrolysis.
Mass, m of metal deposited during electrolysis is directly proportional to charge, Q
mαQ
That is, increasing charge results in a proportional increase in mass of metal deposited. Charge, Q can be
increased by increasing either the current, or the duration of electrolysis (time) or both, since Q =It.
Fig 6.17 is a graph showing the mathematical relationship between mass of metal deposited and size of
charge.
mass of metal
deposited
y
x
Charge Q = It
Fig 6.17 Graph showing
direct variation between
charge and mass of metal
deposited during electrolysis.
Such a graph can be drawn from a series of experimentally obtained values for m and Q. (see exercise 6.5).
The slope, m1 of this line is equal to the constant
slope = m1 =
𝐀𝐫
𝐳𝐱𝐅
𝐀𝐫
𝐳𝐱𝐅
(refer to equations (iii) and (iv))
... (v)
Page
223
This outcome can be used to determine Ar, z or F experimentally, as illustrated in a later example.
Example 1
A constant current of 20.0 A was passed through an electrolytic cell containing CuSO4(aq). The
anode in this cell was made of impure copper, and the cathode was made from pure copper. After 10
hours it was found that 225 g of pure copper had been deposited on the cathode.
(i)
Calculate the following, using appropriate data from the Data Booklet.



(ii)
number of moles of copper produced at the cathode
number of moles of electrons needed to produce this copper
number of moles of electrons that passed through the cell.
Hence calculate the percentage of the current through the cell that was actually used to
discharge the copper metal at the cathode. Explain why your answer is not 100%.
[Adapted from UCLES 9701/ON/2010/41]
Solutions
(i)

Moles of Cu = m/Ar = 225/63.5 = 3.543

amount of electrons needed = charge x moles produced
= 2 x 3.543
= 7.087 moles

Number of coulombs = I x t
= 20 x 10 x 60 x 60
= 720 000 C
Number of moles of electrons passed through the cell
=
=
Total charge
charge on 1 mole of electrons,F
720 000
96 500
= 7.461
(ii)
Of the 7.461 moles of electrons that passed through the cell, only 7.087 were used to form copper.
The percentage of the current used to produce the copper is therefore
7.087/7.461 = 95%
Page
224
The value is not 100% because a small percentage (in this case, 5%) of the current
is used in dissolving the impurities at the anode.
Example 2
Describe a method suitable for the determination of the Avogadro constant, L by an electrolytic
procedure. Explain how the results from your experiment can be used to calculate a value for L.
Solution
A suitable method involves an electrolytic cell using copper electrodes as shown in Fig 6.18


Both the anode and cathode are made of copper metal of known masses.
As soon as the circuit is closed, the time is noted and the current, I, is read off from the ammeter.
The rheostat is adjusted when necessary to ensure that a steady current passes through the cell.
During the electrolytic process, the anode dissolves and loses mass. Meanwhile, copper is deposited at the
cathode, so it gains mass. After some time t , the switch is opened; the anode and cathode are removed,
washed and then reweighed. The aim is to note the gain in mass of the cathode, or the loss in mass of the
anode (the loss in mass by the anode should be equal to the gain in mass by the cathode). Let this mass be
m grams.
Processing results
Amount of electricity passed = charge = I(in amps) x t (in seconds)
Moles of Cu deposited =
m
63.5
m
63.5
moles of copper are deposited by It
Coulombs . Therefore 1 mole of copper would
be deposited by a charge of
63.5 x It
m
This is equivalent to the charge on two moles of
electrons, according to the reduction equation
Cu2+ (aq) + 2e- Ý Cu(s)
That is 2 x e x L =
Fig
6.18
L=
225
... (i)
The expression on the left hand side represents
the charge on two moles of electrons (e is the
charge on one electron and L is the Avogadro constant)
Rearranging equation (i) gives
Page
63.5 x It
m
L =
63.5 x It
2 x m x (1.60 x 10-19)
1.98 x 1020 x It
m
...(ii)
Example 3
When a constant current of 10A was passed through copper(II) sulphate solution for 7
minutes, 1.38g of copper was discharged at the cathode. Use this information to calculate a
value for
(a)
(b)
the Avogadro constant, L
the Faraday constant, F
Solution
(a) Using equation (ii) in example 2 and substituting the given data
L =
L =
1.98 x 1020 x It
m
1.98 x 1020 x 10 x 7 x 60
.
1.38
= 6.026 x 1023
The actual value of the Avogadro constant is 6.022 x 1023. One possible source of error in this method
is the difficulty in maintaining a constant current.
(b)
F = L.e
= 6.026 x 1023 x 1.60 x 10-19
= 96 417 Cmol-1
The Faraday constant is commonly given as 96 500 Cmol-1
Example 4
A current of 12A is passed for 150 minutes through aqueous sodium sulphate using inert electrodes.
What is the volume of gas liberated at the cathode at s.t.p?
A
0.1 dm3
B
2.5 dm3
C
3.1 dm3
D
6.3 dm3
Solution D
OH- ions migrate to the anode where they are discharged as O2 (g) by oxidation
Page
226
4OH-(aq)  2H2O + O2 (g) + 4e- … (i)
From equation (i), 4 mol e- discharges  1 mol O2
Charge on 4 moles of electrons = 4F since 1 mole of e- has a charge given by the Faraday constant, F
= 4 x 96 500 Cmol-1
= 386 000C
Charge that flowed during the process
= Ixt
= 12 x 150 x 60
= 108 000 C
If 386 000 coulombs discharge one mole of oxygen, 108 000 coulombs will
discharge
108 000
386 000
moles = 0.27979 moles of oxygen, which gives a volume of
0.279797 x 22.4dm3 = 6.267dm3
~ 6.3 dm3
Example 5
30 g of a metal X was discharged by electrolysis, according to the equation
X3+(aq) + 3e-  X(s
Given that the electrolysis was done for 24 125 seconds with a constant current of 2.4A, deduce the
identity of metal X
Solution
m=
Ar =
=
𝐐
𝐳𝐱𝐅
m
Q
x Ar ( where Q = I(A) x t(s) = 24 125 x 2.4 = 57 9ooC)
x zF
30 x 3 x 96 500
57 900
= 150
Element X is therefore Samarium, Sm.
Page
227
Example 6
An electrolytic cell was set up invoving copper and silver electrodes immersed in solutions of their
ions. At the end of the process, it was found that 4g of silver metal had been deposited. Calcullate the
mass of copper that was deposited in the same time interval.
Solution
Ag is deposited at the cathode in cell B, whilst in cell A, Cu is deposited
Number of moles of Ag metal deposited =
4
108
Ag+(aq) + e-  Ag(s) ...(i)
½ Cu2+(aq) + e-  ½ Cu(s) ...(ii) (that is, Cu2+(aq) + 2e-  Cu(s))
From equation (ii), it can be seen that the same number of electrons (charge) produces
Ag and Cu metals in the ratio of 1 mole Ag: ½ mole Cu.
Moles of Cu produced is therefore
= ½ moles of Ag
=
Mass of copper produced
4
108
4
108
= 1.18g
228
1
2
= moles x Ar
=
Page
x
x
1
2
x 63.5
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
6.5in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
A student decided to verify the value of the Faraday constant using an electrolytic method.
(Fig 6.19)
Method
• The cathode is cleaned and weighed before being placed in the copper (II) sulphate solution.
• The circuit is completed and the current set at 0.3 A by adjusting the variable resistor.
• The current is maintained at 0.3 A for exactly 40 minutes at which point the circuit is
broken.
• The cathode is removed from the solution and carefully washed with distilled water to
remove any copper(II) sulphate solution.
• Distilled water is removed from the cathode by rinsing it with propanone in which the
water dissolves.
• The cathode is finally dried by allowing the propanone to evaporate from its surface.
• The cathode is reweighed and placed back in the solution.
• A constant current of 0.3 A is passed for a further 40 minutes when the rinsing, drying
and weighing are repeated.
• This procedure is repeated a further 8 times.
The results of the experiment are recorded in Table 6.9.
(a) Use the additional columns of the
Table 6.9 to record the charge
passed and the mass of copper
deposited on the cathode.
You may use some or all of the
columns. Label the columns you
use, including the unit and an
equation to show how the value is
calculated.
(b)
Page
229
Fig 6.19
Present the data calculated in (a)
in graphical form. Draw the line of
best-fit.
A
B
time
/minutes
mass of
cathode
/g
0
C
D
E
F
115.74
40
115.97
80
116.22
120
116.46
160
116.70
200
116.94
240
117.19
280
117.49
320
117.67
360
117.92
400
118.14
Table 6.9
(c)
Draw construction lines on the graph to derive relevant values and use them to calculate
a numerical value for the Faraday constant.
HINT : See section 9.6.2 on how to tackle this problem
[Adapted from UCLES 9701/ON/2008/5]
Page
230
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Page
231
CHAPTER
7
CHEMICAL EQUILIBRIUM
Introduction
When ethanol is mixed with ethanoic acid in a ratio of 1:1 and heated in the presence of a strong acid,
a fruity smell is soon detected due to the formation of an ester. When the contents of the reaction vessel
are analyzed after, say 24 hours, it is found that the product is mixed with unconverted reactants. This
indicates that the reaction has not gone to completion. Even if the vessel is analyzed a few days later, it
is found that some reactant molecules still remain. It is also found that the amounts of reactants and
products remain the same, provided temperature is kept constant. This is an example of a reversible
reaction, in which the products, once formed, are able to react in the reverse direction to regenerate
the reactants.
CH3CH2OH + CH3COOH Ý CH3COOCH2CH3 + H2O
ethanol
ethanoic acid
ethylethanoate(ester)
Now consider what happens when a small piece of sodium metal is placed in water. A reaction occurs
rapidly and all the sodium is soon used up. This is an example of an irreversible reaction, in which the
products can not react backwards to form the reactants.
2Na(s) + 2H2O (l)  2NaOH (aq) + H2 (g)
In this section, Chemical Equilibrium, we explore reversible reactions and try to answer the questions
Page
232
How far does the reaction go? Does it go to completion, for example the reaction between water
and sodium, or is it reversible, for example, the reaction between ethanol and ethanoic acid? If
the reaction is reversible, how far does it go in terms of converting reactants to products?
These are important questions to an industrial chemist. Irreversible reactions result in all reactants
being converted to products. This reduces wastage of raw materials and saves time and costs related to
separation of products from reactants. However, a large number of reactions are reversible, for
example, the Haber process for the manufacture of ammonia. Even so, it is still possible to find ways of
ensuring that as much of the reactants as possible react, and that the yield of the reaction is close to
100%.
7.1
Chemical equilibria
This chapter mainly focuses on reversible reactions. Consider a simple reversible reaction
A+B⇌C+D



There are two opposing reactions, a forward reaction and a reverse reaction. The forward
reaction results in some reactant molecules being converted to products. The reverse reaction results in some product molecules being converted back to reactants.
A 100 % yield of product is not possible because some products are continuously being converted
back to the reactants.
All substances shown in the equation co-exist in the same reaction vessel as a mixture. At
dynamic equilibrium, the concentration of reactants and products remains constant.
Compare with irreversible reactions in which there is 100% yield of product and at the end of the reaction
concentration of one or all reactants drop to zero (if a reactant is present in excess, its concentration does
not decrease to zero).
An important note
In practice most reactions have a degree of reversibility. When we say a reaction goes to completion as shown below
A+B→C
We simply mean that at the end of the reaction, a negligibly small amount of reactant molecules remain, and the yield
of the product is very close to 100%, say 99, 99%.
Page
233
The diagrams below are concentration-time graphs for a reversible and for a non-reversible reaction.
Fig 7.1
Concentration - time graphs for a reversible and non-reversible
reaction.
Notice that for an irreversible reaction, concentration of reactant drops to zero, that is, the graph touches
the x-axis. For a reversible reaction, concentration of reactant drops, but it will not become zero. At
equilibrium, a fixed amount of reactant molecules remain unreacted, giving a fixed equilibrium reactant
concentration, [A]eq.
The graphs below show how the concentrations of product changes with time for a reversible and for a nonreversible reaction.
Notice that for an irreversible reaction, there is
(almost) 100% yield of product. For a reversible
reaction, the yield of product is less than 100%.
concentration
of product
7.1.1
Dynamic equilibrium
Consider the reversible reaction between
hydrogen and iodine to form
time
H₂ (g) + I₂ (g) ⇌ 2HI (g)
Suppose that we start by mixing 1 mole of iodine
and 1 mole of hydrogen gases in a closed flask. At
the beginning, the forward reaction is faster than
the reverse reaction.
This is in accordance with the collision theory of reaction kinetics which states that reactant molecules must
first collide before they react. Since the concentration of reactants is still high, frequency of collisions
between reactant molecules is high and the forward reaction is accordingly fast. At the very beginning,
reactant concentration therefore drops rapidly, as shown by the steep part of the graph in Fig 7.3. As the
reaction proceeds, the concentration of the product increases. The reverse reaction therefore becomes more
important as the frequency of collisions between product molecules (to form reactants) increases.
The reactant concentration does not drop to zero, and the yield of the product does not increase to 100%.
There comes a time when concentration of reactants stops decreasing and concentration of products stops
increasing. At this point, the concentration of reactants and products become constant and the speed of the
forward and reverse reactions are equal. We say the system has achieved dynamic equilibrium.
Fig 7.2 Concentration time graph for a
reversible and for an irreversible reaction
Dynamic equilibrium refers to the situation in a reversible reaction in which the rate of the
forward reaction is equal to the rate of the reverse reaction, and concentration of products and
reactants has stopped changing.
Important note
Page
234
The term dynamic helps to clear a misconception which is likely to arise from the definition given above. The fact that
the concentration of the product and reactant stops increasing may mislead some students into thinking that the
forward and reverse reactions have come to a stop. If a reaction is reversible, it can not come to a stop. There is
continuous conversion of reactants to products and vice versa. At equilibrium, this inter-conversion continues (it is
dynamic), but the rates of the forward and reverse reactions are now equal, so that there is no further decrease in the
concentration of reactants or increase in the concentration of products. The term dynamic equilibrium is appropriate
because it emphasizes that at equilibrium, there is continuous activity among reactant and product molecules.
Fig 7.3 below refers to a situation in which equilibrium is achieved after time T eq. At this point, the
concentration of product and reactant becomes constant.
The equilibrium concentration of the reactants
and products are R and P respectively.
The shapes of the graphs do not always have to
be as in Fig 5.3. Here we have shown a situation
in which the equilibrium concentration of the
reactant is much smaller than that of the
product. In other words, the equilibrium lies in
favour of the product. This is not always the case.
concentration
time
Fig 7.3 When equilibrium has been achieved,
concentration of reactant and product becomes
constant
concentration
time
Fig 7.4 A reversible reaction in which
there are more reactants than products
at equilibrium
At equilibrium, it is found that concentrations of reactants A and B have decreased to certain fixed values
and the concentrations of the products C and D have increased to certain fixed values. The ratio
[C]c[D]d
[A]a[B]b
is therefore found to be constant, at a given
temperature. This constant is known as the
equilibrium constant, Kc .
therefore becomes constant at equilibrium. This ratio is referred to as the equilibrium constant, Kc, that is,
Kc =
[C]c[D]d
[A]a[B]b
Page
235
The square bracket [ ] is a notation which means ‘concentration of...’ . The equilibrium constant Kc is
convenient for reactions that take place in solution, but it can also be used for gas phase reactions.
Example
An equilibrium mixture was found to contain 1 mole of hydrogen gas, 2 moles of iodine gas and 10
moles of hydrogen iodide gas. Calculate the value of the equilibrium constant K c for the reaction
2HI(g) Ý H2(g) + I2(g)
Solution
Concentrations at equilibrium:
[H2] = 1/v , [I2] = 2/V, [HI] = 10/V
[H₂][ I₂]
Kc =
[HI]²
=
( 22 )
V
10
( 2)
V
=
Note

1 2
V V
10
( )²
V
( )( )

= 0.2
In this case, it was not necessary to specify volume.
This is because the volume cancels out when the
expression for Kc is simplified. This is not always the
case. In some equilibria, volume does not cancel out
and so it must be specified.
The value of Kc in this case has no units, since units in
the denominator and numerator cancel out.
7.1.3 Equilibrium constant, Kp
The equilibrium constant Kc applies to both reactions that take place in solution and in gaseous phase.
However, for gaseous phase equilibria, it is more convenient to use the equilibrium constant in terms of
pressure, Kp. This is because for gases it is easier to measure pressure than concentration.
The expression for Kp is written in a similar way to Kc, but this time instead of using concentrations ([ ]) we
use partial pressures (P), for example, the partial pressure of gas A is written PA. The square of the
partial pressure of A is written as PA2. Thus for reaction (II) below
A(g) + B(g) Ý C(g) + D(g) ... (II)
we may write an expression for the equilibrium constant in terms of partial pressures as
Kp =
PCcPDd
PAaPBb
Take note of the following
 Equilibrium constants (Kp and Kc ) apply only to a system that is at equilibrium.
 The units of Kp and Kc vary depending on the specific equilibrium being studied.
Page
236
Recall the following about partial pressures
(see section 4: states of matter)


The partial pressure of a particular gas in a gaseous mixture is the pressure the gas would exert if it
existed alone in the container.
The partial pressure, PA , of gas A in a gaseous mixture is given by
PA = mole fraction of A x Total pressure
= ηA x PT
The mole fraction, ηA, of gas A is given
ηA =
moles of gas 𝐀

The total pressure, PT, of a gaseous
mixture is equal to the sum of the partial
pressures of all the gases present, for
example, if the container contains gases
A, B and C, the total pressure in the
container is given by
Total number of moles of all gases
PT = PA + PB + PC
∴
PA =
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐠𝐚𝐬 𝐀
x PT
𝐓𝐨𝐭𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐚𝐥𝐥 𝐠𝐚𝐬𝐞𝐬
Example 1
Sulphur trioxide was heated in a flask, causing it to decompose to form sulphur dioxide and
oxygen. The equilibrium mixture was to found to contain 360g of sulphur trioxide, 3.80 g of oxygen
and 32g of SO2. Given that the total pressure in the flask was 40KPa, calculate the partial pressure
of each gas in the equilibrium mixture. Hence calculate the value of Kp for the equilibrium and
determine the units.
Solution
moles of each substance present at equilibrium:
PSO2 =
moles of SO3 = 360/Mr = 360/80
= 4.50
0.5
5.119
2SO3(g) Ý 2SO2(g) + O2(g)
moles of O2 = 3.80/32 = 0.119
Kp =
moles of SO2 = 32/64 = 0.5
Total number of moles of gas at equilibrium
=
= 4.50 + 0.119 + 0.50 = 5.119
PSO3 =
=
Page
237
PO2 =
moles of SO₃
Total moles present
4.50
5.119
0.119
5.119
x 40 KPa = 0.928 KPa
Po₂ .P²so2
.
P²so3
0.928 .3.908²
35.2²
= 0.0112
x PT
x 40 KPa = 35.2 KPa
x 40KPa = 0.0977 = 3.908
The units of Kp are
=
KPa.KPa²
KPa²
= KPa
Example 2
Dinitrogen tetraoxide decomposes to form nitrogen dioxide according to the equation
N2O4 (g)Ý 2NO2(g)
When 2 moles of N2O4 were put in a sealed 500 cm3 flask and allowed to reach equilibrium with
NO2, it was found that at equilibrium only x moles of N2O4 had decomposed and the total pressure
in the flask was P atm. Write an expression for Kp in terms of x and P. What are the units?
Solution
initial moles
equilibrium moles
N2O4 (g)Ý 2NO2(g)
2
0
2-x
2x
=
(2+x)
PNO₂ =
(Since the stoichiometry of the reaction
is 1N2O4 : 2NO2, if x moles of N2O4
decompose, 2x moles of NO 2 are formed)
Kp =
Total moles of gas at equilibrium = (2-x) + 2x
xP
2x
(2+x)
xP
P²NO₂
PN₂O₄
= {
= (2 + x)
2x
(2+x)
(2−x)
(2+x)
Equilibrium partial pressures :
PN₂O₄
(2−x)
=
moles of N₂O₄
=
x Total pressure
total moles
2x
x P}2
xP
P
2−x
=
Example 3
At high temperatures, steam decomposes into its elements according to the following equation
2H2O (g) Ý 2H2 (g) + O2 (g)
In one experiment at 1 atm pressure, it was found that 20 % of the steam had been converted into
hydrogen and oxygen.
What are the values of the equilibrium partial pressures, in atm, of the components of this
𝟐𝐱
equilibrium?
=
P
𝟐−𝐱
9701/1/O/N/2004
Solution
2H2O (g) Ý 2H2 (g) + O2 (g)
Page
238
eqlm moles
0.8
0.2
0.1
Total moles at equilibrium = 0.8 + 0.2 + 0.1 = 1.1
PH₂O =
PH₂ =
PO₂ =


0.8
1.1
0.2
1.1
0.1
1.1
x 1 atm = 0.730 atm
x 1 atm = 0.180 atm
x 1 atm = 0.091
moles of steam which dissociated = 20 % = 0.2
∴ moles of steam remaining at equilibrium = 80% = 0.8
Since stoichiometry of reaction is 2 H2O : 2 H2 : 1 O2, moles of hydrogen at equilibrium = moles
of steam that dissociated = 0.2 and moles of oxygen at equilibrium = ½ moles of hydrogen = 0.1
Equilibrium constant of the reverse
reaction
so [C] =
Consider the equilibrium
Equating the right hand sides of equations (i) and (ii)
gives
A+BÝC
The value of the equilibrium constant, Kc, for the
forward reaction is given by
Kc =
[C]
[A][B]
so [C] = Kc x [A][B] ... (i)
The equilibrium constant, K′c for the reverse
reaction
C Ý A + B is given by
K′c =
[A][B]
[A][B]
K̍′ c
... (ii)
Kc x [A][B] =
[A][B]
K′ c
Dividing both sides by [A][B] gives
Kc = 1/ K′c or
K′c =
𝟏
𝐊𝐜
Similarly, it can be shown that K′p =
𝟏
𝐊𝐩
In other words, the value of the equilibrium constant of
the reverse reaction is the inverse of the equilibrium
constant for the forward reaction (see the example
below).
[C]
Example
The ester ethylethanoate undergoes hydrolysis in water to form ethanoic acid and ethanol.
CH3COOCH2CH3 (aq) + H2O (l) Ý CH3CH2OH(aq) + CH3COOH(aq) ... (i)
Page
239
The equilibrium constant for this reaction is 0.25 at 298K. In an experiment, 2 moles of ethanoic
acid were mixed with 2 moles of ethanol in a 250 cm3 flask. The mixture was topped up with
distilled water to the 250 cm3 mark and allowed to reach equilibrium. Calculate the concentrations
of ethanol, ethanoic acid and ethyl ethanoate present at equilibrium.
Solution
CH3CH2OH(aq) + CH3COOH(aq) Ý CH3COOCH2CH3 (aq) + H2O(l) ... (ii)
Initial moles
eqlm moles
2
2-x
2
2-x
0
x
0
x
(since stoichiometry of the reaction is 1 : 1: 1: 1, moles of ethanol which react = moles of ethanoic acid which
react = moles of each product formed)
Kc for reaction (ii) is the inverse of the Kc for reaction (i) = 1/0.25 = 4 (the reactions occur in opposite
directions)
Kc =
4 =
[H₂O][CH₃COOCH₂CH₃]
[CH₃CH₂OH][CH₃COOH]
x x
V V
[(2−x).(2−x)]
𝑉
𝑉
( )( )
=
x²
(2−x)²
(in this case, volume cancels out)
4=
equilibrium concentrations:
[CH3COOCH2CH3] = moles / volume
= x/0.25 dm3 = 1.33/0.25
= 5.32 moldm-3
[CH3COOH] = [CH3CH2OH]
= (2-x)/0.25
x²
4−4x+x²
(2-1.33)/0.25 = 2.68 moldm-3
solving for x using the quadratic
formula gives x = 1.33
7.1.4
Open and closed systems
Consider what happens when CaCO₃ is strongly heated in an open vessel.
CaCO₃(s)  CaO(s) + CO₂ (g)
The reaction fails to be reversible because the carbon dioxide formed immediately escapes. No equilibrium
is achieved and the CaCO₃ will continue to decompose until none of it is left.
However, when a small amount of CaCO₃ is heated in a closed vessel, say at 800⁰ C, it is found that only a
certain amount of the carbonate will decompose. Thereafter no further decomposition occurs; no matter
how long the solid is heated (provided the temperature is kept constant). Similarly, the partial pressure of
CO₂ reaches a certain level (about 25KPa) and will not increase beyond that unless if temperature is
increased. This example serves to illustrate an important point; dynamic equilibrium is only achievable in
a closed system. In fact, many reactions would be reversible if they were carried out in closed vessels (systems).
Page
240
A closed system is defined as one in which no transfer of matter occurs between the system and the
environment.
7.1.5
Achieving equilibrium from the opposite direction
In a closed vessel, the decomposition of calcium carbonate soon reaches equilibrium in which it co-exists
with the products of decomposition
CaCO₃(s) ⇌ CaO(s) + CO₂ (g)
Equilibrium at 800⁰C is achieved when the partial pressure of CO₂ has increased to a constant value of
25 KPa. Since the reaction is reversible, we should get the same result if we start with a certain mass of CaO
and CO₂, say at 50 KPa. These two substances (they are now the reactants) are placed in a closed vessel and
heated at a constant temperature, say of 800⁰C. A reaction occurs to form calcium carbonate. The result is
that the partial pressure of CO₂ and mass of CaO begins to decrease and some calcium
carbonate begins to form. However, CaCO₃ soon starts to decompose back to CaO and CO₂.
CaO(s) + CO₂ (g) ⇌ CaCO₃(s)
The decomposition of CaCO3 causes the partial pressure of CO₂ to increase until it has reached an
equilibrium value of 25KPa. Similarly, the masses of CaO and CaCO3 change until they reach constant
equilibrium values (Fig 7.5).
50Kpa
25KPa
PCO2
mass of CaO
A initial pressure of CO2
B equilibrium pressure of CO2
C initial mass of CaO
mass of CaCO 3
o
time
Fig 7.5 When carbon dioxide and CaO
are heated at 8000C in a closed vessel,
equilibrium is soon achieved in which
the reactants and product are present
in fixed amounts.
7.1.6
D equilibrium mass of CaO
O initial mass of CaCO3
E equilibrium mass of CaCO3
Homogeneous versus heterogeneous equilibria
An equilibrium process is said to be homogeneous if all the substances in the equilibrium are in the same
phase, for example
HI (g) ⇌ H₂ (g) + I₂ (g)
If an equilibrium contains mixed phases it is said to be heterogeneous. An example is the decomposition
of CaCO₃.
CaCO₃(s) ⇌ CaO(s) + CO₂ (g)
Page
241
The two phases present are solid and gas. We may wish to express the equilibrium constant of this
reaction in terms of partial pressures, since we can easily measure the pressure of CO₂.
Suppose that we write the expression for the equilibrium constant as
Keq =
Pco₂ .Pcao
Pcaco₃
Taking the partial pressures of the solids CaO and CaCO3 to be constant, the expression for the
equilibrium constant may then be written as
Keq = K1PCO₂
Dividing both sides by the constant K1 gives
Keq
= K1PCO₂
K1
The term on the left hand side is the constantKp, the equilibrium constant in terms of partial pressure.
Kp = PCO₂

For heterogeneous equilibria, epressions for Kc or Kp do not include solids.

If a liquid in the equilibrium mainly saves as a solvent, it may also not be shown in the equilibrium
expression.
Consider the dissolution of carbon dioxide in water to form a solution containing hydrogen ions and
carbonate ions.
CO₂ (g) + H₂O (l) ⇌ H⁺ (aq) + HCO3⁻ (aq) ... (i)
Only a very small amount of water molecules react with carbon dioxide. Water mainly acts as a solvent
that absorbs and hydrates carbon dioxide
CO2(g)  CO2(aq)
We may treat water as a pure liquid in equilibrium (i), its concentration remains almost constant and so it
may not appear in the expression for the equilibrium constant, that is
Kc =
[H+ ][HCO₃¯]
[CO2 ]
The reader should not conclude that water is always not shown in expressions of equilibrium constants.
One has to consider each situation separately and judge whether water should or should not appear in the
expression of the equilibrium constant. Consider the hydrolysis of an ester in water
CH₃COOCH₂CH₃ + H₂O (l) ⇌ CH₃COOH (aq) + CH₃CH₂OH (aq)
If this was a simple case of a solute dissolving in water, we would ignore the concentration of water in the
expression for Kc. Now, this reaction involves hydrolysis, a reaction in which water is not simply a solvent,
but is a reactant. The concentration of water therefore appears in the expression for the equilibrium
constant.
Kc =
[CH₃COOH][CH₃CH₂OH]
[H₂O][CH₃COOCH₂CH₃]
Page
242
However, in this reaction water also serves as a solvent so it is present in very large amounts and
[H 2O] remains almost constant , so as an approximation, the expression for Kc can be written without
water.
Kc ≈
[CH₃COOH][CH₃CH₂OH]
[CH₃COOCH₂CH₃]
Mathematical interpretation of equilibrium constants
Equilibrium constants provide a mathematical way of quantifying the extent to which a reaction goes to
completion. First, consider a reaction A  B which goes to completion.
Suppose that we treat this equation as if it were reversible, and we write the expression for the equilibrium constant as
Kc =
[𝐵]
[𝐴]
... I
At the end of the reaction, concentration of the reactant, [A] drops to zero, so we have
Kc =
[𝐵]
0
=∞
In other words, if a reaction is irreversible, the value for the equilibrium constant is infinitely larger
than 1. The equilibrium constant is written so that the concentrations of products (raised to their respective
powers) are in the numerator and the concentrations of reactants are in the denominator. For the value of
Kc or Kp to be greater than zero, the numerator must be larger than the denominator, that is, the proportion
of products at equilibrium is relatively high. The greater the numerator, the higher the value of Kc or Kp,
and the greater the proportion of products at equilibrium. We therefore conclude that a very large value of
Kc or Kp implies that the reaction is, to a large extent, irreversible, that is, the tendency of products to react
back to form reactants is relatively small. If the equilibrium constant is greater than 1010, the reaction is
regarded as going to completion (irreversible).
Now consider a reaction whose equilibrium constant is a very small number less than 1, say 10⁻⁵. To get
such a small value, the numerator (representing products) in the expression for K c or Kp must be much
smaller than the denominator (representing concentration of reactants). In this case, the tendency of the
reaction to go to completion is very small. The products, to a large extent, react back to form the
reactants, so that at equilibrium a very small amount of products and a very large amount of reactants is
present. Reactions with an equilibrium constant which is less than 10-10 are regarded as not taking place
at all.
Consider the equilibrium
H₂ (g) + I₂ (g) ⇌ 2HI (g)
Page
243
At 731K, the value of the equilibrium constant, K c, is approximately 47. This value shows that there is a
significant amount of HI present at equilibrium. We may not say straight away how much of the HI is
present, but clearly it is not a negligible amount. It would be different if the value of K c was very small, say
1.0 x 10⁻⁵. In that case we would conclude that the amount of product at equilibrium is very small, that is,
the tendency of the reaction to go to completion is very small.
The equilibrium constant therefore gives a measure of the composition of the equilibrium mixture. Are
there more products than reactants, more reactants than products, or are there comparable amounts of
each?
More worked examples
Kc =
[H
[CH
=
0
Q1
A and B react to form product C
according the equation
A (aq)+ B(aq) Ý C (aq)
500 cm3 of the equilibrium mixture is
found to contain 0.2 moles of A, 0.2
moles of B and 0.4 moles of C at 298K.
Calculate the equilibrium constant,
Kc, of the reaction, and state its
units.
A
Concentrations at equilibrium :
[C] = 0.4 moles / 0.5 dm3 = 0.8 moldm-3
[A] = [B] = 0.2/0.5 = 0.4
Kc =
=
[𝐜]
[𝐀][𝐁]
0.8
0.4 x 0.4
= 5 mol-1 dm3
Q2
Ethanoic acid and ethanol reacts in
aqueous solution and in the presence of an
acid catalyst to form an ester and water. This
reaction is reversible so that at equilibrium
ethanol, ethanoic acid, ethylethanoate and
water are present in the same reaction vessel.
One equilibrium mixture was found to contain
0.117 moles of ethanoic acid, 0.017 moles of
ethanol, 0.083 moles of ethylethanoate and
0.083 moles of water.
Calculate the value of the equilibrium constant
Kc for this reaction.
A
The equilibrium is:
Page
244
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O
Q5
When 1 mol each of ethanol and ethanoic
acid are allowed to reach equilibrium at 60 0C,
what is the number of moles of
ethyl ethanoate formed?
C2H5OH + CH3COOH Ý CH3CO2C2H5 + H2O
Total pressure
That is, PT
1
x
PN2O4 = x/2
PNO2 = x =
ethyl ethanoate
Kc = 4 at 600C
A
C2H5OH + CH3COOH Ý CH3CO2C2H5 + H2O
initial
1
moles
eqlm 1-x
moles
Kc =
1
0
1-x
0
x
x
[ CH₃CO₂C₂H₅][ H₂O ]
[ C₂H₅OH ][ CH₃COOH ]
4 =
x x
V V
1−x 1−x
.
V
V
( )( )
where V is the volume of the container. Upon
simplification, V cancels out and the equation
simplifies to
3x2 -8x + 4 = 0
solving gives x = 2/3 = moles of ethylethanoate
formed.
Q6
At 600C and a total pressure of 1 atm,
N2O4 is 50 % dissociated.
N2O4(g) Ý 2NO(g)
Calculate Kp at this temperature
A
N2O4(g) Ý 2NO(g)
stoichiometry
1
initial partial
pressures
x
eqlm partial
pressures
x/2
:
2
0
atm
x
Page
245
(b)
CH3COOH + CH3CH2OH Ý CH3COOCH2CH + H2O
initial moles
0.1
0.1
eqlm moles
0.1 -x
0.1 -x
0.5
0.5 + x
0.5
0.5 + x
Construct an expression for Kc using the equilibrium
moles (divide the moles by V to convert them to
concentrations, but note that in this case V cancels
out). Equate your expression to 4, which is the
equilibrium constant of the reaction obtained in part
(a). Solve for x using the quadratic formula. You
should get x = 0.1. You can now find the moles of
each substance present at equilibrium.
8
A sample of 1 mole of sulphur dioxide was mixed
with 1 mole of oxygen in a 1 dm3 flask and allowed to
reach equilibrium according to the equation
2SO2(g) + O2(g) Ý 2 SO2(g)
At equilibrium, x moles of oxygen had reacted. What
is the value of the equilibrium constant, Kc, at this
temperature.
A
stoichiometry
initial moles
eqlm moles
2SO2(g) + O2(g) Ý 2SO3(g)
2
: 1
:
2
1
1- 2x
0
0
1-x
2x
If x moles of O2 react, then 2x moles of SO2 react
since the stoichiometry of the reaction is 1O2 : 2SO2.
Similary, 2x moles of SO3 are formed since O2 : SO3 =
1:2.
The equilibrium moles shown are equal to equilibrium
concentrations since the reaction vessel is 1 dm3.
Page
246
Kc =
[ SO₃ ]2
(2x)²
=
[O2 ][SO2 ]²
(1−x)(1−2x)²
Which statement about this equilibrium reaction is correct?
A Decreasing the temperature decreases the value of
Exercise
7.1
4.
The key reaction during the Contact process for
manufacturing sulphuric acid is as follows
the equilibrium constant.
1.
0.20 mol of hydrogen gas and 0.14 mol of iodine gas
were heated at a certain temperature and allowed to
reach equilibrium.
H2(g) + I2(g) Ý 2HI(g)
The equilibrium mixture was found to contain 0.26 mol
2SO2(g) + O2(g) Ý 2SO3(g) , ∆H = - 197 Kjmol-1
When a 2:1 ratio of sulphur dioxide and oxygen at
a total initial pressure of 3 atm is passed over a
catalyst at 4300C, the partial pressure of sulphur
trioxide at equilibrium is found to be 1.9 atm
(i) Calculate the partial pressures of SO2 and O2
B Decreasing the temperature increases the rate of
this reaction.
C
Increasing the pressure increases the value of the
equilibrium constant.
D The reaction is exothermic in the forward direction.
9701/1/M/J/2004
3.
Acetals are compounds formed when aldehydes are reacted
with an alchol and an acid catalyst. The reaction between
ethanal and methanol was studied in the inert solvent
dioxan.
H+
CH3CHO + 2CH3OH Ý CH3CH(OCH3)2 + H2O
ethanal
methanol
acetal A
The equilibrium concentration of the acetal product was
measured. The result is included in the following table.
at
start
at
equilibrium
at
equilibrium
(i)
[CH3CHO]
/moldm-3
[CH3OH]
/moldm-3
[H+]
/moldm-3
[acetal A]
/ moldm-3
[H2O]/
moldm-3
0.20
0.10
0.05
0.00
0.00
(o.20 -x)
x
0.025
Complete the second row of the table in terms of x,
the concentration of acetal A at equilibrium.
(ii) Using the [acetal A] as given, 0.025 moldm-3, calculate
the equilibrium concentrations of the other reactants and
products and write them in the third row of the table.
(iii) Write the expression for the equilibrium constant for this
reaction, Kc, stating its units.
Page
247
(iv) Use your values in the third row of the table to calculate
the value of Kc
9701/42/O/N/2011, adapted
7.2
Factors affecting chemical equilibria
Position of equilibrium
For a system at equilibrium (constant temperature), the value of the equilibrium constant remains the
same, despite changes in factors such as concentration, and pressure. However, the position of the
equilibrium is affected and shifts when these factors are changed.
Factors that affect position of equilibrium without changing the equilibrium constant are

pressure (gases)
 concentration ( solutions or gases)
What about temperature and presence of a catalyst?

Temperature affects both position of the equilibrium and the size of the equilibrium constant.

A catalyst does not have any effect on the position of equilibrium. This is because it speeds up both
the reverse and the forward reactions to the same extent. Consequently, size of the equilibrium
constant does not change.
The term ‘position of equilibrium’ refers to the relative rates of reaction for the forward and for the reverse
reaction. When we say the position of the equilibrium shifts to the right, it means that the forward reaction
becomes faster than the reverse reaction and so the proportion of products increases. Of course, whilst this
is happening, the system is temporarily in disequilibrium and the term equilibrium constant becomes
temporarily irrelevant. However, shifting of the position of equilibrium adjusts quantities of reactants and
products in the equilibrium so that equilibrium is achieved under the new conditions. When this happens,
it is found that the relative amounts of reactants and products at equilibrium have changed but the
equilibrium constant of the reaction has not changed (assuming temperature has remained constant). A
very important concept known as Le Chatelier’s principle is useful in explaining how changing conditions
may affect position of equilibrium.
Le Chatelier’s principle
Le Chatelier’s principle states that:
When a constraint (disturbance) is imposed upon a system at equilibrium, the position of the
equilibrium will shift in that direction which nullifies the constraint and re-establish
equilibrium under the new conditions.
7.2.1
Effect of pressure
Page
248
The position of a gaseous phase equilibrium may be affected by changing pressure. To predict the direction
in which the equilibrium shifts, it is necessary to construct a balanced equation and assess which side has
the greater or the smaller number of moles of gas.
Effect of increasing or decreasing pressure
Suppose that a system is at equilibrium and we impose a constraint by increasing pressure, e.g. by
compressing the gaseous mixture in a syringe . According to Le Chatelier’s principle, the position of the
equilibrium will shift in that direction which will nullify the constraint, that is, reduce pressure.
The position of the equilibrium therefore shifts to that side which forms the smaller number of gas moles,
because a reduction in the number of moles results in a reduction in pressure.
(Pressure is directly proportional to number of moles of gas. The fewer the number of moles, the smaller
the pressure and vice versa).
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 248
Consider the reversible dimerization of NO₂.
2NO₂ (g) ⇌ N₂O₄ (g)
red-brown
colourless
What will be the effect of increasing pressure on this equilibrium? The forward reaction has the smaller
number of moles. Increasing pressure will therefore favour the forward reaction. Reduction in the number
of moles implies a reduction in pressure, so that the constraint initially imposed on the system is removed.
Here we need to explain again the term ‘.. will favour ... the forward reaction.’ Sometimes chemistry
employs terms which are normally used to describe human actions, choices and feelings (For instance, we
may talk about the size of attractive force which an electron in an atom feels from the nucleus). These terms
are perfectly acceptable, but we should understand exactly what they mean in the context of chemistry.
When we say a condition will favour the forward reaction, we simply mean that the forward reaction
becomes temporarily faster, resulting in the formation of more products. After some time, the rates of the
forward and reverse reactions become equal once more. We say the reversible reaction has settled to a new
equilibrium under the new condition imposed, e.g. a higher pressure. In the NO₂/N₂O₄ equilibrium, the
formation of more N₂O₄ is accompanied by the fading of the red brown colour of NO₂, showing that more
of the NO₂ is reacting to form more of the colourless product, N₂O₄ (Fig 7.6). If pressure continues to
increase, the brown colour continues to fade. If pressure is now maintained at a high value, we have a new
equilibrium situation, comprising of more N₂O₄ than NO₂.
Fig 7.6 A syringe can be used to
investigate the effect of changing
pressure on the NO2/N2O4
equilibrium
Now suppose that pressure is reduced, for example, by gradually allowing the compressed gas to expand. It
is found that the brown colour of NO₂ begins to appear. This brown colour becomes denser as the pressure
continues to decrease. In this case, lowering of pressure is now favouring the reverse reaction, which has
the greater number of moles. By forming more moles of gas, pressure of the system increases so
that the constraint of reducing pressure is nullified.
A question which may also arise is this:
What is the effect of increasing or decreasing pressure on the value of the equilibrium constant?
Page
249
The answer is that there is no effect. The value of the equilibrium constant is not affected by changing
pressure. This is because the system responds in such a way that removes the constraint and maintains the
value of the equilibrium constant.
Q
Predict the effect of increasing pressure on the equilibrium
A
Changing pressure has no effect on the equilibrium. This is because the number of moles of gas
2HI(g) Ý H2(g) + I2(g)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 249
is the same on both sides of the equilibrium.
7.2.2
Effect of concentration
Consider the generalized equilibrium
A(aq) + B(aq) Ý C(aq) + D(aq)


Q
Increasing the concentration of a reactant, say A, will cause the equilibrium to shift to the right. By
immediately reacting to form a product (forward reaction), concentration of the reactant A is
reduced, that is, the constraint imposed on the equilibrium is removed.
Similarly, increasing the concentration of a product such as D will cause the equilibrium to shift to
the left so that the added molecules are removed by participating in the reverse reaction.
A reaction A + B Ý C is carried out with an initial concentration of 0.1 moldm-3 of A.
Equilibrium is achieved when the mass of product C has increased to x g.
Within the same axes, draw a graph showing the nature of the product curve when
the experiment is repeated with 2.0 moldm-3 A.
A
The equilibrium mass of C increases. Increasing concentration of a reactant causes the
equilibrium to shift to the right , so that more product is formed.
Page
250
The dichromate - chromate equilibrium
The Cr₂O₇2- (aq)/ CrO₄2⁻ (aq) system can be used to illustrate how changing concentration of a substance
affects the position of an equilibrium. The dichromate ion, which is orange in colour, coexists in an
equilibrium mixture with the yellow chromate ion:
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 250
Cr2O72-(aq) Ý 2CrO42-(aq)
orange
yellow
First consider the dichromate to chromate reaction (forward reaction). This is a hydrolysis reaction in
which a water molecule, acting as a reactant, is split.
O
O
Cr
Cr
O
O
O
2-
2-
O
+ H 2O
2
+ 2H +
Cr
O
O
O
O
O
dichromate- orange
chromate- yellow
In the reverse reaction, hydrogen ions cause two chromate ions to link through an oxygen bridge,
resulting in the formation of water.
2-
O
2H+ +
O
Cr
Cr
O
O
O
Cr
O
O
2-
O
O
+ H 2O
O
O
The dichromate - chromate equilibrium is therefore written in full as
Cr₂O₇2-(aq) + H₂O (l) ⇌ 2CrO₄2- (aq) + 2H⁺(aq)
When solid potassium dichromate is dissolved in water, the solution is mainly orange, showing that in this
equilibrium the predominant species is the dichromate ion. In other words, the equilibrium lies in favour
of dichromate. When NaOH (aq) is gradually added, the solution begins to turn yellow. The OH⁻ ions react
with hydrogen ions in the equilibrium to form water. Lowering of the concentration of a product, in this
case hydrogen ions, causes the equilibrium to shift to the right, to replace the hydrogen ions and remove
the disturbance caused by addition of alkali.
Cr₂O₇2-(aq) + H₂O (l) ⇌ 2CrO₄2- (aq) + 2H⁺ (aq)
Orange
Yellow
2OH-
Page
251
2H2O
When an excess of NaOH has been added, the solution turns completely yellow, showing that chromate ions
are now the predominant species under alkaline conditions. When the alkaline solution is acidified
gradually, e.g. using HCl (aq), the solution remains yellow for some time as the acid neutralizes the excess
alkali. Continued addition of an acid will increase the concentration of hydrogen ions on the product side
of equilibrium 1. This causes the equilibrium to shift to the left, so that the excess hydrogen ions are
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 251
removed. This causes the solution to become orange once more.
Even though the dichromate and the chromate ions coexist at equilibrium in aqueous solutions, the relative
stabilities, and hence amounts of each substance present at equilibrium depends on pH of the
solution. A low pH (acidic) favours the orange dichromate ion. A high pH (alkaline) favours the yellow
chromate ion.
7.2.3
Effect of temperature
To predict the effect of temperature on equilibrium, we need to know which side is endothermic (absorbs
heat energy) and which one is exothermic (releases heat energy)
Increasing temperature (heating) favours the side which is endothermic. This side absorbs the excess
heat applied, thus reducing the constraint of heat.
Decreasing temperature (cooling) favours the side which is exothermic. This side releases heat energy
to counter the constraint of cooling.
If the enthalpy change of reaction is zero, then a change in temperature will not have any effect on the
position of the equilibrium. The equilibrium composition remains unchanged.
We refer once more to the NO₂/N₂O₄ equilibrium.
NO₂ (g) ⇌ N₂O₄ (g) ... ∆Hr < 0
Red brown colourless
The enthalpy change of the forward reaction is exothermic (∆H <0).
An experiment to investigate the effect of temperature on this equilibrium can be carried out using the
apparatus shown in Fig 7.7. Increasing temperature favours the reverse reaction. This reaction, being
endothermic, removes the constraint of increasing temperature by absorbing the extra heat.
 The proportion of NO2 in the mixture increases whilst that of N₂O₄ decreases.
 The colour of the mixture intensifies as more of the red brown NO 2 is formed(tube 2)
Lowering temperature favours the forward reaction. This reaction, being
exothermic, releases heat energy and this nullifies the constraint of cooling.
 The proportion of N2O4 in the mixture increases whilst that of NO2
decreases.
 The brown colour of the mixture fades as more of the red brown NO 2
is consumed and more of the colourless N2O4 is formed (tube 1)
Page
252
Fig 7.7 A set up to show
the effect of changing
temperature on the
NO2/N2O4 equilibrium.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 252
Effect of temperature on the equilibrium constant
Increasing or decreasing temperature causes the value of the equilibrium constant
Fig 7.8 Effect of
temperature on the
value of the
equilibrium constant
to change. Consider a gas phase reaction A(g) + B (g) Ý C (g) which is endothermic
in the forward direction and whose equilibrium constant is K p. Increasing
temperature (T) favours the forward reaction, causing the proportion of products
to increase. In other words, the equilibrium constant K p of the forward reaction
increases whilst that of the reverse reaction decreases (Fig 7.8). Note that
increasing temperature also increases the total pressure in the reaction vessel. The
reaction vessel therefore has a larger proportion of the product C at higher
temperature and pressure.
7.2.4
Effect of catalyst
A catalyst does not affect the position of equilibrium. Neither does it change the
value of the equilibrium constant.
This is because it speeds up both the forward and reverse reactions to the same extent. A catalyst can not
increase yield of a product. What then is the use of a catalyst in equilibrium processes such as the Haber
process for the manufacture of ammonia? A catalyst reduces the time required to achieve equilibrium, that
is, it makes both the forward and the reverse reactions fast. Suppose that without a catalyst it takes 24 hours
for an equilibrium product yield of x g to be achieved. In the presence of catalyst, the same x g of product
will be produced but in a shorter time, say a few sconds.
7.2.5
Le Chatelier’s principle and electrode potentials
You probably recall that standard electrode potentials are quoted under standard conditions



concentrations of aqueous reagents = 1 moldm-3
pressure of gases = 1 atm
temperature = 298K
Standard reduction potentials are equilibria in which the forward reaction is reduction and the reverse
reaction is oxidation, for example
Cu2+(aq) + 2e- Ý Cu(s)
Eθ = +0.34V
The reduction potential of +0.34V strictly refers to 298K and 1 moldm -3 Cu2+(aq). If the concentration of
Cu2+ is increased above 1 moldm-3, the reduction potential becomes more positive than +0.34V. This
observation can be explained in terms of Le Chatelier’s principle. Increasing concentration of Cu 2+ pushes
the equilibrium to the right to remove the excess Cu2+ and restore equilibrium. In other, words, the forward
reaction becomes faster and the electrode potential becomes more positive.
Consider the anode and cathode half reactions in a fuel cell
anode (oxidation): H2(g) + 2OH–(aq) Ý 2H2O(l) + 2e–
Eθ = +0.83V... (i)
Page
253
cathode (reduction): O2(g) + 2H2O(l) + 4e– Ý 4OH–(aq) Eθ = +0.40V ... (ii)

The anode supplies electrons (reaction (i)) and so it is the negative electrode. Increasing
concentration of OH- ions pushes equilibrium (i) to the right, thus releasing more electrons. In
other words, the surface of the electrode becomes more negative. In terms of electrode potentials,
the Eθ value of reaction (i) becomes more positive, that is, the reaction becomes more feasible.
However, note that reaction (i) is an oxidation process which is the reverse of the reduction
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 253


reaction given in the Data Booklet. So if you were to think in terms of reduction potentials, you
would say that the electrode potential of reaction (i) becomes more negative.
The cathode receives electrons from the anode (reaction (ii)). In other words, the cathode is
deficient in electrons relative to the anode. This deficiency in electrons makes the cathode relatively
positive. Increasing concentration of OH- ions pushes equilibrium (ii) to the left so that the excess
OH- ions are removed. The reverse reaction becomes more important and since this reaction
produces electrons, it implies that negative charge begins to accumulate on the cathode. The surface
of the cathode therefore becomes more negative. In terms of electrode potentials, the Eθ value of
reaction (ii) also becomes more negative, that is, the forward reduction reaction
becomes less feasible.
The overall reaction taking place in the fuel cell is obtained by combining equations (i) and (ii).
The net cell potential is not affected by increasing concentration of OH - ions. This is because the
increase in [OH-] makes reaction (i) more feasible and reaction (ii) less feasible to the same
extend. The two effects therefore cancel out (Note that after balancing the number of electrons,
both equations contain 4OH- but on opposite sides of the equations. In that case it is easy to see
that increasing or decreasing concentration of OH - ions should not have any effect on the overall
cell potential).
Now consider the redox processes that take place in the lead-acid battery (‘car’ battery).
anode: Pb(s) Ý Pb2+ (aq) + 2e- ... (i)
cathode: PbO2(s) + 4H+(aq) + 2e– Ý Pb2+ (aq) + 2H2O(l) ...(ii)
The electrodes are lead metal (anode) and lead (IV) oxide (cathode) immersed in aqueous sulphuric acid.
Apart from acting as an electrolyte, sulphuric acid also facilitates reactions (i) and (ii) by combining with
Pb2+ ions to form insoluble lead (II) sulphate. Removal of Pb 2+ ions from equilibrium (i) pushes the equilibrium to the right, resulting in the release of more electrons. As a result, the surface of the lead anode
becomes more negative. The forward oxidation reaction becomes more feasible and the reverse reduction
reaction less feasible. If you think in terms of reduction potentials, you can say that the potential for reaction
(i) becomes more negative. Similarly, removal of Pb2+ ions from equilibrium (ii) forces it to the right and
the reduction potential becomes more positive.
Since both reactions (i) and (ii) are favoured by the removal of Pb2+ ions (these ions are on the same side
of the reactions) we can predict that the overal effect is to make the net cell potential more positive, that is,
the net reaction, obtained by combining equations (i) and (ii) becomes more feasible.
Exercise 7.2
(1) (a)
(b)
State Le Chatelier’s principle.
The key reaction in the Haber process for the manufacture of ammonia is
Page
254
N2(g) + 3H2(g) Ý 2NH3(g), ∆H < 0
(i) Explain which conditions of temperature and pressure (low or high) would favour a high
yield of ammonia.
In practice, a temperature of about 4500C is used, together with a catalyst of finely divided
iron.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 254
(ii) Explain whether or not this temperature agrees with your prediction in (b)(i) above.
(iii) Justify the need of a catalyst in the Haber process.
(2)
A key reaction in the Contact process for the manufacture of sulphuric acid is
SO2(g) + O2(g) Ý SO3(g), ∆H = -197 Kjmol-1
(a)
Predict which conditions of pressure and temperature (high or low) would favour a high yield of
sulphur trioxide.
(b)
In practice, the reaction is carried out at about 450 0C and 1 atm, in the presence of vanadium
(V) oxide as catalyst.
Suggest why a very low pressure is used
(3)
In relation to the following equilibria
Equilibrium I:
H2O (g) + C(s) Ý H2 (g) + CO(g); ∆H = +131Kj
Equilibrium II: 2CrO42-(aq) + 2H+ (aq) Ý Cr2O72- (aq) + H2O (l)
use Le Chatelier’s principle to predict and explain the effect of
(i)
increasing the pressure on equilibrium I
(ii)
increasing the temperatsingure on equilibrium I
(iii) increasing [H+] on equilibrium II
7.3
Ionic equilibria
We now focus on those equilibrium processes which involve participation of ions in aqueous solution
7.3.1
Acid - Base equilibria
Defining acids and bases
There are different ways of defining the terms acid and base. The most widely used definition is the
Brønsted-Lowry theory.
Page
255
According to the Brønsted-Lowry theory, an acid is a proton (hydrogen ion) donor and a base is a proton
acceptor.
All Brønsted-Lowry bases bear one or more lone pairs of electrons which are used to accept one or more
protons from an acid. Most neutralization reactions can be described using the Brønsted-Lowry theory.
Consider the reaction between HCl and NH3 to form ammonium chloride.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 255
NH₃ (g) + HCl (g) → NH₄Cl(s)
HCl donates a proton to ammonia, thereby acting as an acid. NH₃ uses its lone pair of electrons to accept a
proton from HCl. NH₃ therefore acts as a base. One advantage of the Brønsted-Lowry definition of acids
and bases is that it is not confined to aqueous solutions. For example, in the reaction above, both ammonia
and HCl are gases. A white ‘smoke’ of ammonium chloride is formed. This substance is a solid, but owing
to the very small size of its particles, it will appear as fumes which can settle as a white powder.
The Brønsted-Lowry definition includes substances which are usually not considered to be acids or bases.
Consider the reversible reaction between ammonia and water.
NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq) ... (i)
A water molecule donates a proton to an ammonia molecule. Water therefore acts as a Brønsted acid, and
ammonia as a Brønsted base. In most cases we would not consider water to be an acid.
Conjugate acid base pairs
A reversible Brønsted neutralization frequently contains an acid and a base on either sides of the
equilibrium. On the left hand side of equation (i) above, ammonia is a base and water is an acid. On the
right hand side, an ammonium ion donates a proton to the OH⁻ ion, thus promoting the reverse reaction.
The ammonium ion therefore acts as a Brønsted acid and the hydroxide ion as a Brønsted base.
NH₃ and its ion, NH₄⁺, form one conjugate acid-base pair. H₂O and its ion, OH⁻ forms the second conjugate
acid-base pair.
forward reaction: am m onia is base and water is acid
..
OH
-
reverse reaction : am m onium is acid and OH- is base
....
..
Page
256
The conjugate acid-base pairs involved are shown in Fig 7.9 .
conjugate acid-base pair
Another example of a Brønsted-Lowry
neutralization reaction is the reaction between
water molecules (Fig 7.10).
A
B
B
A
H₂O + H₂O ⇌ H₃O⁺ + OH⁻
NH3 + H2O
NH4+ + OH reaction,
which
is the autodissociation
Prepublication manuscript © L. MWANAWENYU 2011. Please This
do not
photocopy
or reproduce
Page 256of
water, is sometimes simplified as
A acid
conjugate acid-base pair
H₂O ⇌ H⁺ + OH⁻ ... I
B base
But the proton, being very reactive, will quickly
Fig 7.9
Combining equations I and II gives the overall equation
H₂O + H₂O ⇌ H₃O⁺ + OH⁻
In this reaction, one water molecule acts as a Brønsted acid, and the other as a Brønsted base. Also notice
the presence of conjugate acid base pairs.
conjugate acid-base pair
A
H2O +
B
H2O
B
OH - +
Q
A
H3O+
What are the two statements of the valence shell
electron pair repulsion theory? What changes in
shape and bond angle occur when H₂O is
converted to the oxonium ion?
conjugate acid-base pair
Fig 7.10
Acidity of the ammonium ion
Study again the reversible reaction between water and ammonia
H₂O (l) + NH₃ (aq) ⇌ NH₄+ (aq) + OH⁻ (aq)
We have seen that the ammonium ion is a Brønsted acid (it is the conjugate acid of ammonia). What this
means is that if we start with a solution of ammonium ions in water, they should show an acidic character.
Indeed, it is found that when ammonium chloride is dissolved in water the resulting solution is not
neutral (as might have been expected for a salt) but it is slightly acidic .
NH₄Cl (s) → NH₄+(aq) + Cl⁻ (aq)
pH < 7
The ammonium ion is a stronger acid than water, so it donates a proton to a water molecule, that is, water
acts as a base. A Brønsted-Lowry neutralization takes place as shown in Fig 7.11.
conjugate acid-base pair
Page
257
A
B
NH4+ + H2O
B
A
NH3 + H3O+
conjugate acid-base pair
The concept of pH and pOH
pH is defined as the negative logarithm of hydrogen ion
concentration.
pH = - log10[H⁺]
(In subsequent discussions, we will ignore the subscript
(base) 10 and simply assume its presence.)
Fig 7.11 acidity of the ammonium ion
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 257
It is therefore possible to calculate the hydrogen ion concentration when pH of the solution is known.
[H+] = 10-pH
pH is a measure of how acidic or alkaline a solution is. The concept of pH was developed using the
assumption that acidity is caused by hydrogen ions and alkalinity by hydroxide ions.
The pH scale runs from 0 to 14. On this scale, a substance with a pH of 7 is neutral. In such a solution, the
concentration of hydroxide ions is equal to the concentration of hydrogen ions.
pOH is defined as the negative logarithm of hydroxide ion concentration.
pOH = -log [OH⁻]
The concentration of OH- ions in a solution can therefore be calculated.
[OH-] = 10-pH
Note
The letter p can be interpreted as ‘...negative logarithm of...’. Thus pE means -log E (where E is the electrode potential)
An important relationship between pH and pOH is
pH + pOH = 14 (at 298K)
Example 1
Calculate
(a) The pH of 0.01 moldm-3 HCl
(b) The hydrogen ion concentration that gives a pH of 0
(c) The pOH of a solution whose pH is 4
(d) The hydroxide ion concentration in a solution in which [H+] = 3.0 x 10-6 moldm-3
Page
258
(e) The hydrogen ion concentration in a solution whose pOH is 11.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 258
Solutions
(a)
HCl  H+ + Cl0.01 0.01 0.01
(d)
pH = -log(3.0 x 10-6) = 5.5
Since HCl fully dissociates in water,
[HCl] =
[H+
pH = -log
[H+] = 3.0 x 10-6
pOH = 14 – 5.52 = 8.48
(aq)] = 0.01
[OH-] = 10-POH
[H+]
= 10-8.48
= -log 0.01
= 3 x 10-9 moldm-3
=2
(e)
(b)
pH = -log
[H+]
=
[H+]
pOH = 11, so pH = 3
[H+] = 10-3
10-pH
= 100
= 1 moldm-3
(c)
pH + pOH = 14
pOH = 14 - pH
= 14 - 4
= 10
Important note
Consider a solution whose pH = 1. This solution has a large concentration of H + ions and is therefore strongly acidic.
However, this does not mean that there are no hydroxide ions. When we say a solution is acidic, we simply mean that
the concentration of hydrogen ions is greater than the concentration of hydroxide ions. Similarly, in an alkaline
solution, the concentration of OH- ions is greater than that of hydrogen ions. In a neutral solution, [OH-] = [H+].
7.3.2
The ionic product of water
Water slightly autodissociates according to the equation
H₂O (l) Ý H⁺ (aq) + OH⁻ (aq) ... I
We may write the equilibrium constant for this autodissociation as
Keq =
[𝐻 + ][𝑂𝐻 − ]
[𝐻₂𝑂]
Since water is weakly dissociated, equilibrium I lies much to the left, that is, a very small amount of
water is ionized and its concentration therefore remains fairly constant. We may therefore write
Page
259
Keq =
[𝐻 + ][𝑂𝐻 − ]
𝐾
Where K = [H2O] = constant. This simplifies to
KeqK = [OH⁻][H⁺]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 259
The product of the two constants on the left hand side is another constant, known as the ionic product of
water, Kw.
Kw = [OH⁻][H⁺] = 1.0 x 10-14 at 298K
From this expression we can work out the units for Kw.
Units: moldm⁻3 x moldm⁻3 = mol2dm⁻6
Kw, like any other equilibrium constant, is affected by temperature. At 298K (25⁰C), the value of
Kw is 1.0 x 10 ⁻14 mol2 dm⁻6. How does Kw change when temperature is increased or decreased? The
enthalpy change for the reaction
H₂O (l) ⇌ H⁺ (aq) + OH⁻ (aq)
is endothermic. By Le Chatelier’s principle, the value of Kw is expected to increase as temperature increases,
that is, the extent of dissociation of water increases with increasing temperature. This is supported by the
following data.
Temp/0C
Kw
x 10-14
Table 7.1
Q
0
5
10
15
100
0.114
0.186
0.293
0.452
51.3
The effect of temperature on the
ionic product of water
Show that the pH of pure water is 7 at 298K.
x=
A
H₂O (l) ⇌ H⁺ (aq) + OH⁻ (aq)
Let [H+] = x = [OH-]
Kw = [OH⁻][H⁺] = 1.0 x 10-1
Page
260
1.0 x
10-14
=
x2
10-14mol2dm-6
= [H+] = 10-7
pH = -log[H+]
= -log (10-7)
= 7
If you try to calculate the pH of pure water at 100 0C (refer to Table 7.1), you will find that it is 6.1. The
statement, ‘the pH of pure water is 7’ is therefore not necessarily true, because the pH of water depends on
its temperature.
Can we therefore conclude that water is more acidic at 10o 0C than at 250C? By no means! A solution would
be acidic if it had a greater concentration of hydrogen ions than hydroxide ions. No matter what the
temperature of water is, the concentration of hydrogen ions is always equal to the concentration of
hydroxide ions, that is, pure water is always neutral. When temperature increases, the degree of
dissociation of water increases. The concentration of H + and OH- increase by the same amount. Since both
[H+] and [OH-] increases, the value of Kw also increases.
The relationship between pH, pOH and Kw
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 260
We know that
kw = [OH⁻][H⁺]
Taking the logarithms of both sides:
LogKw = log [OH⁻] + log [H⁺]
Multiplying throughout by -1 gives
- LogKw = -log [OH⁻] - log [H⁺]
But ‘-log’ is denoted by p, so
pKw = pOH + pH
But Kw = 10-14 at 298K, so
-log (10-14) = pH + pOH
Simplifying the left hand side gives
14 = pH + pOH
A problem with the pH scale
It is taught in junior school that the pH scale runs from 0 to 14. However, when you try to work out pH using some
values of [H⁺], you get unexpected answers. For example, let us calculate the pH of a solution whose [H⁺] is 2.0
moldm⁻3.
pH = -log 2 = -0.30
Here we have obtained a negative answer, which does not fit into the pH scale. However, notice that mathematically
the answer makes sense. If a number is greater than 1, then the negative of its logarithm would be a negative
number. Can we therefore conclude that the definition of the pH scale is wrong? To clear this paradox, we need to
take a brief look at the history of the pH scale.
In 1909, Soren Sorenson, a Danish scientist working for the Carlsberg beer manufacturing company, devised a
method that would allow him to work conveniently with very small concentrations of hydrogen ions. The
concentrations which Sorenson was working with where so small (for example, 0.000000035) that they were tedious
when used in calculations. He invented the pH scale in which he took the logarithm of the hydrogen ion concentration
and multiplied it by -1. For instance, instead of working with a hydrogen ion concentration of 10⁻⁵ mol/dm3 he took
the logarithm of the number, which gave -5. Multiplying the answer by -1 made the number positive. Since he was
working with very small concentrations of hydrogen ions, his method always gave him positive numbers between 0
and 14. These became the limits of his pH scale.
When we use the pH scale, we should therefore know that it applies to very small concentrations which are not greater
than 1. In routine calculations we may encounter hydrogen ion concentrations which are greater than 1. Taking the
negative logarithm of such numbers will give a negative pH value.
7.3.3
Acid strength
The strength of an acid refers to the extent to which it dissociates in water.



A strong acid is one which fully ionizes in water, that is, once it forms ions, the ions can not
recombine in the reverse reaction to reform the acid.
A weak acid is one whose dissociation in water is partial. Equilibrium is established between
the acid and its ions.
Let the notation HA represent a weak acid.
Page
261
HA ⇌ H⁺ + A⁻
At equilibrium, there are three species present; H⁺ ions, the anion A⁻ and undissociated acid molecules
HA.
An example of a weak acid is ethanoic acid, which undergoes partial dissociation in water according
to the equation
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 261
CH₃COOH (aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)
ethanoate ion
A common misconception
The acid dissociation constant, Ka
Ask several people this question:
Which one is the stronger of the two acids A
and B given below?
Acid
A
B
pH
2
4
Chances are that most people will think that
acid A is the stronger acid because it has the
lower pH. In fact, this is what is often taught in
junior science courses. The answer to the above
question is that we can not know which of the
two is stronger, unless if we know the acid
dissociation constants of both acids. pH is not a
measure of acid strength. It is simply a measure
of the concentration of hydrogen ions in the
solution. If the concentration of hydrogen ions
is very small, the pH of the solution will be quite
high. For instance, the pH of 1.0 x 10-5 moldm-3
HCl is 5. This is a relatively high pH, even
though HCl is a strong acid.
The acid dissociation constant, Ka
This is a measure of the strength of an acid.
The strength of an acid is determined by its Ka value, and not
by its pH.Consider the weak acid HA, partially dissociating in
water
HA Ý H⁺ + A⁻
We may write the equilibrium constant Ka as
Ka =
[𝑯+ ][𝑶𝑯− ]
[𝑯𝑨]
Ka is the acid dissociation constant of the acid, measured in
moldm-3 . If an acid is fully ionized, e.g. HCl, then at
equilibrium [HA] = 0 and the value of Ka is infinitely large.
The larger the value of Ka, the stronger the acid. Strong acids
have large Ka values which are well over 1, for example, the Ka
for nitric acid is 40. However, the fact that there is a
numerical value for the Ka of nitric acid implies that it is not
exactly fully ionised; there is a degree of reversibility to the
reaction. On the other hand, HCl is infinitely dissociated in
water, so no Ka value can be assigned for it.
Table 7.2 shows the Ka values of some acids. The acids have
been arranged in order of decreasing acid strength. As the
acids become weaker, the Ka values become smaller, but the
pKa ( = -log Ka) values become larger.
Dissociation of polyprotic (polybasic) acids
A polyprotic acid, also known as a polybasic acid, is capable of releasing more than 1 protons in solution,
for example, a diprotic acid such as H2SO4 is capable of releasing two protons (Table 7.3)

A polyprotic acid dissociates in steps, for example, sulphuric acid , which is diprotic, dissociates in
two steps.
H2SO4  HSO4- + H+
HSO4- Ý H+ + SO42-
Page
262
Each step has its own dissociation constant. The dissociation constant for the first step is the first
dissociation constant, Ka₁, and that for the second step is the second dissociation constant, Ka₂.

Ka₁, Ka₂ , K₃ (e.t.c) decrease in that order. Also
note that the second, third (etc) dissociations
of a strong
acid are
Take sulphuric
Prepublication
manuscript
© reversible.
L. MWANAWENYU
2011. Please do not photocopy or reproduce
acid as an example. The second dissociation step,
HSO4- Ý SO42- + H+ is reversible, showing that
HSO4- (hydrogen sulphate) is a weak acid.
Page 262
Acid
Ka/
moldm-3
pKa
HNO3
40
-1.60
H2SO3 (sulphrous
acid)
1.6 x 10-2
1.80
CH2ClCOOH
1.3 x 10-3
2.89
CHCl2COOH
3.29
HNO2 (nitrous
acid)
4.7 x 10-4
3.33
HCOOH(methanoic
acid)
1.78 x 10-4
3.75
C6H5COOH(benzoic
acid)
6.31 x 10-5
4.20
CH3COOH
1.74 x 10-5
4.76
NH4+
5.62 x 10-10
9.25
constants
Table 7.3 Dissociation nature
of some acids
Determination of Ka values
The Ka value of an acid (at a certain temperature ) can be worked out if both pH and concentration of the
acid are known. At 298K, 0.1 moldm-3 methanoic acid has a pH of 2.4. The acid dissociation constant can
be calculated as shown below.
moles at beginning
HCOOH Ý HCOO- + H+
0.1
0
0
moles at eqlm
Ka =
[HCOO-][H+]
[HCOOH]
x2
0.1 - x
x
x
...(i)
Page
263
=
0.1 - x
Assuming a volume of 1dm3,
concentration of HCOOH at the
beginning of the experiment is
equal to number of moles present.
Let x be the number of moles of
HCOOH that dissociate. Since the
reaction is 1:1:1, x moles of H+ and
x moles of HCOO- are produced.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 263
(3.98 x 10¯3 )²
0.1−(3.98 x 10¯3 )
but x = [H+ ] = 10-pH
(3.98 x 10¯3 )²
0.1
(since 3.98 x 10-3 ≈ 0)
= 10-2.4 = 3.98 x 10-3
Ka ≈ 1.58404 x 10-4 moldm-3
Substituting in equation (i) gives
Q
≈
Calculate the pH of
(i) 0.1 moldm-3 HCl
(ii) 0.1 moldm-3 CH3COOH (Pka = 4.76)
Explain why your answers to part (i) and (ii) are different.
A
(i)
HCl  H+ + Cl- ... (i)
[ ] after dissociation
0
0.1
0.1
pH = -log [H+]
= -log 0.1
=1
(ii)
CH3COOH Ý CH3COO- + H+ ... (ii)
0.1
0
0
initial moles
in 1 dm3
moles at eqlm
Ka =
0.1 -x
[H+ ][CH₃COO¯]
[CH₃COOH]
1.74 x 10-5 =
x
x
(where Ka = 10-Pka = 10-4.76 = 1.74 x 10-5)
x²
0.1−x
Since Ka is very small, x ≈ 0, so 0.1 - x ≈ 0.1, that is
1.74 x 10-5 =
x²
0.1
x = √1.74 𝑥 10¯5 𝑥 0.1
= 1.32 x 10-3 = [H+]
pH = -log [H+]
= - log (1.32 x 10-3)
Page
264
= 2.88
The pH of 0.1 moldm-3 HCl is lower than that of 0.1 moldm-3 CH3COOH. This is because CH3COOH is
a weak acid whereas HCl is strong. Equation (i) shows that HCl undergoes full dissociation, releasing
the maximum possible number of moles of H+ ions in solution. Equation (ii) shows that CH3COOH
dissociates only partially, so the concentration of H + in solution is smaller. Consequently, pH would
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 264
be higher.
Q
(a) Apple juice has a pH of 3.5.
Calculate the molar concentration of H+ in apple juice
(b)
A 25.00 cm3 sample of apple juice was exactly neutralized by 27.50 cm3 of 0.10
moldm-3 NaOH using phenolphthalein indicator.
Assuming that a single monobasic acid is present, calculate the molar concentration
of the acid in the juice.
(c)
(i) How can you explain the difference between the results in (a) and in (b)?
(ii) What constant can be determined from these results?
(iii) Calculate the numerical value for this constant.
9701/2/M/J/1990
A
(a)
pH = -log [H+] = 3.5
Conc of acid = moles / volume
[H+] = 10-pH = 10-3.5
=
= 3.16 x 10-4 moldm-3
0.00275
25
1000
= 0.11 moldm-3
(b)
moles of OH- = moles of acid
=
0.10 x 27.5
1000
= 0.00275
(c)(i)
Apple juce is a weak acid, that is, it only partially dissociates so the concentration of of H +
ions is small. During neutralization, the acid dissociates completetly, so a larger
concentration of H+ is obtained.
(ii/iii)
Ka =
[H+ ][A¯]
[HA]
but [H+] = [A-] since acid is monobasic
so Ka =
(3.16 x 10¯4 )²
0.11
[HA]eqlm ≈ 0.11 since the acid is weak.
= 9.07 x 10-7 moldm-3
Page
265
Q
1.0 moldm-3 of a monoprotic acid HA has a pH of 1. Which of the following statements are
true about the acid ?
1. It is a strong acid
2. The acid dissociates completely in water whose pH is greater than 7
3. The acid only partially dissociates in water whose pH is 7
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 265
A
2 and 3 are correct.
1. The acid is not strong. If it were strong, it would dissociate fully in water and its pH would be
-log 1.0 = 0. The fact that the pH of the acid is higher shows that it does not dissociate fully in
water, that is, it is a weak acid.
2. Presence of alkali (pH >7) causes the acid to dissociate fully. In pure water (pH =7), the acid
partially dissociates
HA Ý H+ + AIn the presence of an alkali, H+ ions are removed from the right hand side of the equilibrium as
water. By Le Chatelier’s principle, this causes more of the acid HA to dissociate to replace the H+
ions.
Degree of ionization of acids, α
The degree of ionization (dissociation) of an acid, α is given by
α=
amount ionized
initial amount
... (i)
This can be converted to % by multiplying by 100.
For a strong acid initial amount ≈ amount ionized, so α ≈ 1 (≈ 100%). For a weak acid,
initial amount > amount ionized, so α is a fraction less than 1 (0< α<1).
Q
A weak acid HA dissociates in water according to the equation
HA (aq) Ý H+(aq) + A- (aq)
2 moles of the acid were dissolved in water and allowed to reach equilibrium with its ions.
If the acid is only 60% dissociated, calculate the acid dissociation constant of the acid.
A
initial moles
eqlm moles
α=
x
Page
266
2
7.3.4
HA Ý H+ + A2
0
0
2-x
x
x
amount ionized
initial amount
60
60
= 100
[H+ ][A¯]
Ka =
[HA]
=
=
x²
2−x
1.2²
2−1.2
= 1.8
= 100 solving gives x = 1.2
Weak bases
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 266
A weak base is one which only partially dissociates in water. Ammonia is a typical weak base. When added
to water, a reversible reaction takes place.
NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)
The equilibrium mixture contains a large concentration of unreacted ammonia molecules. Compare with a
strong base such as NaOH. In water it dissociates completely and the concentration of NaOH drops to
zero.
NaOH(s) → Na⁺ (aq) + OH⁻ (aq)
Compare the pH of 0.1 moldm⁻3 ammonia and 0.1 moldm⁻3 sodium hydroxide. For NaOH, the calculation
is straightforward
NaOH(s) → Na⁺(aq) + OH⁻(aq)
initial moles 0.1
moles at end
0
0
0
0.1
0.1
At the end, all the NaOH has dissociated, and the concentration of OH⁻ ions is equal to 0.1.
pOH = -log[OH⁻]
= -log 0.1 = 1.
pH = 14-pOH = 14 – 1 = 13.
For ammonia
initial moles
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
0.1
0
0
equilibrium moles 0.1-x
x
x
In this case [OH-] at equilibrium is not equal to initial [NH3] because not all of the ammonia dissociates.
For us to calculate the pH of the solution, we need to know the value of x, that is, the amount of ammonia
molecules that actually dissociate in water, which is equal to the amount of OH - present at equilibrium. We
can not possibly know the value of x unless we know the extent to which the ammonia dissociates, which is
given by the base dissociation constant.
The base dissociation constant, Kb
The base dissociation constant, Kb, gives a measure of the extent to which a base dissociates in water. In
other words, it is a measure of base strength. The larger the value of Kb, the stronger the base. A strong
base such as NaOH has an infinitely large Kb.
Consider the weak base ammonia
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ for which the equilibrium constant Kc is given by
Page
267
Kc =
[NH4+][OH-]
[NH3][H2O]
...(I)
But the concentration of water remains fairly constant since a very small amount reacts with ammonia.
Treating [H₂O] as a constant and transferring it to the left hand side of equation I, we get a new constant
on this side, which is the base dissociation constant, Kb.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 267
Kb =
[NH4+][OH-]
[NH3]
Kb is sometimes converted to pKb, just for convenience.
pKb = -log Kb
When you use pKb you must remember that the larger the value of pKb, the weaker the base.
Weak bases such as ammonia have very small Kb values. This implies that at equilibrium, a large amount
of the base remains undissociated. The Kb value for ammonia is 1.67 x 10⁻⁵ moldm⁻3. Its pKb value is given
by -log (1.67 x 10⁻⁵) = 4.78.
Example
Calculate the pH of 0.1 moldm-3 ammonia.
Solution
Start
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
0.1
0
0 moles
Equilibrium
0.1-x
Kb =
x
x
x = moles of ammonia that dissociate
= moles of NH4+ and OH- that are formed
[NH4+][OH-]
[NH3]
1.67 x 10⁻⁵ = x2/0.1-x
≈ x2/0.1
judging by the very small value of Kb, x is very small
so that 0.1 -x ≈ 0.1
x = √ (1.67 x 10⁻⁵ x 0.1) = [OH⁻]
that is [OH⁻] = 1.29 x 10-3
pOH = -log[OH-] = - log (1.29 x 10-3)
= 2.89
pH = 14-pOH = 14-2.89
Page
268
= 11.11
The relationship between Ka, Kb and Kw
In any aqueous solution, the product of [H+ ] and [OH-] is constant at a particular temperature and is
known as the ionic product of water.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 268
Kw = [H+ ] [OH-]
It can be shown that Ka x Kb = [H+] [OH-] That is,
K a x K b = Kw
Example
Given that pKa (NH4+) = 9.25 at 298K, calculate the value of Kb for ammonia.
Solution
5.62 x 10-10 x Kb(NH3) = 10-14
ka(NH4+) = 10-9.25
= 5.62 x 10-10
ka(NH4+) x Kb (NH3) = Kw
Kb(NH3) =
10-14
5.62 x 10-10
= 1.78 x 10-5 moldm-3
7.3.5 Buffer solutions
A buffer is a solution which resists change in pH when a little acid or base is added. Buffers are designed to
keep pH nearly constant in some processes which are sensitive to changes in pH.
Acidic versus alkaline buffers
An acidic buffer is designed to maintain an acidic pH, that is, pH < 7. It consists of a weak acid and its
soluble salt in aqueous solution.
pH of the buffer depends on the exact proportion of acid to salt.
An alkaline buffer is designed to maintain an alkaline pH, that is, pH > 7. It consists of a weak base, such
as NH3, and its soluble salt, for example,NH4Cl, in aqueous solution.
pH of the buffer depends on the exact proportion of base to salt
The pH of an acidic buffer
Consider the buffer made up of the weak acid HA and its salt Na+A-. Since the Na⁺ ions will exist as
spectator ions, we ignore them and take the anion A⁻ to represent the salt.
The weak acid dissociates partially
HA ⇌ H⁺ + A⁻
Page
269
Now, Ka =
[𝐻 + ][𝐴⁻]
[𝐻𝐴]
Making [H⁺] subject of the formula
[H⁺] = Ka
[𝐻𝐴]
[𝐴− ]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 269
Taking the logarithms of both sides
[𝐻𝐴]
log [H⁺] = log{ Ka
[𝐴− ]
}
log [H⁺] = logKa + log
- log[H⁺] = - logKa - log
pH = pKa - log
[𝐻𝐴]
[𝐴− ]
[𝐻𝐴]
[𝐴− ]
and multiplying throughout by -1.
and using the definition p = -log...
[𝐻𝐴]
[𝐴− ]
[𝐴𝑐𝑖𝑑]
That is, pH = pKa - log
OR
[𝑆𝑎𝑙𝑡]
pH = pKa + log
That is, pH = pKa + log
... I
[𝐴⁻]
[𝐻𝐴]
[𝑆𝑎𝑙𝑻]
[𝐴𝑐𝑖𝑑]
... II
The highlighted characters are meant to provide a
mnemonic. When you use equation (II), which has a +
sign, then the salt should be in the numerator.
Equations I and II are different mathematical versions of the same formula, known as the
Henderson – Hasselbach equation. You may use either of the formulae to calculate the pH of a buffer
when the ratio of salt concentration, [A⁻] to acid concentration [HA] is known. The concentrations [Salt]
and [Acid] can be expressed either in moldm-3 or as gdm-3.
Example 1
What should be the ratio of ethanoic acid to sodium ethanoate in a buffer whose pH is to be
maintained at 6? (pka of ethanoic acid = 4.73)
Solution
Page
270
You may quote the Henderson - Hasselbach equation without further proof.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 270
PH = pKa + log
That is , pH -pka = log
[𝑆𝑎𝑙𝑡]
[𝑆𝑎𝑙𝑡]
Let
[𝐴𝑐𝑖𝑑]
[𝑆𝑎𝑙𝑡]
[𝐴𝑐𝑖𝑑]
be x
1.24 = log₁₀x
[𝐴𝑐𝑖𝑑]
x = 101.24 = 17
But pH = 6 and pka of ethanoic acid is 4.76
[𝑆𝑎𝑙𝑡]
6 - 4.76 = log
[𝑆𝑎𝑙𝑡]
[𝐴𝑐𝑖𝑑]
= 17
[𝐴𝑐𝑖𝑑]
That is, [Acid]: [Salt] = 1:17
Example 2
A buffer solution is prepared as follows :
53.5 g of NH4Cl are dissolved in 400 cm3 of 15.0 moldm-3 NH3 and the mixture is diluted to 1.00 dm-3.
(a)
Calculate
(i) [NH4+(aq)]
(ii) [NH3(aq)] of the prepared buffer solution.
(b) Use one of the following values to calculate the pH of the prepared buffer solutin :
NH4+ , Ka = 6.00 x 10-10 moldm-3
NH3(aq) , Kb = 1.67 x 10-5
9701/3/M/J/1995
Solution
(a)(i)
moles of NH4Cl = m/Mr =
53.5
53.5
(ii) moles of NH3 = C x V =
= 1.oo
∴ [NH4+] = 1.00 moldm-3
(b)
Using NH4+ :
The ammonium ion behaves as an acid
according to the equation
+
Ka =
[NH₃][H+ ]
[NH₄]⁺
1 000
= 6.00 moles
∴ [NH3] = 6 .00 moldm-3
rearranging
[H+] = Ka x
[NH₄⁺]
[NH₃]
= 6 x 1o-10 x
=
1
6
10-10
pH = -log (10-10) = 10
Page
271
NH4 Ý NH3 +
H+
15 x 400
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 271
Alternatively
Using NH3 :
Aqueous ammonia contains the base
ammonium hydroxide
[OH-] = Kb x
[NH₃]
[NH₄⁺]
= 1.67 x 10-5 x
NH3(aq) + H2O(l) Ý NH4OH(aq)
NH4OH(aq) Ý NH4+(aq) + OH-(aq)
Kb =
[NH₄⁺][OH¯]
[NH₃]
(NH3 = NH4OH])
= 1.002 x
1
6
10-4
pOH = -log (1.002 x 1o-4) = 4
but pH + pOH = 14,
so pH = 14 - 4 = 10
Example 3
Calculate the pH of a buffer solution prepared by mixing 500cm3 of an aqueous solution containing
34g of sodium ethanoate and 500cm3 of a solution containing 2g of ethanoic acid (pKa = 4.76).
Solution
Moles of salt (CH3COO-Na+)
= m/Mr
= 34/82
= 0.415
[salt] =
0.415
0.5
When the two solutions are mixed, volume is doubled
and so concentration of acid and salt are
halved, that is
[salt] = 0.830/2 = 0.415
[acid] = 0.0667/2 = 0.03335
pH = pKa +log
= 0.830
Moles of acid (CH3COOH)
= 2/60
= 0.0333
[acid] =
0.0333
0.5
= 0.0667
[salt]
[acid]
pH = 4.76 + log
[0.415]
[0.03335]
= 5.85
= 4.76 + log 12.45
= 5.86
The buffer is acicid so its pH should be less than 7.
Calculate the pH of a solution whose concentration with respect to ethanoic acid and with respect
to sodium ethanoate is 0.40 moldm-3 and 0.20 moldm-3 respectively.
Page
272
Example 4
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 272
Solution
Ratio of salt to acid is 0.2 : 0.4 = 1:2.
Substituting into the Henderson-Hasselbach equation,
pH = 4.76 + log(1/2) = 4.46
How does an acidic buffer work?
A buffer resists changes in pH when a little acid or base is added to it. The mode of action of an acidic buffer
can be illustrated by the CH₃COOH/CH₃COO⁻Na⁺ system.
The first step is to write the equations of the processes which exist in the buffer before an acid or an alkali
is added.
Ethanoic acid is partially dissociated, and the salt, sodium ethanoate fully ionizes.
CH₃COOH (aq) ⇌ CH₃COO⁻ (aq) + H⁺ (aq) ... I
CH₃COONa
→ CH₃COO⁻ (aq) + Na⁺ (aq) ...
II
When the concentration of hydrogen ions increases (that is, when a little acid is added), the ethanoate
ions in reaction II remove the excess hydrogen ions.
CH₃COO⁻ (aq) + H⁺ (aq) → CH₃COOH ... III
The salt CH₃COONa is the important source of ethanoate ions since it fully ionizes in solution. There are
also ethanoate ions from equation I, but they are present in very small amounts (ethanoic acid dissociates
weakly) and are therefore not significant in removing the hydrogen ions. Reaction III is a buffering reaction. Such a reaction should not readily regenerate the species it has removed from solution. In this case,
the removed species is H⁺ ions. They are not readily returned to the solution because the ethanoic acid
formed is a weak acid; it does not readily release a proton. Hydrogen ions are thus effectively removed from
the solution and this prevents the pH of the solution from decreasing.
When pH of the solution starts to increase (that is, upon addition of a little alkali), the excess hydroxide
ions are removed by the protons in reaction I, to form water.
H⁺ (aq) + OH⁻ (aq) → H₂O (l)
This is an effective buffering equation because the water formed only slightly dissociates and so does not
readily give back the hydroxide ions to the solution. pH of the solution is thus prevented from increasing.
Alkaline buffers
The mode of action of an alkaline buffer can be illustrated using the NH₃/NH₄Cl buffer system.
The reactions which exist in the buffer are
NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq) ... I
Page
273
NH₄Cl (aq) → NH₄⁺ (aq) + Cl⁻ (aq) ... II
When excess hydrogen ions are introduced into the system, they are neutralized by the OH⁻ ions present
in reaction I.
H⁺ (aq) + OH⁻ (aq) → H₂O (l)
When excess hydroxide ions are introduced into the system, they are removed by the ammonium ions in
equation II.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 273
NH₄⁺ (aq) + OH⁻ (aq) → NH₃ (aq) + H₂O (l)
Applications and uses of buffers
Some chemical processes are sensitive to changes in pH. Such reactions should therefore take place in a
buffered environment to reduce wide fluctuations in pH.

Citric acid and a little NaOH are added to some mixtures to produce a ‘pH adjuster’.
The NaOH partially neutralizes the citric acid to produce the salt sodium citrate. A buffer is thus
formed, containing a weak acid (citric acid), and its salt, sodium citrate. The structure of citric acid
and its salt are shown below. On average, only one –COOH group is neutralized (not necessarily
the one shown) since NaOH is added in very small amounts.
CH2COOH
CH2COOH
HO
C
COOH
HO

COO-Na+
CH2COOH
CH2COOH
citric acid
C
sodium citrate
Baby lotions are buffered to prevent or minimize nappy rash. This skin condition arises when
bacteria present in faeces and urine convert some chemicals, for example urea in urine, to
ammonia.
CO(NH2)2(aq) + H2O(l)  2NH3(aq) + CO2(aq)
It is the ammonia produced in this reaction that irritates the skin, causing a rash to appear. The
production of ammonia increases pH on the surface of the skin, which causes the bacteria to
proliferate, since they grow well in a slightly alkaline environment. Baby lotion is buffered at
around pH 6, which kills the bacteria or prevents them from multiplying.

HCO3- in the form of soluble salts such as NaHCO3 , plays a vital role in the buffering of blood. This
salt actually works with its corresponding weak acid, carbonic acid, H 2CO3.
NaHCO3  Na+(aq) + HCO3- ...(I)
H2CO3 Ý H+(aq) + HCO3-(aq) ...(II)
When the concentration of H+ ions begin to increase, for example in rapidly respiring tissues,
HCO3- helps to prevent lowering of pH by removing the excess hydrogen ions.
HCO3- + H+  H2CO3
This prevents poisoning of tissues by acid (acidosis).
Page
274
When pH becomes too alkaline , the excess OH- ions are removed by the hydrogen ions in
equilibrium (II).
7.4
Acid - Base titrations
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 274
The technique of titration can be used to find the quantity or concentration of acid or base that reacts during
neutralization .
7.4.1
Acid – base indicators
Acid-base indicators are substances which change colour according to the pH of the solution in which they
are. They are frequently used to identify the end point in acid -base titrations.
Most indicators are weak organic acids in which the dissociated form (anion) and the undissociated form
have different colours. For example, methyl orange is red in its undissociated form and yellow in its
dissociated form.
undissociated form of methyl orange (red)
Fig 7.12
dissociated form (yellow)
The dissociated and undissociated forms of methyl orange
We will represent the undissociated form as HMe and the dissociated form as Me⁻. In solution, the following equilibrium exists:
HMe ⇌ H⁺ + Me⁻ ... (I)
red
yellow
When the pH is low (high [H⁺]), the reverse reaction is favoured (Le Chatelier’s principle) and so the
solution would be red due to the presence of HMe. This is the colour of methyl orange in an acicic
environment. Now suppose that an acidic solution containing methyl orange is titrated with an alkali such
as NaOH. Some hydrogen ions are removed from the right side of the equilibrium (I). This causes the
equilibrium to shift to the right, resulting in the production of more of the dissociated form of methyl
orange. At the equivalence point, the colour of the indicator is halfway between the red undissociated form
and the yellow dissociated form, that is [HMe] = [Me⁻]. The colour at the equivalence point is therefore
orange (mixture of red and yellow). When one more drop of NaOH is added beyond the equivalence point,
the solution in the titration flask becomes alkaline and the yellow form becomes predominant. For an
indicator to be useful, it must give a rapid colour change when one more drop of titrant (from burette) is
added to the analyte (in titration flask) past the equivalence point. Methyl orange, phenolphthalein and
bromothymol blue are frequently used as acid- base indicators because they give sharp colour changes at
the end-point of the titration.
Page
275
Q
A
What is the ratio of [Me⁻] to [HMe] in solution at a pH of 2 (pKa for methyl orange
is 3.7)?
Let the undissociated form of the indictor
be HIn.
2 = 3.7 + log
[In− ]
[HA]
[In− ]
log
= 2 – 3.7 = -1.7
[HA]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
[𝐈𝐧− ]
[𝐇𝐀]
=
10-1.7
= 0.02 (2/100)
Page 275
HIn (aq) Ý H+ (aq) + In-(aq)
Using the Henderson-Hasselbach
equation
pH = pKa + log
[In¯]
[HIn]
...(I)
This calculation confirms that for methyl orange, the ratio of the dissociated form, In⁻ to the undissociated
form, HMe, is very small in acidic medium. In other words, at equilibrium, there is a very small amount of
the dissociated form and more of the undissociated acid. Thus at low pH, methyl orange would be red in
colour, since the red form, HMe, is more predominant.
Q
A
At what pH does methyl orange change colour during an acid-base titration? (Pka for
methyl orange is 3.7)
Colour change takes place approximately at the equivalence point. At this point, [HA] =
[A-]. The ratio
[In− ]
[HA]
is therefore equal to 1 at the equivalence point.
Using the Henderson-Hasselbach equation,
pH = pKa + log
[In¯]
[HIn]
pH = Pka + log 1
pH = pKa = 3.7
In other words, the pH at which an indicator changes colour is given by the pKa value for that indicator.
Range of an indicator
The calculation above shows that methyl orange will change colour from the red form to the yellow form,
or vice versa, when the pH of the solution is 3.7. In fact, 3.7 is the mid-point of the range of pH over which
methyl orange will change colour.
Page
276
The range of an indicator is the range of pH over which it will change colour.
The three common indicators mentioned above have a pH range of about two units. That is, as a rule of
thumb, the visible colour change takes place about 1 pH unit on either side of the Pka value. For methyl
orange we calculated that the mid point of its range is 3.7(=pKa). pH therefore starts to change at a pH of
about 3.7 -1 = 2.7, and will continue to change until the pH is about (3.7 + 1) = 4.7. The range for methyl
orange is therefore approximately 2.7 – 4.7. The actual range is shown in Table 7.4. Since we know the pKa
values for the other indicators, we can estimate their ranges.
For the indicator to be useful in an acid-base titration, its pH range must be in the vertical region of sharp
pH change on the titration curve. The indicator must also be able to change colour rapidly at the end point. Different indicators have different ranges of pH because they have different acid-dissociation
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 276
constants. However, different indicators may be suitable for the same titration if they both have ranges
within the region of sharp pH change of the titration curve.
An important note
Some readers may suppose that all indicators change colour when the pH of the solution is 7. This is not true. Of the
indicators given in Table 7.4, only bromothymol blue changes colour at around pH 7. Methylorange changes colour at
a pH well below 7 and phenolphthalein changes colour at a pH well above 7.
Colour change
Indicator
pKa
Range(from pKa-1
to pKa + 1)
Acid
Alkali
Yellow
Methyl orange
3.7
3.1-4.4
Red
Bromothymol blue
7.0
6.0-8.0
Yellow
Phenolphthalein
9.0
8.0-10.0
Colourless
Blue
Red
Table 7.4 The ranges and colours of three common indicators. Note that
the colourless form of phenolphthalein persists even when the solution
becomes alkaline; it will only start to change at a pH of about 8 and is
completely changed to red when pH becomes 10.
7.4.2
Titration curves
Titration curves help to visualize what exactly happens during the course of an acid-base titration. A
titration curve is obtained from a plot of pH of solution in flask (analyte) against volume of titrant from
burette. As the titrant is gradually added to the conical flask, neutralization takes place between the titrant
and the analyte, and this causes a change in pH of the solution in the flask. The exact trend of this change
in pH gives a titration curve. The shape of the titration curve depends on the relative strength of the acid
and the base used in the titration, but all titration curves generally have a sigmoid shape (roughly S-shaped).
At the equivalence point of the titration, the two forms of the indicator are present in equal amounts,
giving a composite colour, for example, orange for methyl orange and green for bromothymol blue.
In all of the titrations below, an alkali is added from the burette to a flask containing an acid. The titration
curve therefore starts at a low pH but gradually rises until it is above pH 7.
During the titration, it is essential that the rapid change in colour be caused by a single drop of titrant;
otherwise the end point would be overshot. For example, at the end point using phenolphthalein, the colour
should be faint pink and not red. A red colour shows that too much alkali has been added from the burette.
There are four types of acid-base titrations.
Page
277
Type I : Strong acid against strong base
Titrant: 0.1 moldm-3 NaOH (aq)
Analyte: 50 cm3 of 0.1 moldm-3 HCl (Fig 7.13)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 277
Fig 7.13
Titration set up used to obtain a titration curve
The initial pH
This is found using concentration of HCl before the titration is carried out.
pH = - log [H⁺] = -log 0.1 = 1
The titration curve will therefore begin at a pH of 1 (Fig 7.14). Titration curves are usually drawn so that
they start from a low pH and increase towards alkaline pH values. This is the reason why an acid, and not
an alkali, was put into the flask. However, there is nothing wrong with doing the titration the other way
round, that is, placing the acid in the burette and the alkali in the titration flask. Of course, this will invert
the shape of the titration curve; it now starts at a high pH above 7 and decreases as acid is added to the
flask.
During the titration: Before the equivalence point is reached
The alkali is added gradually to the acid in the conical flask. A neutralization reaction occurs, to form
water and salt.
NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)
When spectator ions are ignored, the reaction is found to be simply
OH⁻ (aq) + H⁺ (aq) → H₂O (l)
We therefore expect pH to remain fairly constant during this stage of the titration, because the sole product
of the reaction, H2O, is neutral.
At the equivalence point
The number of moles of HCl in the flask is given by
Page
278
C x V = 0.1 x 50/1000 = 0.005
Since the reaction is 1:1 with respect to the acid and the base, the equivalence point is when
exactly 0.005 moles of HCl have reacted with exactly 0.005 moles of NaOH.
The volume of NaOH discharged from the burette to achieve this equivalence point is given by
V = n/C = 0.005/0.1
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 278
= 0.05 dm3 = 50cm3
At the equivalence point, the solution in the flask is the neutral salt NaCl, in which [OH⁻] = [H⁺]. At this
point, pH is equal to 7. However, one or two more drops from the burette beyond 50 cm3 will suddenly
cause the concentration of hydroxide ions to be higher than the concentration of hydrogen ions. This will
cause the indicator to change colour rapidly as the pH increases sharply. This sharp increase in pH corresponds to the steep part of the titration curve (Fig 7.10). The final volume of titrant which causes a
sudden colour change of the indicator is the end point volume. This volume is slightly larger than the equivalence volume because it is achieved by adding one or two drops of titrant past the equivalence point. The
end point volume is therefore a good approximation of the volume of the titrant required for complete
reaction to take place.
The end point pH is found from the mid point of the region of sharp pH change on the titration curve.
Continued addition of titrant
This will make the solution even more alkaline so that pH continues to increase. However, the increase in
pH beyond the equivalence point is only gradual because addition of the titrant to the flask also causes
dilution of the solution. Dilution implies that the increase in the concentration of OH⁻ ions is very small
and so pH should remain fairly constant.
pH when 55.00cm3 of NaOH has been added
The alkalinity of the solution at this point is due to the 5.00cm3 of NaOH that has been added beyond the
end point volume of 50.00cm3. The number of moles of NaOH present in these 5.00cm3 of solution is given
by C x V = 0.1 x 0.005 dm3 = 0.000 5
The total volume of solution at this stage (volume of NaOH + volume of HCl) is now 55 cm 3 + 50cm3
= 105 cm3 = 0.105dm3. The concentration of OH⁻ ions is then given by
C = n/V = 0.000 5/ 0.105
= 4.762 x 10⁻3
Calculating pOH at this point of the titration:
pOH = -log[OH⁻] = -log (4.762 x 10-3)
= 2.32
pH = 14-2.32 = 11.67
Page
279
Thus
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 279
Fig 7.14 Titration curve for the titration of 50cm3 of 0.1 moldm -3 HCl and 0.1 moldm-3
NaOH. Take note of key points which are discussed in the main text; the equivalence volume,
the end-point pH and the pH when 55cm3 of titrant have been added.
When we calculate the pH at higher volumes, say at 70.00 cm3 and 80.00 cm3, we find that the pH is only
slightly higher than 11.67. This is shown by the leveling out of the graph in Fig 7.10
The pH should gradually approach the pH typical of the solution of a strong base, NaOH.
However, as mentioned, the graph does not really reach pH 14 with practical volumes of titrant, because of
continued dilution of the solution in the flask.
Choice of indicator
Any indicator which changes colour within the region of sharp pH change can be used. Methyl orange,
bromothymol blue and phenolphthalein all have pH ranges within this region, so any one of them may be
used.
Page
280
Type II : Strong acid against weak alkali
Suppose that we start with 50.00 cm3 of 0.1 moldm-3 HCl in the flask and titrate it with 0.1 moldm-3
ammonia (a weak alkali). The titration curve obtained is similar to the one above, but there are important
differences (Fig 7.15)
The reaction that takes place before the equivalence point has been reached is
NH₃ (aq) + HCl (aq) → NH₄⁺ (aq) + Cl⁻ (aq)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 280
The end point volume of NaOH (aq) is 50.00cm⁻3 as before. There are two important differences of the
titration curve to the one obtained for a strong acid and a strong base.
The highest pH achieved is lower and the region of sharp pH change is shorter. After the end point an excess
of a weak base, ammonia, is being added to the titration flask. The highest pH that can be achieved is
therefore lower than what would be obtained with a strong base. The decrease in the height of the region of
sharp pH change has two important consequences.


Phenolphthalein, whose pH range is now outside the region of sharp pH change, becomes useless
as an indicator. This is because it will not start to change colour until about pH 8.
The pH of the solution at the equivalence point is now slightly less than 7 because the NH₄Cl
solution present at this point is slightly acidic. The ammonium ion, being the conjugate of a weak
base (NH₃), is slightly acidic in nature, as shown by the equation below.
NH₄⁺ (aq) + H₂O (aq) → NH₃ (aq) + H₃O⁺
It is the presence of the oxonium ions (H₃O⁺) that causes the solution to be acidic.
Choice of indicator
Fig 7.15
Titration curve for a strong acid - weak base titration
Page
281
You have noticed that the decrease in the height of the region of rapid pH change puts phenolphthalein out
of range and therefore renders it useless. If you check with Fig 7.15, you will notice that methyl orange and
bromothymol blue are still in the region of sharp pH change, so they can be used as indicators in the titration
of a strong acid with a weak base.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 281
Type III: weak acid against strong base.
The titration curve in Fig 7.16 below was obtained when 50.00cm3 of 0.1 moldm⁻3 ethanoic acid (weak acid)
was titrated with 0.1 moldm⁻3 NaOH.
Fig 7.16
Titration curve for a weak acid and strong base
Calculating the initial pH
The following equilibrium exists in the flask before the base is added.
CH3COOH (aq) Ý CH3COO-(aq) + H+ (aq)
At equilibrium, a certain amount of the acid, say, x, moles, has dissociated. The amount of the acid present
at equilibrium is therefore 0.1-x. The amount of ethanoate and hydrogen ions on the product side at
equilibrium is each equal to x. Thus
Ka = x2/ (0.1-x)
Page
282
1.7 x 10⁻⁵ = x2/(0.1-x)
The very small value of Ka shows that the acid is very weak and x is approximately equal to zero, that is,
0.1-x ≈ 0.1.
∴ 1.7 x 10⁻⁵ ≈ x2/0.1
solving gives x ≈ 1.30384 = [H⁺]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 282
pH = -log 1.30384
= 2.88
Notice that the initial pH is higher than in the two previous cases where the acid used was strong.
During the titration
The reaction taking place is
CH₃COOH (aq) + NaOH (aq) → CH₃COO⁻Na+ (aq) + H₂O(l)
As soon as the alkali hits the acid in the flask, it is neutralized to form the salt sodium ethanoate and
water. Before the equivalence point, pH increases only slightly because of the buffering effect of a mixture
of ethanoic acid and sodium ethanoate.
The titration curve obtained would be similar to the one obtained for titration of a strong acid against a
strong base. The key difference is that the lower limit of the curve has been raised and the height of the
region of sharp pH change has been reduced. This causes methyl orange to go out of range so that it is
useless as an indicator. It would change colour well before the equivalence point has been reached. Also
notice that the end point pH is now above 7. That is, the solution present at the end point is slightly alkaline.
Ethanoate ions which are present at the equivalence point are slightly alkaline, giving a pH greater than 7
in solution. The phenomenon which causes a solution of the salt sodium ethanoate to be
acidic is known as salt hydrolysis. The ethanoate is a conjugate of a weak acid, so it has significant
alkalinity due to the hydrolysis reaction
CH₃COO⁻ (aq) + H₂O (l) → CH₃COOH (aq) + OH⁻ (aq)
It is the presence of hydroxyl ions that causes the solution to be slightly alkaline.
The final pH
Beyond the equivalence point, there is an excess of the strong base, sodium hydroxide in the flask, so the
final pH would be high, but it will not increase to 14 because of dilution.
Choice of indicators
The fact that the minimum pH of the curve is raised renders methyl orange useless. However, both
phenolphthalein and bromothymol blue have pH ranges within the vertical region of sharp pH change. Any
one of these may be used.
Hint
Just remember the titration curve for a strong acid and a strong base. In this case, all three indicators work, the order
from the bottom (lower pH) is methyl orange, bromothymol blue, and phenolphthalein. When titrating a strong acid
with a weak base, the maximum pH that can be attained is reduced. The graph is lowered and phenolphthalein becomes
useless whilst the other two remain effective. When titrating a weak acid with a strong base, the starting pH is raised,
so that methyl orange becomes useless, whilst the other two remain effective.
Page
283
Determination of Ka
When exactly half of the weak acid in the titration flask has been neutralized, the flask now contains
equimolar amounts of the acid, CH3COOH and its salt, CH3COO-Na+. This happens at ½ of the equivalence
volume (= 25 cm3). The mixture in the flask is therefore a buffer and it causes pH to change only gradually
when a small volume of alkali is added from the burette. The K a value of the weak acid can
be found using the Henderson-Hasselbach equation
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 283
pH = pKa + log
[Salt]
[Acid]
But at ½ of the equivalence point [Salt] = [Acid], so
[Salt]
[Acid]
=1
∴ pH = pKa
The value of Ka can therefore be found from the pH at this point.
pH = -logKa
and Ka = 10-pH
Type IV: Weak acid against weak base
Let us use the titration curve of a strong acid with a strong base as a reference point. The initial pH would
be very low because the acid, being strong, dissociates fully, supplying the maximum possible amount of
hydrogen ions to the solution in the flask. Similarly, the final pH is high because the base, which is now in
excess, is strong and is able to dissociate fully, supplying a large amount of hydroxide ions to the solution.
Since the vertical region of sharp pH change spans values of pH from very low to very high, all three
indicators have their ranges included in this vertical region, and any one of them can be used.
When the acid is weak and base strong, the initial pH is higher, and this has the effect of reducing the height
of the region of sharp pH change. We have already seen that methyl orange then becomes useless as an
indicator.
When the base is weak and the acid strong, the maximum pH is lowered and this also causes the region of
sharp pH change to be shorter. This renders phenolphthalein useless.
Now consider what happens when both the acid and the base are weak. The initial pH is raised, while the
final pH is lowered. Height of the vertical region of the curve is therefore much reduced. In fact, it is so
reduced that none of the three indicators is effective because an indicator requires a vertical portion of the
curve which is at least two pH units to give a sharp colour change.
The titration curve in Fig 7.17 was obtained for 0.1 mol/dm3 ammonia and 50 cm3 of 0.1 mol/dm3 .
ethanoic acid. The reaction taking place before the equivalence point is
CH₃COOH (aq) + NH₃ (aq) → CH₃COO⁻ (aq) + NH4⁺ (aq)
The pH at the end point is approximately 7. This is because the two ions present, CH₃COO⁻ and NH₄⁺ have
opposite effects which almost cancel each other out. CH₃COO⁻ hydrolyses in water, producing OH⁻ ions.
NH₄⁺ ions hydrolyze in water producing oxonium ions. The oxonium ions and hydroxide ions then
neutralize each other so that overally, the solution is neutral.
Page
284
H₃O⁺(aq) + OH⁻(aq) → 2H₂O(aq)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 284
Fig 7.17
7.4.3
Titration curve obtained for a weak base and a weak acid
Titrations involving polybasic acids
A polybasic acid can yield more than one hydrogen ions in solution. In pure water, only the first proton can
easily be released. Subsequent protons find it difficult to dissociate from the acid and enter solution because
they are attracted to the negative charge that now resides on the partially dissociated acid after loss of the
first proton. Consider the dissociation of H2SO4. The first proton is released according to the
equation
H2SO4  HSO4- + H+
The second proton is released according to the equation
Page
285
HSO4- Ý SO4 2- + H+ ... (i)
This process is difficult because it requires loss of a positively charged particle (H +) from a negatively
charged ion (HSO4-). However, in the presence of an alkali such as NaOH, the OH - ions neutralize H+ ions
and remove them from the right hand side of equilibrium (i) as water. This promotes further dissociation
until the acid loses all of its hydrogen atoms. There is therefore step wise neutralization of the acid by the
base, and this causes the titration curve to have more than one regions of sharp pH change. The number of
such regions is equal to the number of dissociable hydrogen atoms on the acid. For sulphuric acid, there are
two regions of sharp pH change, and for phosphoric acid, H3PO4, there are three. For phosphonic acid,
H3PO3, there are only two regions of sharp pH change, because the third hydrogen atom is not bonded to
oxygen but to phosphorous (H3PO3 is diprotic) .This hydrogen atom can not be neutralized.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 285
Consider what happens when 50cm3 of 0.1 moldm-3 phosphoric acid, H3PO4 is titrated with 0.1 moldm-3
NaOH (aq). This acid has three acidic protons, so the titration curve has three distinct regions of sharp pH
change, corresponding to three separate neutralization reactions (Fig 7.18). These reactions are labeled (i),
(ii) and (iii) in the following discussion.
The first hydrogen atom is neutralized according to the equation
H3PO4 (aq) + OH-(aq)  H2PO4- (aq) + H2O (l) ... (i)
The reaction is 1:1, so
moles of NaOH required to reach end-point = moles of acid that react
= Cacid x Vacid
= 0.1 x
End point volume =
=
50
1000
= 0.005
moles of NaOH
concentration of NaOH
0.005
0.1
= 50 cm3
When 50cm3 of NaOH have been added, end point of reaction (i) is reached and the titration curve rises
sharply. The end point pH for reaction (i) is acidic because the product formed in the flask (H2PO4-) is still
acidic. Methyl orange, whose pH range is on the acidic side of the pH scale, would be suitable as an indicator
for the first step of the neutralization.
After the first end point, H3PO4 (aq) has been used up and only H2PO4-(aq) remains. A second
neutralization reaction begins, according to the equation
H2PO4-(aq) + OH-(aq)  HPO42-(aq) + H2O (l) ... (ii)
The end point volume for reaction (ii) is also 50 cm3. By the time the second end point is reached, 100cm3
of NaOH would have been added to the titration flask. The end point of this second step is alkaline
because the product formed, HPO42- (aq) undergoes salt hydrolysis, producing OH- ions.
HPO42- + H2O  H2PO4 - + OHPhenolphthalein, whose pH range is between 8 and 10, is a suitable indicator for the second neutralization
reaction (The two indicators, methyl orange and phenolphthalein, are put together in the titration flask
before the titration). Note that the region of sharp pH change associated with the second step is shorter,
because the acid that is reacting, H2PO4-, is weak.
At the end of the second reaction, the product that remains in the flask is HPO 42-. This is also acidic, so it
reacts with more alkali according to the equation
A further 50 cm3 of alkali is required to reach end point. The third region of sharp pH change on the titration
curve is very short and not so well defined because the acid being neutralized, HPO42-, is very weak. No
indicator is suitable here. After reaction (iii) is complete, the graph levels out because the final product,
PO43-, has no further reaction with NaOH. Addition of more NaOH from the burette does not cause any
significant increase in pH because of dilution.
Page
286
HPO42-(aq) + OH-(aq)  PO43- + H2O (l) ... (iii)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 286
(i) First end point pH.
Methyl orange turns from red to orange
(ii) Second end point pH.
Phenolphthalein changes from colourless to light pink.
(iii) Third end point is not well pronounced. No
indicator is suitable for this stage of the titraton
Fig 7.18 Titration
of a triprotic acid
with a strong acid
Q
The pH of 0.200 mol dm-3 NH3 is 11.3. 20.0 cm3 of this solution was titrated with 0.100
moldm-3 HCl. Sketch a graph to show how the pH changes during the titration. Mark clearly
the key points of your graph.
The initial pH (11.3) is that of the
solution in the titration flask
(ammonia). Recall that the solution
whose concentration is known (HCl)
is put in the burette. The pH in the
flask gradually decreases due to
addition of an acid.
A
The end point pH is slightly less than
7 because the ammonium ion present
at the end point is slightly acidic due
to salt hydrolysis.
Page
287
Note that there is enough information
in the question to allow calculation of
the end point volume (shown by the
broken vertical line).
7.5
Solubility and solubility product
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 287
7.5.1 Solubility product
We may wish to assess the extent to which an ionic compound dissolves in water. Consider a generalized
ionic compound AxBy dissolving in water reversibly to give its ions, A n⁺ and Bp⁻
AxBy (s) ⇌ xAn⁺ + yBp⁻
We may write the equilibrium constant, Keq, as
Keq =
[An+]x [Bp-]y
[AxBy(s)]
Keq =
[An+]x [Bp-]y
K
but
[AxBy(s)] is constant
Taking the constant K to the left hand side, we see that this side now has two constants multiplying each
other. This gives another constant, which we call the solubility product, K sp.
Ksp = [An+]x [ Bp-]y
For instance, the solubility product of Al₂O₃ refers to the equilibrium
Al₂O₃ (s) ⇌ 2Al3⁺ (aq) + 3O2- (aq)
Ksp = [Al3⁺ ]2[O2- ]3 ... I
The solubility product, Ksp, is a measure of the extent to which an ionic compound dissolves in water. Ksp
is a true equilibrium constant and it is valid only for a system in which equilibrium has been achieved, that
is, when rate of dissolution of the salt is equal to the rate at which the ions combine to form the salt.



The larger the value of Ksp, the greater the concentration of ions of the salt present in solution at
equilibrium and the more soluble the salt.
Ksp is constant for a particular salt at a specified temperature. The value of Ksp changes when
temperature changes.
The units of Ksp depend on the salt which is being studied. For Al₂O₃ above, the units may be
worked out from equation I above:
units: (moldm⁻3)2(moldm⁻3)3 = mol5dm⁻15

The value of Ksp depends on temperature. In most cases, the solubility product increases as temperature increases. In other words, most partially soluble salts become more soluble when temperature
is increased. This is because the dissolution of most partially soluble salts is endothermic, and so is
favoured by an increase in temperature (Le Chatelier’s principle)
Page
288
Table 7.5 shows the solubility products, Ksp, of some compounds at 298K.
Suggest reasons for the differences in the solubilities
Prepublication
manuscript
©pairs
L. MWANAWENYU
Please do not photocopy or reproduce
of the
following
of substances2011.
.
(a) BaSO4 and CaSO4
(b) Ca(OH)2 and Mg(OH)2
Page 288
Compound
Q
Solubility product
BaSO4
1.3 x 10-10
CaSO4
2.4 x 10-5
Ca(OH)2
5.5 x 10-6
Mg(OH)2
1.1 x 10-11
QAgCl
1.8 x 10-10
AgBr
7.7 x 10-13
Table 7.5 Solubility products of
some compounds
A
Ksp for PbI₂ is 7.0 x 10⁻⁹ mol3 dm-9 at 25 ⁰C. This very small value shows that PbI₂ is almost insoluble in
water at 25⁰C. In other words, there are very few ions in solution formed from the dissolution of the lead
iodide.
Determination of solubility product
This involves finding the concentration of the ions of the salt present at equilibrium, as illustrated in the
example below.
Example
A 20.0 cm3 sample of saturated, aqueous Ca (OH)2 required 18.0 cm3 of 0.050 moldm-3 HCl for
complete neutralization. Calculate
(a) The pH of the saturated solution
(b) The solubility product of calcium hydroxide, stating the units .
Solution
Being saturated, the solution may be treated
as being at equilibrium.
Page
289
Ca (OH) 2(s) Ý Ca2+ (aq) + 2OH-(aq) ... (I)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 289
(a)
Moles of HCl used = C x V
= 0.05 x
18
[H+] =
Kw
[OH¯]
=
10¯¯¹⁴
0.0450
1000
= 2.222 x 10-13
= 9.0 x
10-4
pH = -log[H+]
= -log (2.222 x 10-13)
Since HCl reacts with OH- in
the ratio of 1:1,
= 12.65
moles of OH- = moles of HCl
= 9.0 x 10-4
[OH-] = n/V
=
9 x 10¯⁴
(b)
From equation (I)
[Ca2+] = ½ [OH-]
= ½ x 0.0450
20/1000
= 0.0225 moldm-3
= 0.0450
These are the moles of OH- ions
present in the saturated solution.
Now Kw = [OH-][H+] = 10-14
Q
7.5.2
Ksp = [Ca2+][OH-]2
= 0.0225 x (0.0450)2
= 4.56 x 10-5 mol3 dm-9
State a use of calcium hydroxide which depends on its solubility in water .
Solubility
Solubility is defined as the amount of solute that dissolves in 1 dm3 of solvent, usually water, at a given
temperature. Suppose that for a particular solute, only 2g can dissolve in 1 dm3 of water at 298K. The
solubility of the salt at this temperature is therefore 2 gdm-3. Alternatively, the solubility can be given in
moldm-3. It does not matter how much of the solute is added to 1 dm3 of water, only 2g will dissolve, provided
the temperature is the same. The amount of solute that can dissolve in a solvent is affected by the volume
of solvent used, for example, more sugar will dissolve in 1 dm3 of water than in 50 cm3 . This is why it is
important to define solubility with reference to a fixed volume of solvent. The following are important
factors that affect solubility (for a fixed volume of solvent).
Page
290
Temperature
For most partially soluble salts, solubility increases with increasing temperature. Of course, this also
increases the solubility product by increasing the number of moles of ions present in solution.
Presence of ions in common with the salt
Consider PbI2 as an example. It will dissolve better in pure water than in a solution that already contains
some Pb2+ or some I- ions. The decrease in solubility of PbI2 in the presence of Pb2+ or I- is an example of
the common ion effect.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 290
7.5.3
The Common Ion Effect
This is the observation that the solubility of an ionic compound in a solvent is reduced by the presence of
ions which are common to that ionic compound.
Consider again the dissolution of PbI2 in water
PbI2(s) Ý Pb2+ (aq) + 2I-(aq) ... (I)
Now suppose that the water already contains I- ions. Presence of these ions will, by Le Chatelier’s principle,
cause equilibrium (I) to shift to the left, thus removing the extra iodide ions which were already present in
solution. In this way, the solubility of PbI2 is reduced and the equilibrium constant, that is, Ksp, is kept
constant.
Example
Calculate the solubility of AgCl in
(a)
pure water
(b)
in 0.1 moldm-3 NaCl(aq) [Ksp(AgCl) = 1.8 x 10-10)
Explain why the value you obtained in (b) is different from that obtained in (a).
Solution
Initial
AgCl(s) Ý Ag+ (aq) + Cl-(aq)
1
0
0
moles/dm3
Eqlm moles
1-x
x
x
(x are moles of AgCl that dissolve)
Ksp = [Ag+][Cl-]
1.8 x 10-10 = (x) (x+0.1) ... (I)
1.8 x 10-10 = x2 + 0.1x
x2 + 0.1x - 1.8 x 10-10 = 0
(a)
Ksp =
[Ag+][Cl-]
Let
[Ag+]
=x=
Solving for x using the quadratic formula
[Cl-]
at equilibrium
1.8 x 10-10 = x2
x = √( 1.8 x 10-10) = 1.34 x 10 -5
Solubility = 1.34 x 10-5 moldm-3
(Since x is moles of AgCl that dissolve)
x =
=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−0.1± √(−0.1)2 −[4 𝑥(−1.8 𝑥 10¯10 )]
= 3.6 x
2
10-9
Solubility = 3.6 x 10-9 mol2dm-6
AgCl(s) Ý Ag+(aq) + Cl-(aq)
Initial moles/dm3 1
0
0.1
Eqlm moles
1-x
x
x+0.1
Page
291
(b)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 291
Alternatively
(b)
1.8 x 10-10 = (x)( 0.1)
AgCl(s) Ý Ag+(aq) + Cl-(aq)
Initial moles 1
0
0
/dm3
Eqlm moles 1-x
x
x+0.1
Ksp = [Ag+][Cl-]
1.8 x 10-10 = (x) (x + 0.1)
Since the Ksp of AgCl is very small, the value of x is
very small, so small that (x + 0.1) is approximately
equal to 0.1, that is
7.5.4
x = (1.8 x 10-10)/ 0.1
= 1.8 x 10-9 = solubility of AgCl
This is a good and acceptable estimation
of x (= solubility of AgCl).
The solubility of AgCl in 0.1 moldm-3 HCl is
therefore many times smaller than the
solubility in pure water. This is caused by the
common ion effect
Saturated solutions and precipitation processes
A saturated solution is one that has dissolved so much solute that it can not dissolve any more. You have
probably used limewater to test for the presence of carbon dioxide. Limewater is in fact a saturated solution
of Ca(OH)2. This solution is not capable of dissolving any more Ca(OH)2. In this solution, the
product of ions [Ca2+][OH-]2 is exactly equal to Ksp.
Ca(OH)2(s) Ý Ca2+(aq) + 2OH-(aq) ...(I)
Ksp = [Ca2+][OH-] 2 = 5.5 x 10-5 mol3 dm-9
Suppose that some Ca2+ ions are added to the saturated solution. The product [Ca 2+][OH-] 2 is now greater
than Ksp. This causes the equilibrium (I) to shift to the left, removing the excess Ca2+ ions and resulting in
the formation of a precipitate of Ca(OH)2. In this way Ksp remains constant.
Precipitation of an ionic compound occurs if the product of its ions in solution, raised to their
respective powers, exceeds the value of Ksp .
Example
Equal volumes of 0.001 moldm-3 Pb(NO3)2 and 0.002 moldm-3 KI are mixed. Predict whether or
not a precipitate would be formed (Ksp of PbI2 = 1.0 x 10-9).
Solution
Page
292
PbI2 is the solid expected to form. When the two volumes are mixed, concentration of both Pb(NO 3)2 and
KI are halved.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 292
[Pb2+] =
[I-] =
0.01
2
0.02
2
= 0.0050
= 0.01
PbI2(s) Ý Pb2+ (aq) + 2I-(aq) ... (i)
[Pb2+] [I-]2 = 0.0050 x 0.01 = 5.0 x 1o-5
Since [Pb2+] [I-]2 > Ksp, a precipitate will be
formed)
Formation of a precipitate reduces the concentration
of ions on the left side of equilibrium (i), thus
keeping Ksp constant.
Ksp (PbI2) = [Pb2+][I-]2
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise 7.3
solubility and solubility product
text box.]
1.
The solubility of barium hydroxide, Ba(OH)2 , at 250C is 0.24gdm-3.
(a) (i)
Calculate the molar concentration of the aqueous solution
(ii) If the solute is completely ionized, calculate the hydroxide ion concentration of the
solution.
(iii) Calculate the pH of the saturated aqueous barium hydroxide
(b)
(i) Write an expression for the solubility product of barium hydroxide.
(ii) Calculate the solubility product of Ba(OH)2, stating its units
(c)
Bottles containing aqueous barium hydroxide need to be kept firmly stoppered or a white
deposit forms on the surface.
(i) Name the white deposit.
(ii) Write an equation to show how the white deposit is formed.
9189/2/M/J/2011
2
(a)
Barium ions are poisonous. Patients with digestive tract problems are sometimes given
an X-ray after they have swallowed a ‘barium meal’, consisting of a suspension of BaSO4 in
water. The [Ba2+(aq)] in a saturated solution of BaSO4 is too low to cause problems of toxicity.
Page
293
(i)
Write an expression for the solubility product, Ksp, for BaSO4, including its units.
(ii) The numerical value of Ksp is 1.30 × 10–10. Calculate [Ba2+(aq)] in a saturated
solution of BaSO4.
(iii) The numerical value of Ksp for BaCO3 (5 × 10–10) is not significantly higher than that for
BaSO4, but barium carbonate is very poisonous if ingested. Suggest a reason why this
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 293
might be so.
(b)
A useful commercial source of magnesium is sea water, where [Mg2+ (aq)] is 0.054
moldm–3. The magnesium is precipitated from solution by adding calcium hydroxide.
Mg2+ (aq) + Ca(OH)2(s)  Ca2+ (aq) + Mg(OH)2(s)
(i)
Write an expression for the Ksp of Mg(OH)2, including its units.
(ii)
The numerical value for Ksp is 2.00 x 10–11. Calculate [Mg2+ (aq)] in a saturated
solution of Mg(OH)2.
(iii)
Hence calculate the maximum percentage of the original magnesium in the seawater
that this method can extract.
9701/04/M/J/2003
Page
294
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 294
CHAPTER
8
Reaction Kinetics
Introduction
The central question in Reaction Kinetics is
How fast...?
In this topic we study reaction rates and the factors that affect them. We also study quantitative
methods of determining reaction rates. In Chapter 5 we discussed the energetic point of view of chemical
processes, in which the key question is ‘Is the reaction feasible?’
In this chapter we look at another point of view, the kinetic point of view. A reaction is said to be
energetically feasible if it has a negative enthalpy change, ∆H. Whether or not the reaction actually
takes place is beyond the scope of chemical energetics; this is a kinetic issue. Think of a boulder resting
on top of a hill. Is it energetically feasible for the rock to roll downward? The answer is clearly, yes. But
how many boulders do we see rolling down hills? In practice the event might not take place because of
the presence of an energy barrier. For example, the boulder might be stuck to the ground, and to initiate
the rolling movement, a certain amount of energy must be applied first to loosen the rock. Similarly, a
reaction might be energetically feasible but in practice it may fail to take place, or will take place very
slowly, because of the presence of a kinetic barrier. The minimum energy required to overcome this
kinetic barrier and initiate the reaction is known as activation energy.
The concept of activation energy is very important in chemistry and its applications. If a reaction has a
very high activation energy, then a large amount of energy is required to force it to take place. In
industry, this implies large consumptions of energy, therefore increasing production costs. The use of
high temperatures might also weaken equipment such as pipes, leading to high maintanance costs and
increased risks of explosions or leakages. The use of large amounts of heat to drive reactions can be
avoided by using catalysts, which work by lowering the activation energy of a reaction.
Page
295
8.1
Rate of reaction
This is the decrease in reactant concentration or the increase in product concentration in a given time. The
term ‘time’ is essential in this definition. Rate is always how much a certain quantity changes in a given
time.
The unit of rate of reaction is moldm-3 s-1.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 295
Rate of reaction =
𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐞 𝐢𝐧 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭 𝐜𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧
𝐭𝐢𝐦𝐞
=
𝐦𝐨𝐥𝐝𝐦¯³
𝐬
= moldm-3 s-1
OR
Rate of reaction =
8.1.1
𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐞 𝐢𝐧 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭 𝐜𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧
𝐭𝐢𝐦𝐞
=
𝐦𝐨𝐥𝐝𝐦¯³
𝐬
= moldm- 3s-1
concentration/time graphs
Fig 8.1 below shows the nature of a concentration/time graph. Graph (a) is for the consumption of a
reactant during the reaction, and graph (b) is for the formation of a product.
Fig 8.1


Graphs showing how concentration of reactant decreases and how
concentration of product increases with time.
Concentration of reactant decreases with time as the reactant is being consumed. The slope of the
graph is therefore negative, that is, the graph is down-sloping. However, note that the decrease in
concentration is not linear, but curvilinear. The slope of the curve, which gives the rate of
reaction, is not constant. Rate of reaction depends on concentration of reactant.
At the beginning of the reaction, reactant concentration is high, and so the reaction is fast and
concentration of reactant drops rapidly. This explains why the slope of the curve is initially steep.
As the reaction progresses, concentration of reactant drops and this makes the reaction slower.
Conversion of reactants to products becomes slower and the slope of the curve becomes less steep.
Eventually, the concentration of reactant approaches zero, provided the reaction is
irreversible. A large number of reactions have a degree of reversibility if carried out in a closed
system. In such reactions, concentration of reactant does not drop to zero.
Concentration of product increases during the reaction. Again, the curve is not linear. The slope of
the curve represents the rate of reaction. The steeper the slope, the faster the reaction. At the
beginning, the reaction is fast since the concentration of reactants is high. Products are thus rapidly
formed. This explains why the first part of the graph is steep. As the reaction progresses, rate of
reaction becomes slow as reactants are used. The curve therefore begins to level out.
In a zeroth order reaction, rate of reaction is not affected by changes in reactant concentration. In other
words, rate remains constant when concentration of reactant is increased or decreased. The slope of
the concentration time/graph is therefore constant, that is, the graph is linear, as shown in Figure 8.2.
Page
296
Zeroth order reactions
Graphical determination of rate of reaction
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 296
The rate of a reaction can be determined from a suitably
drawn concentration/time graph. However, for most
reactions, rate of reaction changes with time, as already
explained, so we talk about the instantaneous rate of
concentration
of reactant
time
Fig 8.2
Concentration / time graph
for a zeroth order reaction
Rate of reaction at time, t
concentration
of reactant
= slope of tangent at that time
x
=
y
t
time
=
𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐲
𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐱
change in reactant concentration
change in time
Fig 8.3 Finding rate of reaction at
time t using a concentration/time.
Alternatively, the rate can be obtained from a product concentration/ time graph.
8.1.2
Rate / time graph
The shapes of the graphs above have been explained in terms of the decrease in rate of reaction with time.
For any reaction, the rate/time graph is therefore as shown in Fig 8.4.
rate
time
Fig 8.4
graph
rate/time
Initially the rate of reaction is high since the concentration of reactants is
still high. Rate of reaction depends on how frequently reactant particles
collide. When concentration of reactants is high, frequency of collisions is
also high and the reaction is correspondingly fast. The rate of reaction
decreases with time, that is, the rate/time graph is down sloping. This is a
result of the decrease in reactant concentration. However, this decrease is
not uniform, so the graph is not a straight line but is curvilinear. The
consumption of reactants is rapid at the beginning of the reaction, so at
approximately time zero, the curve is very steep. The curve becomes less
steep with time because the speed at which reactant molecules are
consumed decreases.
Rate α
1
t
Suppose that reaction A occurs in 10 seconds and reaction B in 40 seconds, then
Page
297
The inverse (concave) nature of the graph also illustrates an important point. Rate of reaction is inversely
related to time. The larger the time it takes for the reaction to take place, the slower the
reaction. Conversely, a reaction which occurs within a very small amount of time is fast, that is
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 297
Rate of reaction A α
1
10
= 0.1
Rate of reaction B α
1
40
= 0.025
Reaction A is therefore four times faster than reaction B.
8.2
Factors that affect rate of a reaction
When we discuss factors that affect the speed of a reaction, we should make a clear distinction between one
chemical reaction carried out under different conditions and two different chemical reactions. Different
reactions proceed at different speeds, even when the reaction conditions are the same. This observation can
be explained in terms of the different activation activation energies of the reactions.
A specified reaction will proceed slowly or rapidly, depending on the following factors:





Concentration of reactant or pressure of a gaseous reactant
Temperature
reactant particle size(surface area)
Presence or absence of a catalyst
Light(for some reactions)
A concept known as the collision theory is useful in explaining these factors.
8.2.1 The collision theory
The collision theory sates that for reactant molecules to react, they must first collide with sufficient
energy and with the correct orientation.
For a reaction to take place, reactant particles must first collide, but the collisions must be energetic enough
to overcome the activation energy of the reaction and form a product. A collision which actually results in
a product being formed is known as an effective collision. The speed of reaction depends on the frequency
of effective collisions. The more frequent they are, the faster the reaction.
..
..
Page
298
The orientation factor is also important. This is about the direction in which reactant particles
approach each other. Orientation factors are more significant if the reactant molecules are large. Such molecules may have certain reactive centres which must approach each other if the collision is to be successful.
Consider the condensation reaction between two amino acids, which are the building blocks in all proteins.
For the reaction to take place, the –COOH group on one amino acid must collide with the
–NH2 group on the second amino acid. This is the orientation that favours a reaction (see case 1 below)
Case 2 is an example of how the two amino acids may collide with a wrong orientation. In this case, the two
amino acids actually repel each other because of the lone pair of electrons on each N atom.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 298
..
..
Effect of temperature
The speed of most reactions increases as temperature increases.
The effect of temperature can be explained in terms of the collision theory. Increasing temperature results
in an increase in the kinetic energy of the reactant particles, and this increases the frequency of effective
collisions.
Effect of concentration and pressure
Increasing the concentration of a reactant will also increase the speed of reaction. The higher the
concentration, the greater are the chances of effective collisions. Increasing pressure of reactant in a gas
phase reaction also increases rate of reaction. When pressure of the gas is high, particles are close and they
collide frequently. Note that increasing concentration or pressure does not make collisions effective; it
simply increases the chances of collisions. For the collisions to be effective, temperature must be increased
to make the collisions more energetic.
Effect of particle size
Reducing the particle size of a reactant exposes a greater total surface area for reaction. This increases the
speed of reaction. A well known example are explosions which were often reported in flour mills. The very
fine nature of the flour particles greatly increased the speed of combustion reactions (An explosion is a very
rapid reaction which results in a sudden increase in the volume of gaseous products). If reactant particles
are course or large, only particles on the surface can react. In other words, the total surface area of the
reactant is reduced.
8.3
The concept of activation energy
The collision theory says that for reactant particles to react they must possess enough energy.
When particles collide, they must do so with energy greater than or equal to the activation energy, otherwise
the reaction will not take place.
Activation energy may be visualized as an energy barrier to a reaction. The higher the activation energy of
the reaction, the higher the energy barrier, and the slower the reaction.
Page
299
Activation energy is the minimum energy which reactant particles must possess in order
to collide and form a product.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 299
Energy profile diagrams
An energy profile shows both energetic and kinetic information about a chemical reaction. Energetic
information is shown in terms of relative energy levels of reactants, and kinetic data is shown in terms of
the magnitude of the activation energy, Ea. Contrast with an energy level diagram which shows only
energetic data. Fig 8.5 shows an energy profile for an exothermic and for an endothermic reaction.
The concept of activation energy helps to explain
why some reactions are very slow or do not take
place at all, despite being energetically feasible. An
example is the reaction of peroxodisulphate ions
with iodide ions.
energy
Ea
reactants
S₂O₈²⁻ (aq) + 2I⁻ (aq) → SO₄2-(aq) + I₂ (aq)
 H(negativ e)
products
progress of reaction
energy
Ea
reactants
products
H(positiv e)
Despite being energetically feasible, the reaction is
so slow that a mixture of the two ions can stay for a
very long time without showing any appreciable
reaction. This is due to the high activation energy of
the reaction. It is so high that very few particles can
achieve it. As a result, the number of particles that
have sufficient energy for reaction at any given time
is very small. The high activation energy of this
reaction is expected since the two reactants are both
negatively charged and so they naturally
repel. A very large amount of energy is
required to force the ions to collide and react.
progress of reaction
Fig 8.5 Energy profile for (a) an exothermic
reaction (b) an endothermic reaction
The Maxwell-Boltzmann curve
The distribution of molecular speeds in a gaseous reactant can be shown by the Maxwell-Boltzmann curve
(Fig 8.6). In the sample, the particles possess a range of kinetic energies, from lowest to highest. Only
molecules whose energy is greater than or equal to activation energy are able to form products. These are
the particles whose collisions are energetic enough. In fig 8.6, Ea2 is the activation energy for a relatively
slow reaction. Since the activation energy is relatively high, very few particles, represented by area y, can
afford the energy required to react. For a faster reaction, the activation energy is smaller. Consequently,
more particles, represented by the area (x + y), can achieve the energy required for reaction.
The highest point on the curve corresponds to the most probable (mean) energy of the particles, labeled A
on the diagram. The Boltzmann curve can be used to explain the effect of temperature and a catalyst on
the speed of a reaction.
Page
300
8.4
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 300
Effect of temperature
proportion of
particles
x
y
A Ea 1 E a2
Energy
Fig 8.6 The Boltzmann curve showing
the distribution of molecular energies
in a gaseous reactant.
a gas sample
proportion of
of particles
T1
n1
n2
T2
(T2 > T1)
y
x
A1 A2 E
a
energy
Fig 8.7 shows the effect of increasing temperature
(from T1 to T2) on the position and shape of the
Boltzmann curve.
 Increasing temperature lowers the maximum
point of the curve and shifts it to the right. As
a result, the average energy of the particles
increases from A1 to A2 but the proportion of
particles that can achieve this energy decreases from n1 to n2.
 The activation energy of the reaction is not affected by changing temperature. At the lower
temperature T1, the proportion of particles
with sufficient energy to react is x. When
temperature is increased to T2, the proportion
of particles whose energy is equal to or greater
than Ea increases to (x+y). More particles
therefore have sufficient energy to react, and
the speed of reaction increases.
Increasing temperature increases rate of reaction by
increasing frequency of collisions and also by increasing the number of particles that have energy
equal to or greater than activation energy.
Fig 8.7 The effect of changing temperature on the distribution of molecular
energies in a gaseous sample.
8.5
Catalysts and catalysis
A catalyst is a substance that speeds up a reaction by providing an alternative
reaction route with lower activation energy, but is recovered unchanged in nature or in
quantity at the end of the reaction.
Page
301
Catalysis is the speeding up of a chemical reaction by a catalyst.
The Boltzmann curve (Fig 8.8) can be used to explain the effect of a catalyst on the speed of a reaction.
 A catalyst does not alter the kinetic energies of the molecules. The shape of the curve is therefore
not altered.
 Activation energy for the unanalyzed reaction is Ea2. At this higher activation energy, fewer
particles can achieve the energy required for effective collisions. The Region marked x represents
the particles whose energy is greater than or equal to E a2 . These are the particles that can react
when activation energy = Ea2.
 A catalyst creates an alternative route with lower activation energy (Ea1). More particles can
therefore achieve the energy required for effective collisions. Note that the proportion of particles
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 301
that can react increases from x to (x + y) and so the reaction becomes faster at the lower activation
energy.

proportion of
particles
x
y
Ea1 Ea2
Energy
Fig 8.8
A catalyst lowers activation
energy and increases the proportion of
particles that can react.
A catalyst is the only factor that alters the
magnitude of activation energy. Other
factors such as temperature and concentration do not affect the size of the
activation energy. They simply change the
frequency or energy, or both, of
collisions.
The effect of a catalyst can further be explained
using an energy profile (Fig 7.9). This diagram
shows the lowering of activation energy by a catalyst.
Properties of catalyst




Fig 8.9 Energy profile showing the
lowering of activation energy by a
catalyst
that does not take part in a chemical reaction. For it to speed up a reaction, a catalyst must play
an active role. In fact, a catalyst undergoes a temporary chemical change during a reaction.
However, it is regenerated in its original chemical form, and in its original quantity, at the end of
the reaction.
In a reversible reaction, a catalyst speeds up the forward and the reverse reaction to the same extent. In other words, a catalyst does not alter the position of equilibrium, but simply allows the
equilibrium to be achieved faster. The value of the equilibrium constant is therefore not altered by
the presence of a catalyst.
Page
302


Generally increases speed of a reaction.
Creates a new reaction route (mechanism) whose activation energy is lower
than for the uncatalyzed reaction.
Does not change the enthalpy of reaction.
This is expected from Hess’ law. The
enthalpy change of reaction
remains the same when the route of a
reaction is changed, provided that the
relative energy levels of reactants and
products are the same.
A catalyst does not make an impossible
reaction possible. It simply makes a
feasible reaction faster.
A catalyst takes an active role during a
chemical reaction. An erroneous definition is often made by students in which
they claim that a catalyst is a substance
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 302
Q
During the Haber process for the manufacture of ammonia, hydrogen and nitrogen
combines according to the equation
3H2(g) + N2(g) Ý 2NH3(g) ∆H = -93 Kjmol-1
Which statements are true about this process?
1. The rate of the reaction is increased by lowering temperature
2. The yield of ammonia is increased by lowering temperature
3. The rate of reaction increases when temperature is increased
2 and 3 are correct.
A
1 and 2.
Since the reaction is exothermic, lowering temperature favours the formation of ammonia
(Le Chatelier’s principle). In other words, the percentage of ammonia in the equilibrium
mixture increases. However, the rate of reaction decreases, that is, it takes a longer time for
the maximum yield of ammonia to be achieved. For this reason, very low temperatures are
avoided in the Haber process
3. Increasing temperature increases rate of reaction, that is, equilibrium is rapidly achieved.
However, the percentage of ammonia in the equilibrium mixture decreases because the
equilibrium now lies more to the left. For this reason, very high temperatures are avoided in
the Haber process. The actual temperature used (about 450 0C) is a compromise between
low temperature (which favours high yield of ammonia but reduces rate of reaction) and
high temperature (which increases rate of reaction but decreases the yield of ammonia).
8.5.1
Homogeneous versus heterogeneous catalysis
Homogeneous catalysis is the speeding up of a reaction by a catalyst which is in the same phase as the
reactants, e.g. both catalyst and reactants could be in solution.
Heterogenous catalysis is when the catalyst and reactant are in different phases. Well known examples
of heterogeneous catalysis include the use of solid iron in the Haber process for the manufacture of
ammonia, and the hydrogenation of unsaturated oils during the manufacture of margarine.
Example of homogenous catalysis: The catalysis of the reaction between peroxodisulphate and iodide ions
Peroxodisulphate can be reduced to sulphate by iodide ions. Meanwhile, the iodide ions are oxidized to
iodine.
This is a redox reaction because the oxidation number of sulphur in peroxodisulphate decreases, whilst that
of iodine increases. We may wish to assess the energetic feasibility of this reaction. A large and positive E θ
value implies a feasible reaction. Reaction 1 above can be resolved into its half equations, either by
inspection or by using the data booklet.
Page
303
S₂O₈²⁻ (aq) + 2I⁻ (aq) → 2SO₄²⁻ (aq) + I₂ (aq) … (i)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 303
Reduction half equation : S₂O₈²⁻ (aq) + 2e⁻ → 2SO₄²⁻(aq)
Eθ/V
+2.01 ...(ii)
Oxidation half equation : 2I⁻(aq) →I₂(aq) + 2e⁻
-0.54 ...(iii)
net equation
+1.47
S₂O₈²⁻(aq) + 2I⁻(aq)  2SO₄²⁻(aq) + I₂(aq)
Since the Eθcell of reaction (i) is large and positive, it can be concluded that the reaction is energetically
feasible. Now, a clear distinction has to be made between energetic and kinetic feasibility. If a reaction is
energetically feasible (large and positive Eθcell value or negative enthalpy change) it does not
necessarily mean that it will happen spontaneously. If the reaction has a very high activation energy, it may
not take place. It is said to be kinetically unfeasible. Such a reaction can be made to take place by using
harsh conditions (high temperature, high pressure/concentration) or by using a catalyst.
The reaction between peroxodisulphate and iodide has a very high activation energy since the ions
participating in the reaction have like charges and so they repel each other. A very large amount of energy
is required to force the reacting ions to collide.
Introducing a catalyst such as Fe2+(aq) or Fe3+(aq) speeds up the reaction by creating an alternative reaction
mechanism with a lower activation energy.
Eθ/V
S₂O₈²⁻ (aq) + 2e⁻ → 2SO4²⁻ (aq)
+2.01
… (ii)
Fe²⁺ (aq) → Fe³⁺ (aq) + e⁻
- 0.77 … (iv)
2I⁻ (aq) →I₂(aq) + 2e⁻
-0.54
…(iii)
If Fe²⁺ is used as the catalyst, it will reduce peroxodisulphate to sulphate, a role which is played by iodide
ions in the uncatalyzed reaction. In turn, Fe2+ is oxidized to Fe3+. By combining equations (ii) and (iv):
S₂O₈²⁻ (aq) + Fe²⁺ (aq) → 2SO₄²⁻ (aq) + Fe³⁺ (aq) … (v)
Eθ = 2.01 + (- 0.77) = +1.24V
This reaction is energetically feasible since it has a positive Eθ value. It is also kinetically possible since it
involves ions of opposite charge which naturally attract each other. The activation energy of this reaction is
therefore very low, and so it is very fast. However, at this stage, we can not conclude that iron (II) has acted
as a catalyst. We can only make this conclusion after the Fe(II) has been successfully recovered from Fe(III).
In reaction (vi) below, the Fe (III) formed in reaction (v) is reduced back to Fe (II) by iodide ions.
Meanwhile the iodide ions are oxidized to iodine. Note that in this reaction, Fe 3+ replaces S2O82- as the
oxidizing agent in the conversion of iodide to iodine. Reaction (vi) is obtained by combining equations (iii)
and (iv) above.
2I⁻ (aq) + Fe³⁺ (aq) → Fe²⁺ (aq) + I₂ (aq) ... (vi)
Eθ = 0.77 + (-0.54) = +0.23V
Page
304
Combining equations (v) and (vi) gives the original reaction (i). The Eθcell for reaction (i) is also obtained
by adding the Eθcell values for reactions (v) and (vi).
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 304




S₂O₈²⁻ (aq) + Fe²⁺ (aq) → 2SO₄²⁻ (aq) + Fe³⁺ (aq)
Eθ/V
+1.24
2I⁻ (aq) + Fe³⁺ (aq) → Fe²⁺ (aq) + I₂ (aq)
+0.23 ... (vi)
S₂O₈²⁻ (aq) + 2I⁻ (aq)  2SO₄²⁻ (aq) + I₂ (aq)
+1.47 as before
...(v)
Fe2+ has acted as a homogeneous catalyst. It created a new reaction mechanism in which the
reacting particles have opposite charges which naturally attract (equations (v) and (vi)). A very
small amount of energy is therefore needed to bring about energetic collisions, that is, the activation
energy is very small. The reaction is very fast since the attraction between ions of opposite charge
is rapid.
The mechanism described above emphasizes the fact that a catalyst takes an active role in the
reaction and often undergoes temporary chemical changes during the reaction.
It is also possible to use Fe3+ instead of Fe2+ as the catalyst. First, Fe3+ oxidizes I- to I2. Meanwhile,
the Fe3+ is reduced to Fe2+. The catalyst is regenerated when Fe2+ reduces S₂O₈²⁻ to SO42-.
Meanwhile, Fe2+ is oxidized back to Fe3+.
Transition elements and their ions or compounds often act as catalysts because they have the ability
to change their oxidation states during the catalysis, for example, iron can easily change form the
+2 to the +3 state and back.
Heterogeneous catalysis
Well known cases of heterogeneous catalysis involve a solid catalyst and gaseous reactants. It is believed
that the catalyst adsorbs reactant particles on its surface, bringing them together in the correct orientation
required for a reaction to take place. The catalyst must be able to bind the reactant particles by forming
weak bonds. This usually requires the catalyst to have available orbitals to accept electrons from the reactant
particles when the weak bonds are formed between the catalyst and the reactant particles. This explains the
ability of transition metals and their compounds to participate in catalysis. They have vacancies in the dorbitals to accommodate electrons from the reactant molecules. Formation of weak bonds between the
catalyst and reactants weakens bonds in the reactant molecules. The activation energy of the reaction is
therefore lowered since only a small amount of energy is required to break bonds in the reactants.
A large number of industrial processes involve heterogeneous catalysis. Examples are the use of iron in the
manufacture of NH3 from H2 and N2 (Haber process) and the use of vanadium (V) oxide in the Contact
process for the manufacture of sulphuric acid.
A heterogeneous catalyst usually works by bringing reactant molecules closer together with the correct
orientation, and by weakening bonds in reactant molecules.
Page
305
Some cases of heterogeneous catalysis involve a liquid reactant and a solid catalyst. A well known example
is the decomposition of hydrogen peroxide in the presence of solid manganese dioxide. On its own,
hydrogen peroxide decomposes very slowly to form water and oxygen. When a little MnO2 (black powder)
is added, a vigorous reaction takes place and there is brisk effervescence as oxygen gas is rapidly released.
This catalysed decomposition of hydrogen peroxide is a convenient way of preparing oxygen in the lab.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 305
8.5.2 Enzymes
These are biological catalysts which speed up biochemical reactions. Like inorganic catalysts, enzymes work
by providing an alternative reaction pathway with a lower activation energy. However, enzymes have certain
properties which make them unique (See section 11.1 for further details). Enzymes allow biological
processes to take place under relatively mild conditions of temperature, pressure and pH. In the absence of
enzymes, most biochemical reactions would require stringent conditions, for example, high temperature.
Such conditions would destroy cells and kill the organism.
8.5.3 Autocatalysis
This is a phenomenon in which a product of a reaction speeds up the reaction.
rate of
reaction
Fig 8.10
time
A well known example is the hydrolysis of ethyl ethanoate to form
ethanoic acid and ethanol. This reaction requires H + as a catalyst.
When the reaction is carried out in the absence of an acid, it is very
slow at the beginning, but after some time the rate of reaction
suddenly increases (Fig 7.10).
At the beginning, the reaction is very slow because of the absence
of a catalyst. Ethanoic acid, which is produced during the reaction,
acts as a catalyst by providing H+ ions.
This explains the increase in the rate of reaction. After some time, the rate of formation of products begins
to decrease again and soon levels out. This is because the hydrolysis of ethyl ethanoate is reversible.
CH3COOCH2CH3(aq) + H2O(l) Ý CH3COOH(aq) + CH3CH2OH(aq)
When the concentration of the products, ethanoic acid and ethanol, begins to increase the reverse reaction
becomes more significant and this reduces the rate of formation of products.
8.6
Quantitative kinetics
8.6.1 The rate law
You now know that altering concentration of a reactant causes the rate of reaction to change. It is important
to be able to quantify the effect of changing reactant concentration on the rate of reaction.
The rate law, also known as the rate equation, is an experimentally determined equation linking
the rate of reaction to the concentration of reactants.

Page
306

The rate law gives a quantitative description of the effect of changing concentration on the rate of
reaction. For example, the rate law may tell us that doubling concentration of a reactant causes the
rate of reaction to double as well.
The rate law can only be established experimentally. It is not possible to determine the rate
equation by simply inspecting the stoichiometrically balanced equation.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 306
Order of reaction
The order of reaction with respect to a particular reactant is the power to which the concentration
of that reactant is raised in the rate law.
The overall order of reaction refers to the sum of the orders of the different reactants shown in the
rate law.
Consider the reaction
aA + bB → cC + dD
The rate law may be written as
R = k [A]m[B]n
m is the order of reaction with respect to reactant A. n is the order of reaction with respect to reactant B.
The overall order of reaction is (m+ n).
K is the constant of proportionality (rate constant) between the rate of reaction and the concentration terms
in the equation.
8.6.2
First order reactions
If the reaction aA + bB → cC + dD is first order with respect to reactant A, the rate law is
R = k[A]
This is clearly a linear equation of the form Y =mX; the graph of R against [A] is a straight line passing
through the origin (Fig 8.11).
This graph shows that when the concentration of A is
changed, the rate of reaction changes proportionally, for
example, doubling concentration also causes the rate of
reaction to double.
The slope of this line gives the value of the rate constant,K
-3
Rate/moldm
Fig 8.11
-3
[A]/moldm
K =
Change in rate of reaction
change in concentration
= slope of line
This is the same result that could have been obtained by making k subject of the formula in the rate law
R = k[A].
8.6.3 The concept of half life
Page
307
The half life of a reaction refers to the time it takes for the concentration of reactant to be reduced to half
of an initial value.
The half life of a first order reaction is constant, as illustrated in Fig 8.12. Let the initial concentration be
[A]0. The time required for this concentration to be halved to ½ [A] 0 is the first t1/2. The time needed for
further halving of the concentration to ¼[A]0 is the second t1/2. Further halving of the concentration reduces
it to 1/8 [A]0, giving the third t1/2.
For a first order reaction, the first, second, third (etc) half lives are constant. In the graph below,
t1/21 = t1/22 = t1/23 . This is a good method of recognizing a first order reaction.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 307
Determination of t1/2
concentration
For a first order reaction, t1/2 can be determined
graphically, as shown in Fig 7.12, or by formula
[A]0
t1/2 =
1
2 [A]0
1
4 [A] 0
t
Q
A
2
1
1 /2
t
3
1 /2 t 1 /2
time
𝐥𝐧 𝟐
𝐊
where K is the rate constant
Fig 8.12
The half life of a first order
reaction is constant
The rate/concentration graph of the reaction A  B is shown below
(a)
Deduce the order of the reaction with respect to
reactant A.
(b)
Determine the half-life of the reaction (assume
the units of rate are moldm-3s-1)
(a) The order = 1 (since the graph of rate/[A] is linear, that is, it has a constant slope)
(b)
K = slope of rate/[A] graph = 2 units
t1/2 =
𝐥𝐧 𝟐
𝐊
=
𝐥𝐧 𝟐
𝟐
= 0.35 seconds
8.6.4 Second order reactions
If the reaction aA + bB → cC + dD is second order, say, with respect to reactant B, the rate law in terms
of B is written as
R= k [B] 2
This is a quadratic expression of the form Y = mX2. The graph of R against [B] is therefore the positive half
of a parabola whose minimum point is at the origin (Fig 8.13(a)). The graph can be linearized for
easier analysis:
Page
308
R= k[B]2
Let [B]2 be X, then
R= kX
A linear graph is then plotted for R against X, as shown in Fig 7.13(b). The slope of this linear graph gives
the value of K, the rate constant.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 308
Concentration - time graph for a second order reaction
Rate
The graph has the same shape as that of a first order reaction, and for the
same reason (Fig 8.14). Concentration of reactant decreases with time, so
the graph is down-sloping, but the rate of decrease of concentration is not
uniform, so the slope of the curve becomes less and less steep with time,
that is, the curve has an inverse shape.
[B]
concentration
[B] 0
R
Fig 8.14
X( = [B]2)
t
Fig 8.13
1
1 /2
t
2
1 /2
time
What distinguishes between the graph of a second order and that of a first order reaction is the half life. For
a first order reaction, the graph has a constant t1/2. For a second order reaction, the t 1/2 successively becomes
larger with time, for example, in Fig 8.14 above, t1/22 > t1/21 .
8.6.5 Zeroth order reactions
It sometimes happens that the rate of reaction is unaffected by changing the concentration of a reactant.
The reaction is said to be of zeroth order with respect to that reactant. If a reaction is zeroth order with
respect to reactant A, we may write
R = k[A]0
Which simplifies to R=k, that is, rate of reaction remains constant when concentration of the reactant is
altered. The graph of R against [A] is shown in Fig 8.15(a). The corresponding concentration/time graph is
shown in Fig 8.15(b). Since the rate of reaction is constant, the rate of decrease of [A] is also constant, that
is, the concentration/time graph is linear.
Rate
K
[A]
Page
309
Fig 8.15
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 309
8.6.6 Finding orders of reaction by the initial rates method.
Rate of reaction has already been defined as the rate of decrease of the concentration of a reactant, that is
decrease in concentration of reactant
Rate =
Time
The rate of reaction at a particular time, that is, the instantaneous rate of reaction, can be found by taking
the slope at the appropriate time on the concentration /time graph. When the slope is taken at time zero,
the rate obtained is known as the initial rate of reaction (Fig 8.16).
The initial rate of reaction therefore refers to the rate of reaction when an infinitesimally small amount
of reactant has reacted, that is, it is the rate of reaction at approximately time zero.
concentration
of reactant
Tangent drawn at time zero . Slope of this tangent gives the
initial rate of reaction.
time
0
Fig 8.16 Finding the initial rate of
reaction by graphical means
Consider the reaction aA + bB → cC + dD. Suppose that we wish to find the order of reaction with respect
to reactant A. The concentration of A is varied in several experiments (runs), whilst that of B is held
constant. The initial rate of reaction at each new concentration of A is then determined. Similarly, to find
order of reaction with respect to reactant B, the concentration of B must be varied whilst that of A is held
constant. The initial rate when B is varied is then recorded.
Consider a simple reaction A + B  C. Suppose that when the reaction is carried out using an initial
concentration [A]0 the corresponding rate of reaction is R1.
The rate law can then be written as
R1 = K [A]1m ...(i)
The experiment is now repeated using a higher concentration of A, [A]2. Let the corresponding rate of
reaction be R2.
The rate law then becomes
R2 = K [A]2m ...(ii)
Note that changing concentration changes rate of reaction, but does not change the order of reaction.
Dividing equation (ii) by (i) (or vice-versa).
R2
R1
m
=
[A]2
m
[A]1
Page
310
m
R2
R1
=
[A]2
[A]1
The value of m, which is the order of reaction with respect to reactant A can then be calculated.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 310
Example 1
At 7000C, nitrogen monoxide and hydrogen reacts as follows.
2NO(g) + 2H2(g)  N2(g) + 2H2O
The results of some investigations of this reaction are shown below.
Experiment
Initial concentration
of NO/moldm-3
Initial
concentration of H2
/moldm-3
Initial rate
of reaction
/moldm-3s-1
1
0.0020
0.012
0.0033
2
0.0040
0.012
0.013
3
0.0060
0.012
0.030
4
0.012
0.0020
0.020
5
0.012
0.0040
0.040
6
0.012
0.0060
0.060
Use the data given in the table to deduce the rate law for the reaction between nitrogen monoxide
and hydrogen gas. Hence deduce the value for the rate constant, including its units.
9701/3/M/J/1993, adapted.
Solution
The rate equation will have the general form
R = K [NO]m[H2]n
First, the orders of reaction m and n must be found
Finding m
Using lines 1 and 2
The concentration of NO is doubled, but that of H 2 remains constant. Any change in rate of
reaction is therefore due to the change in concentration of NO. Using the formula
m
R2
[A] 2
[A] 1
Page
311
R1
=
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 311
0 .0 1 3
0 .0 0 3 3
=
0 .0 0 4 0
m
0 .0 0 2 0
The rate law is therefore
R = K [NO]2[H2]
Overally, the reaction is third order (recall that
overall order is the sum of the orders of all the
4 = 2
reactants shown in the rate law). Such a reaction
4 = 2m ⇒ m = 2
is said to be termolecular. It requires the collision
Therefore order of reaction with respect to NO is 2
of three molecules to occur, which is quite unusual.
(In other words, doubling the concentration of
The rate constant K is found by making it subject
NO causes the rate of reaction to increase fourfold). of the formula, then substituting values
from any run, for example, run 6.
Finding n
R
Using any lines in which [H2] changes but [NO]
K=
[NO]2[H2]
remains constant, for example,
runs 4 and 5
0.060
=
= 6 9444
n
0.0122 x 0.0060
m
0.040
0.020
=
0.0040
0.0020
2 = 2n
⇒n = 1
The units are
moldm-3s-1
(moldm-3)2(moldm-3)
= mol-2dm6s-1
Order of reaction with respect to hydrogen is 1.
Example 2
Ethylethanoate slowly hydrolyzes in water, according to the equation
CH3COOC2H5 + H2O  CH3COOH + C2H5OH
HCl
The reaction is catalyzed by a strong mineral acid such as HCl. The reaction was followed twice with
two different concentrations of HCl. The results obtained are recorded in Table 8.1.
Use a graphical method to deduce the rate law. Hence calculate the value of the rate constant, giving
its units.
Page
312
Why is it not possible to determine the order of reaction with respect to water in this experiment.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 312
Time/min
With[ HCl] = 0.1moldm-3
[ethylethanoate]/moldm-3
With [ HCl] = 0.2 moldm-3
[ethylethanoate]/moldm-3
0
0.200
0.200
25
0.152
0.115
50
0.115
0.067
75
0.088
0.038
100
0.067
0.022
125
0.051
0.013
9701/1/O/N/1999, adapted
Solution
The aim here is to find the rate of reaction with respect to ethylethanoate and HCl. This can be done
graphically. The two graphs are shown in Fig 8.17.
The initial rate of reaction is given by the slope of the tangent at time zero. For graph A (where [HCl =
0.1], the initial rate is
=
(0.200−0.130)
30
= 2.3 x 10-3
Graph B is the graph of concentration of ethylethanoate against time, with a doubled concentration of
HCl. Notice that the curve is steeper, showing that the reaction is faster.
From this graph, the initial rate of reaction
=
(0.2−0.1)
25
= 4.0 x 10-3
Both graphs have a constant half-life (check), showing that the reaction is first order with respect to
ethylethanoate.
The rate law may be written as
R = K[ethylethanoate][HCl]m
To find order with respect to HCl, use the relationship
m
R1
=
[A] 2
[A] 1
Where R2 is the initial rate of reaction when [HCl] = 0.2 moldm-3, R1 = initial rate of
reaction when [HCl] = 0.1 moldm-3, and A = HCl.
Page
313
R2
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 313
Page
314
Fig 8.17
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 314
4.0x10-3
=
2.3 x 10-3
0.2
0.1
m
The rate law is therefore
R = K [ethyl ethanoate][HCl]
1.74 = 2m
m=
log 1.74
log 2
Order with respect to HCl = 1
≈1
In this reaction, it is not possible to determine the
order of reaction with respect to water. This is
because water is manly present as a solvent and so
its concentration remains fairly constant.
Notes
8.7

The rate law includes the catalyst, HCl, indicating that the catalyst participates in the slower (rate
determining ) step of the hydrolysis mechanism. The rate law is always written using the slower or the
slowest step in the mechanism. If a catalyst appears in the rate law, the reaction rate is affected by altering
the concentration of the catalyst. In this case, the reaction is first order with respect to the catalyst.
Increasing or decreasing concentration of the acid therefore results in a proportional increase or decrease in
the rate of reaction, for example, doubling the concentration of HCl doubles the rate of reaction.

If a catalyst does not appear in the rate law, it is because it participates in a faster (non rate determining)
step. In such a situation, changing concentration of the catalyst has no effect on the rate of reaction. The
reaction is said to be zeroth order with respect to the catalyst.

A catalyst, whether it appears in the rate law or not, usually alters the rate law by changing the orders of
reaction. This is because the catalyst creates an alternative pathway which is has a different rate law. The
value of the rate constant K in the new rate law increases, indicating that the reaction becomes faster.

t1/2 is a good measure of reaction rate, especially where the purpose is to compare the speeds of two
reactions. Graph B above has a steeper slope and its half-life is half of that in graph A (check). This means
that the reaction is two times faster when the concentration of HCl is doubled.
Reaction mechanisms
Reactions are usually written in the format aA + bB → cC + dD. This gives the impression that the
reaction is a single step process. In practice, reactions rarely proceed in a single step, but involve one or
more steps. The sequence of small steps leading from reactants to products is known as the mechanism of
the reaction.
Consider a reaction
A→C
I.
A→ X
slower
II. X→C
faster
Page
315
Suppose that the reaction proceeds in two steps, first by forming an intermediate X, which reacts further
to form the product C. Also assume that the first step is the slower step.
The mechanism of the reaction can then be shown as
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 315
The overall reaction can be obtained by vertical summation of reactions I and II. Note that the intermediate
X cancels out in this summation. If summation of all steps of the mechanism does not give the expected
equation, then the mechanism is wrong (or the equations in the mechanism are not balanced). However, if
summation of the steps gives the expected equation, it can not be concluded straightaway that the
mechanism is correct. All we can say at this stage is that the mechanism is sensible. The actual
mechanism can only be determined experimentally.
I.
II.
A→ X
X→C
AC
A complete energy profile for such a reaction can be shown as in Fig 8.18 below.
energy
Ea2 activ ation energy of the first step
E a1
Ea 2 X
E a1 activ ation energy of the second step
A
H
B
progress of reaction
Fig 8.18 Energy profile of an exothermic two step reaction



The profile has two ‘humps’; the first hump represents the activation energy of the first step, and
the second hump represents the activation energy of the second step. If the reaction had, say, three
steps, the energy profile would also have three humps.
This profile has been drawn on the assumption that the reaction is overally exothermic, so the
final product B is at a lower energy level than the reactant A.
The first step is endothermic since it results in the formation of an unstable intermediate. In other
words, the intermediate X is higher in energy than the reactant A. The instability of the
intermediate allows it to react rapidly to form the product. However, the intermediate can also
react in the reverse direction to form the reactant once again. From an energetic point of view, both
the forward reaction (forming the product B) and the reverse reaction in which the
intermediate reforms the reactants are feasible.
A Ý XB
This is because the reactant and the product are both stable relative to the intermediate.
Page
316

The highest point in the energy profile gives the activation energy of the overall reaction. This has
been labeled Ea2 in the diagram above. Note that the activation energy of the second step (E a1) is
smaller. This is because the intermediate has weak bonds, making it unstable. A relatively small
amount of energy is required to break these bonds and convert the intermediate into a product. The
step with the smaller activation energy is the faster. Note that the product(X = intermediate) of the
first step is the reactant of the second step. The activation of step II is therefore measured relative
to its reactant, X.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 316

The rate of reaction is determined by the slower or slowest step. In the mechanism above, reaction
I has been arbitrarily made the slower reaction, so it is the one from which we may write
the rate law:
R=k[A]m

We may not know the actual value of m until we carry out an experiment to determine it.
Energy profiles are not always as shown in Fig 8.19. The exact shape of the profile depends on the
mechanism of the reaction and on whether the reaction is overally exothermic or endothermic.
Fig 8.19 below shows the energy profile of a two step reaction which is overally endothermic.
energy
Ea 1
E a2
products
H
Fig 8.19 Energy profile
for a two step endothermic
reaction
reactants
progress of reaction

Some mechanisms are said to be concerted. In such a reaction, the intermediate is so unstable that
its existence is extremely brief. Such a reaction can therefore be considered as taking place in one
step. The energy profile of such a reaction is usually drawn with only one hump. The activation
energy (hump) of the step involving the intermediate is so small that it is close to zero. This hump
is therefore omitted on the energy profile. An example of a reaction which has a
concerted mechanism is the reaction of chloroethane with NaOH (aq).
CH3CH2Cl + OH-(aq) CH3CH2OH + Cl-
CASE 1
The alkaline hydrolysis of a chloroethane: A concerted reaction.
Halogenoalkanes are organic compounds which contain the C-Hal bond, where Hal represents a halogen
atom. In alkaline hydrolysis of halogenoalkanes, the Hal atom is replaced by an OH group from a strong
base, forming an alcohol. There are two different mechanisms involved, depending on the nature of the
halogenoalkane. These mechanisms have been termed SN1 or SN2 depending on the nature of the halogenoalkane.
Hydrolysis of chloroethane proceeds by SN2 (nucleophilic substitution bimolecular- see section3.2 :
mechanisms of organic reactions)
The overall equation is
CH3CH2Cl + OH-(aq) CH3CH2OH + Cl-
Page
317
The reaction can be understood by splitting it into two steps.
I. CH3CH2Cl + OH-(aq)  [CH3CH2 (OH) Cl]‡ slower
II. [CH3CH2 (OH) Cl]‡  CH3CH2OH + Clfaster
NB The symbol ‡ shows a transition state, which is less stable than an intermediate.
The first step is an addition reaction, in which the reactants combine to form a single product. This is the
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 317
rate determining step. So we may write the rate law as
R = k[CH₃CH₂Cl]m[OH⁻]n
It has been found experimentally that the values of m and n are both 1, that is, the rate of reaction is first
order with respect to each reactant. So the rate law is
R = k[CH₃CH₂Cl][OH⁻]
The reaction is first order with respect to both reactants. Changing the concentration of either reactant will
cause the rate of reaction to change proportionally, as illustrated in Fig 8.19 below.
Rate
Rate
-
[OH ]
Fig 8.19
[CH 3CH2Cl]
The reaction between chloroethane and NaOH (aq) is first order with respect
to both reactants.
Overally, the order of reaction is 2, so the mechanism of this reaction is termed SN2. The
number 2 represents the order of reaction, that is, the rate determining step involves the collision of two
reactant particles. SN stands for nucleophilic substitution. If you look carefully at the reaction, you will see
that it is a substitution reaction in which the OH⁻ group replaces the Cl group. A nucleophile is an electron
rich species (in this case, OH-) which participates in some types of organic reactions.
The energy profile of the reaction has only one hump (Fig 8.20) because step II is very rapid. The
intermediate, CH3CH2 (OH) Cl, which should be more properly termed a transition state, is very unstable.
Its conversion to the products has a very low activation. In practice, the reaction can therefore be considered
to take place in one step. This is an example of a concerted mechanism. The structure of the transition state
is shown in Fig 8.20. It is very unstable because one of the carbon atoms has too many bonds around it (five
instead of four)
H H
energy
CH 3 C
OH
Cl
The transition state occupies the
highest energy level on the energy
profile diagram.
CH3CH2Cl
Page
318
CH 3CH 2OH
Fig 8.20 Energy profile for the
reaction between chloroethane
and aqueous sodium hydroxide
progress of reaction
CASE 2
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 318
The alkaline hydrolysis of 2-chloro-2-methylpropane: a non-concerted mechanism
2-chloro-2-methylpropane, (CH₃)₃CCl is also a halogenoalkane. It is attacked by the OH⁻ ion in a
nucleophilic substitution reaction, but the mechanism is different.
I. (CH3)3CCl  (CH3)3C+ + ClII. (CH3)3
C+
+
OH-
 (CH3)3COH
slow
fast
The rate determining step is the first step, so we may write
R= K[(CH₃)₃CCl]m
The value of m has been experimentally determined to be 1, so the order of reaction is 1 and the mechanism
is appropriately named SN1 (nucleophilic substitution first order). Note that if the rate law is written to
include OH⁻, then [OH-] would be raised to power zero, since OH - ions are involved in a
non - rate determining step:
R= K[(CH₃)₃CCl][OH⁻]0
he reaction is therefore zeroth order with respect to OH⁻ ions. Increasing or decreasing the concentration
of NaOH has no effect on the rate of reaction. The graphs in Fig 8.21 illustrate the kinetics of this reaction.
Rate/concentration tim e graphs
Rate
Rate
First order
zeroth order
[NaOH]
(CH 3)3CCl
concentration/time graphs
[NaOH]
[(CH 3)3CCl]
graph is linear, that is slope = rate of
reaction is constant
time
time
Page
319
Fig 8.21 Graphs illustrating the kinetics of the reaction between [(CH₃)₃CCl and
NaOH.
The intermediate in this reaction is a carbocation, (CH3)3C+. This carbocation is relatively stable and it
exists for some time in the reaction vessel before it is converted to the product. The mechanism therefore
has two distinct steps, that is, it is non-concerted. The energy profile has two humps, (Fig 8.22)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 319
energy
(CH 3)3C
(CH 3)3CCl
(CH 3)3COH
progress of reaction
Fig 8.22 Energy profile for the reaction between (CH3)3CCl and NaOH
Orders of reactants in multistep reactions
In a multistep mechanism, the faster steps are reversible, wheras the slowest step in the mechanism can be
regarded to be irreversible. This can be explained in this way:
 The rapid steps involve unstable intermediates, which can therefore react in both directions, that
is, some intermediate species react in the forward direction to form products,whilst some react in
the reverse direction to form reactants. This outcome is expected since both the products and the
reactants are lower in energy than the intermediates. Conversion of the intermediate to both
reactants and products is therefore energetically feasible.
 The slowest step involves conversion of an unstable intermediate to a stable product. The reaction
therefore tends to be unidirectional, that is, it proceeds only in the direction reactant  product.
Q
The Esson - Harcourt reaction involves the reaction between hydrogen peroxide and iodide
ions in the presence of an acid
H2O2 + 2I– + 2H+ Ý 2H2O + I2
This reaction is considered to proceed by the following steps.
step (i)
H2O2 + I–  IO– + H2O
step (ii)
IO– + H+ HOI
step (iii)
HOI + H+ + I–  I2 + H2O
The general form of the rate equation is
rate = k[H2O2]a[I–]b[H+]c
(a) Show that the mechanism is consistent with the given reaction
Page
320
(b) Determine values for the orders a, b and c in the rate equation for a case in which the
second step is the slowest.
A
(a) Intermediates that are on opposite sides
of equations cancel out. In steps (i) and
(ii), IO- cancels. In equations (ii) and (iii),
HOI cancels out. Combining the three
[IO–] =
Kf [H₂O₂][I¯]
Kb [H₂O]
Substituting this expression for [IO–] in
equation (I) and simplifying gives
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 320
R = K{
Kf [H₂O₂][I¯]
Kb [H₂O]
}[H+] ... (V)
but K and K are constants, and [H O] can be
equations after cancelling out the two
intermediates gives the expected overall
equation.
(b) Being slowest, the second step is the rate
determining step, that is
rate = K[IO–][H+] ... (I)
We will treat the first step (i), which is
fast, as being reversible
H2O2 + I– Ý IO– + H2O
Rate equation for the forward reaction:
Rf = Kf [H2O2][I-] ... (II)
Rate equation for the reverse reaction:
Rr = Kr [IO–][H2O] ... (III)
Step (i) should attain dynamic equilibrium
since step (ii) is much slower and does not
easily remove the products of the first step.
When step (i) achieves dynamic equilibrium,
rate of forward reaction equals rate of the
reverse reaction, that is
Kf [H2O2][I-] = Kr [IO–][H2O] ... (IV)
Page
321
Comparing equations (I) and (IV), we notice
that [IO–] is a common factor. Making this
factor subject of equation (IV) gives
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 321
Another worked example
Q
A dilute solution of hydrogen peroxide can be used to
bleach hair. It decomposes slowly in aqueous solution according
to the following equation:
2H2O2 (aq)  2H2O(l) + O2(g)
A solution with an original concentration of 3.0 moldm-3 was
placed in a bottle contaminated with transition metal ions,
which act as catalysts for the decomposition. The rate of
decomposition was measured by withdrawing 10 cm3 portions at
various times and titrating with acidified 0.1 moldm-3
KMnO4(aq) (5 moles of peroxide react with 2 moles of KMnO4 ).
The following results were obtained.
Time/min
0
5
10
15
20
25
30
Volume of 0.1 moldm-3
KMnO4 (aq)/cm3
30.0
23.4
18.3
14.2
11.1
8.7
6.8
(a) Confirm that the reaction is first order with respect to the
peroxide.
(b) Write an expression for the rate equation and calculate the
rate constant and half life
(c) Calculate the concentration of the hydrogen peroxide at the
time the first portion was withdrawn. Hence estimate how
long the solution had been in the contaminated bottle.
(d) Suggest a method whereby the shelf life of hydrogen
peroxide colud be increased.
9701/03/O/N/1991
A
The concentration of hydrogen peroxide is directly
Page
322
proportional to the volume of KMnO4 used, that is, the greater
the concentration of hydrogen peroxide, the greater the volume
of hydrogen peroxide required for reaction. Voume of KMnO4
can therefore be used to represent concentration of hydrogen
peroxide (Fig 8.23)
(a) The graph has a constant half-life. This shows that the
reaction is first order with respect to peroxide.
(b) R = k[H2O2]
Fig
8.23
From the graph, t1/2 ≈ 14 minutes
Rate constant, K =
ln 2
t1/2
=
ln 2
14
= 4.95 x 10-2
(c) We need to find the concentration, C of H2O2
when the first 10 cm3 sample is withdrawn .
moles of H2O2 = conc x V = C x 10/1000
= 0.01C
moles of KMnO4 = conc x V = 0.1 x 30/1000
= 0.003
(The first sample of H2O2 is titrated with 30.00
cm3 of KMnO4)
moles of H₂O₂
moles of KMnO₄
0.01C
0.003
=
=
5
2
5
2
(refer to question)
⇒ C = 0.75 moldm-3
The original concentration of the peroxide was
3.0 moldm-3. When the titration with KMnO4
was peformed, [H2O2] had dropped to 0.75
moldm-3. There are two half-lives between 3.0
moldm-3 and 0.75 moldm-3.
From 3.0 to 3/2 = 1.5
from 1.5 t0 1.5/2 = 0.75
first half-life
second half life
Since each half-life is 14 minutes, two half-lives
give 28 minutes. This is the time the peroxide
has been in the bottle.
(d) Possible methods include

Using dark coloured bottles to reduce
exposure of the peroxide to light.

Keeping the peroxide in a cool place.

Adding chelating agents such as EDTA4-
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 322
to trap transition metal ions which catalyse the
decomposition.
Exercise 8.1
1
Acetals are compounds formed when aldehydes react with
an alcohol and an acid catalyst. The reaction between ethanal
and methanol was studied in the inert solvent dioxan.
expt
[(CH3)2CO]
[H+]
[CN-]
relative initial
rate /moldm-3
sec-1
1
2
3
4
0.020
0.020
0.020
0.025
0.060
0.050
0.050
0.050
0.060
0.050
0.060
0.050
1.00
0.833
1.00
1.042
(ii) Hence write a rate equation for this reaction.
Two different mechanisms have been suggested
for this reaction
H+
CH3CHO + 2CH3OH Ý CH3CH(OCH3)2 + H2O
When the initial rate of this reaction was measured at various
starting concentrations of the three reactants, the following
results were obtained.
experiment
[CH3CHO]
/moldm-3
[CH3OH]/
moldm-3
[H+]
/moldm-
Relative
rate
3
1
2
3
4
(i)
0.20
0.25
0.25
0.20
0.10
0.10
0.16
0.16
0.05
0.05
0.05
0.10
1.00
1.25
2.00
3.20
Use the data in the table to determine the order with
respect to each reactant.
(ii) Use your resullts from part (i) to write the rate equation for
the reaction.
Mechanism A:
(CH3)2C=O + H+  (CH3)2COH+
(CH3)2COH+ + CN–  (CH3)2C(OH)CN
Mechanism B:
(CH3)2C=O + CN–  (CH3)2C(O–)CN
(CH3)2C(O–)CN + H+  (CH3)2C(OH)CN
(iii) Which mechanism is consistent with the rate
equation you deduced in (ii), andwhich step in
this mechanism is the slower (rate determining)
step? Explain your answer.
9701/04/M/J/2005
3
(iii) State the units of the rate constant in the rate equation.
(iv) Calculate the relative rate rate of reaction for a mixture in
which all three reactants are 0.2o moldm-3
H2O2 + 2I– + 2H+  2H2O + I2
9701/04/O/N/2011
2 (a)
(b)
This reaction is considered to go by the following
steps.
What do you understand by the term order of reaction?
Cyanohydrins can be made by reacting ketones with an
acidified solution of sodium cyanide.
(CH3)2C=O +
H+
+
CN–
 (CH3)2C(OH)CN
step 2
IO– + H+  HOI
step 3
HOI + H+ + I–  I2 + H2O
Suggest how the appearance of the solution
might change as the reaction takes place.
Page
323
H2O2 + I–  IO– + H2O
rate = k[H2O2]a[I–]b[H+]c
(a)
Use the data in the table to deduce the order of
the reaction with respect to propanone, hydrogen
ions andcyanide ions
step 1
The general form of the rate equation is as
follows.
In a series of experiments, the reaction was carried out
with different concentrations of the three reagents, and
the following relative initial rates were obtained.
(i)
In the late 19th century the two pioneers of the
study of reaction kinetics, Vernon Harcourt
and William Esson, studied the rate of the
reaction between hydrogen peroxide and iodide
ions in acidic solution.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 323
(b)
Suggest values for the orders a, b and c in the rate equation for each of the following cases.
Case
numerical value
a
b
c
step 1 is the slowest overall
step 2 is the slowest overall
step 3 is the slowest overall
8.8
methods of determining reaction rates
8.8.1 Titrimetric analysis
This method is only possible for a quantitative reaction that takes place in solution.
Samples are withdrawn from the reaction container at timed intervals and analyzed, for example, by
volumetric analysis (titration) to ascertain how much of the reactant has been consumed. It is then possible
to find rate of reaction, that is, how fast the concentration of reactant decreases. One major source of
inaccuracy of this method is that when the sample has been withdrawn, the reaction continues in the sample,
and this will affect the time factor in the analysis of reaction rate. One method of preventing this error is
quenching. A substance is added to the withdrawn sample to stop the reaction. Such a substance is known
as a quenching agent. An example is the iodination of propanone, which is
acid catalyzed as shown
CH₃COCH₃ (aq) + I₂ (aq) → CH₂ICOCH₃ (aq) + HI (aq)
H⁺
To follow the rate of reaction by titrimetric analysis, a sample is withdrawn from the reaction vessel and
pipetted into a solution of sodium hydrogen carbonate, which neutralizes the acid catalyst, thus quenching
the reaction. This prevents further changes in concentration before the titration is carried out. The
quenched mixture can be analyzed by titrating the unreacted iodine against a standard solution of
sodium thiosulphate.
I2 + 2S2O32-(aq)  2I-(aq) + S4O62-
8.8.2 Monitoring rate of decrease of mass of reactant
Page
324
The reaction vessel can be placed on a scale and the decrease in mass over time is noted. The decrease in
mass is caused by the reactant being used up. This particularly applies when gaseous products are formed
and escape from the reaction vessel. The rate of decrease in mass of the container is equal to the rate of
disappearance of reactant, which in turn is equal to the rate of reaction. Particular care should be taken that
no reactant molecules are lost, for example through evaporation and liquid sprays.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 324
8.8.3 Monitoring rate of increase in volume or pressure of a gaseous
product.
If the reaction produces a gaseous product, rate of reaction can be conveniently found by measuring the
increase in volume of the product over regular time intervals, for example, the gas can be trapped in a
syringe. Similarly, the reaction vessel can be connected to a pressure gauge. The rate of reaction is them
measured in terms of how fast pressure of the product increases. This method can also be used for reactions
in which a gaseous reactant is used up. In this case, the pressure gauge can be used to measure the rate of
decrease of pressure of the reactant.
8.8.4 Colorimetric analysis
(From the word ‘color’ – students often confuse between the terms calorimetry [energetics] and colorimetry.)
This method is useful and convenient for a reaction which is accompanied by a change in colour, for
example, the reaction of iodine with propanone. The analysis involves measuring colour intensity at timed
intervals. From these measurements, concentration of the coloured substance can be calculated at
different times. Fig 8.24 below shows a photoelectric colorimeter. The mode of operation of this colorimeter
is explained in Fig 8.25.
Fig 8.24 A
colorimeter
source of
light
light in
filtered light
Page
325
Filter
Fig 8.25
light
sam ple
photocell
m eter
A schematic diagram illustrating how a photoelectric colorimeter
works.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 325
This method depends on measuring the intensity of light that is transmitted through the reaction mixture,
which must be in solution form. First the light is passed through a filter that will select the most appropriate
wavelength. The intensity of light transmitted by the solution depends on the depth of colour of the solution.
If at the start of the reaction, the solution is colourless, then a large amount of light is transmitted through
the solution. This light strikes a photoelectric cell which converts light energy to electrical energy. The
greater the intensity of light passing through the solution, the larger the electric signal generated. A meter
records the amount of current and relates it to the concentration of the solution. Suppose that a coloured
product is produced during the reaction. As the concentration of the product increases, the depth of colour
of the reaction mixture increases and less light is transmitted through the solution. As a result, a smaller
current is generated. The rate of reaction is given by the rate at which the electric signal diminishes in
strength. In turn, this is proportional to the rate of formation of the coloured product.
8.8.5 Conductrimetric analysis
This is appropriate for a reaction which involves ions. The electrical conductivity of the solution decreases
when ions are being used up and increases when ions are produced. The rate of change of conductivity is
proportional to the rate of reaction.
All the methods discussed above do not give the rate of reaction directly. Instead they give concentration/time data which can then be plotted on a graph. The Rate of reaction at a particular time is
obtained by taking the slope of the curve at that time.
8.8.6 UV –Vis spectroscopy
This technique makes use of the absorption of ultraviolet and visible radiation by some types of compounds,
for example, complexes of transition elements. See section 9.5 for details.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise 7.3
miscellaneous problems
text box.]
1 (a) What do you understand by the terms (i) order of reaction (ii) rate constant?
(b) Explain with the aid of suitable diagrams, why a relatively small increase in temperature causes a
large increase in the rate of a chemical reaction.
(c)
The kinetics of the acid-catalyzed reaction of propanone with iodine
CH3COCH3(aq) + I2(aq)  CH2ICOCH3(aq) + HI(aq)
Page
326
can be investigated experimentally by varying the concentrations of the three substances involved
and detrmining the time for the colour of iodine to disappear. In this method the rate of reaction
is measured in terms of the rate at which the iodine concentration changes, that is,
volume of aqueous iodine used
rate of reaction α time for colour of iodine to disappear
The following results were obtained in such an experiment.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 326
Deduce the order of the reaction with respect to (i) propanone (ii) iodine (iii) hydrogen ions.
Write a rate equation for the reaction.
What conclusions about the mechanism can you draw from the rate equation?
Volume of
propanone/cm3
8
2
Volume of
iodine/cm3
4
Volume of
sulphuric acid/cm3
8
Volume of
water/cm3
0
Relative time for
colour of iodine to
disappear.
1
8
4
4
4
2
4
4
8
4
2
8
2
8
2
0.5
Zinc is an essential element for plant and animal life. It is often administered in the form
of a chelate, which is a complex between a metal ion and a polydentate ligand.
The rate of the reaction between zinc ions and the ligand 4-(2-pyridylazo)resorcinol, PAR, has
been studied.
Both PAR and its zinc complex absorb radiation in the UV-visible region. Figure 8.26 shows
their absorption spectra.
Devise a suitable experimental technique for studying
how the rate of this reaction varies with [Zn2+(aq)].
Fig 8.26
Page
327
9701/04/M/J/07
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 327
3
The ester 4-nitrophenyl ethanoate hydrolyses in alkaline solution according to the following
equation.
(a)
Suggest, and briefly describe, a suitable experimental technique for studying the rate of this
reaction.
(b)
The reaction rate was studied using two solutions of different hydroxide ion concentrations.
run A: [OH–] = 0.20 mol dm–3
run B: [OH–] = 0.40 mol dm–3
The following graphs show how the concentration of the ester, 4-nitrophenyl ethanoate,
varied over time in the two runs.
Page
328
(i)
By drawing tangents on the graphs, measure and calculate the initial rates of reaction during
the two runs. Give the units in each case.
(ii)
By using your results, calculate the overall order of reaction with respect to [OH –].
(iii)
From the curve of run B, determine the order of reaction with respect to [ester].
(iv)
Explain how you arrived at you answer in (iii).
(v)
Write a rate equation for the reaction.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 328
(vi)
Use your rate equation and the initial rates to calculate a value for the rate constant,
including units.
9701/04/M/J/02
4 (a) What do you understand by the term rate of reaction?
(b) The mechanism of a reaction is as shown below
H2O2 + I- H2O + IOH+ + IO-  HIO
slow
Fast
HIO + H+ + I-  H2O + I2
fast
(i) Deduce the overall equation for the reaction and give its rate expression.
(ii) Give the overall order of the reaction
(c) The following data was obtained on studying the decomposition of a gas X.
time/s
pressure of
gasX/atm
(i)
0
300
600
900
1 200
0.367
0.208
0.130
0.072
0.043
Plot a graph of pressure of X against time
(ii) Given that pressure of X is directly proportional to the concentration of gas X, show that
the order of the reaction with respect to X is one.
(iii) Calculate the rate constant for the reaction.
(d) The following reaction is slow and has a high activation energy. It can be catalysed by Fe 2+
ions.
S2O82- + 2I-  2SO42- + I2
(i)
Suggest a reason for the high activation energy of the uncatalysed reaction.
(ii)
Show by means of equations how Fe2+ can catalyze the reaction.
(iii)
What property of Fe2+ makes it a suitable catalyst for the reaction?
9189/01/O/N/2004
5 (a) Define the term order of reaction.
Page
329
(b) Hydrogen peroxide reacts with iodide ions in acidic conditions according to the following
equation.
H2O2 (aq) + 2I-(aq) + 2H+(aq)  I2(aq) + 2H2O(l)
The rate of this reaction can be determined by mixing known amounts of the three reactants,
periodically sampling the mixture and immediately adding manganese (IV) oxide, MnO 2 before
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 329
titrating with sodium thiosulphate.
Why would adition of AgNO3 instead of MnO2 not be appropriate for this reaction?
(c)
Experiments to determine the order of the reaction gave the following results
experiment
1
2
3
4
Initial
[I-]/moldm-3
0.01
0.01
0.01
0.03
Initial
[H+] /moldm-3
0.1
0.1
0.2
0.1
Initial
[H2O2]/moldm-3
0.005
0.010
0.005
0.005
Initial rate/
mol dm-3s-1
0.083
0.166
0.332
0.249
(i) Deduce the order of the reaction with respect to each of the three reactants.
Page
330
(ii) Calculate the rate constant, stating its units
9189/01/O/N/2010
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 330
Section 2
Page
331
Inorganic Chemistry
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 331
9
THE PERIODIC TABLE AND
PERIODIC TRENDS
9.1 CHEMICAL PERIODICITY: PERIOD
3 ELEMENTS
Introduction
Periods of the Periodic Table show well defined trends in chemical and physical properties. Properties
which are observed in one period apply to the other periods. This recurrence of physical and chemical
properties is known as periodicity. The discovery of periodic patterns has greatly simplified the study
of inorganic chemistry.
In this section we study the trends in chemical and physical properties of the Period 3 elements and
apply the knowledge to other periods. As you go through this section, make sure you see how these
trends link with the concepts you learnt in physical chemistry. In fact, if your understanding of physical
chemistry is good, you will find inorganic chemistry easy and straightforward. For, example, when we
discuss the trends in melting points of the Period 3 elements, you will notice the importance of
understanding the concepts of bonding learnt in section 4.
9.1.1
Periodicity of physical properties
Na
Atomic
number
11
Mg
Al
Si
12
13
14
15
Giant
covalent
bonding
(very
strong)
Simple covalent bonding
(relatively weak )
metals
metalloid
S
16
Cl
17
Ar
18
non metal
Page
332
Metallic bonding
(relatively strong)
P
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 332
1.1.1 Periodicity in bonding and structure
There is a change from metallic to non-metallic character across the period. Na, Mg and Al are metals. Si is
a semi-metal. The remaining elements are non-metals.
Explanation
Metallic nature is favoured by large atoms with a relatively small number of protons, or atoms that have a
large number of shells. Elements to the left hand side of the periodic table, that is, Na, Mg and Al, have large
atoms with a small number of protons in the nucleus. The electrons in the valence shell are therefore weakly
held to the atom. They are quite easily detached and contributed into a common pool which characterizes
metals. Going towards the right hand side of the periodic table, atoms shrink in size due to increasing
nuclear charge. The valence shell electrons are therefore closer to the nucleus. These electrons are more
strongly held to the atom since they are close to the nucleus and since the nucleus has more positive charges.
These valence electrons are not easily detached and contributed into a common pool. The non– metallic
elements therefore tend to exist as covalently bonded substances.
The gradation in bonding and structure across the period has a profound effect on physical properties of
the elements, such as boiling points and densities.
1.1.2 Boiling and melting points
Na
Mg
Al
Si
P
S
Cl
Ar
Mpt/0C
98
650
660
1407
44
119
-101
-189.2
Bpt/0C
890
11o7
2467
2357
280
445
-35
-185.6
Table 9.2
Melting points and boiling points of Period 3 elements.
The trend in melting points along Period 3 elements is shown in Fig 9.1 below.
Page
333
Fig 9.1

Trend in the melting points of the Period 3 elements
The metals, Na, Mg and Al have high melting and boiling points because of their metallic nature.
The bonds in these elements are strong electrostatic attractions between positive metal cores and a
sea of delocalized electrons. Atoms are therefore strongly held to the structure and it is difficult to
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 333



pull them from each other during melting or boiling. A large amount of energy is needed to bring
about boiling and melting.
The melting and boiling points of Na, Mg and Al increase in that order. This shows increasing
strength of the metallic bonds. The increase in size of positive charge on the metallic core (from 1
in Na , 2 in Mg and 3 in Al) and the increase in the number of electrons delocalized in the lattice
results in stronger electrostatic attractions between the positive metal cores and the delocalized
sea of electrons. The fact that nuclear charge increases and atomic size decreases also implies that
there are stronger attractions between the nucleus and the sea of delocalized electrons. All these
factors help to hold the atoms firmly to the lattice.
Si has the highest melting point in Period 3. This can be explained in terms of the type of bonding
present. Silicon has a giant molecular structure, similar to that in diamond. A very large amount
of energy is required to break the Si to Si bonds, which are very strong and numerous.
P, S, Cl and Ar are all simple covalent substances with weak Van der Waals forces between the
molecules. Only a small amount of energy would be required to break these forces during melting
and boiling. The differences in the boiling points of these elements are a result of differing strength
of VDW forces. Sulphur, which exists with S₈ molecules, has the largest number of electrons per
molecule, and hence the highest melting point. The melting point of sulphur is higher than that of
phosphorous because phosphorous exists as P4 molecules. Each P4 molecule has less electrons than
an S8 molecule (compare Mr values). Van der Waals forces are therefore stronger in sulphur than
in phosphorous.
Bond angle in P4 is approximately 1070, as in NH3 (3 bond pairs and one lone pair around each P atom).
In S8, the bond angle is approximately 1040, as in water. Notice that there are 2 bond pairs and 2 lone
pairs around each sulphur atom.
The type of bonding present in a substance affects its physical properties such as melting and boiling points. Melting
and boiling are physical processes. They do not change the chemical identity of a substance. When a substance melts
or boils, lattice forces holding atoms or molecules together are broken, separating the atoms or molecules from each
other. The size of melting or boiling point depends on the strength of these lattice forces. In metals, the lattice forces
are strong, so metals generally have high melting and boiling points. When simple covalent substances such as
phosphorous and chlorine melt or boil, it is only weak intermolecular forces, not covalent bonds, that are broken.
Simple covalent substances therefore have relatively low melting and boiling points. The stronger the intermolecular
forces, the higher the melting or boiling point. However, note that in giant covalent substances such as silicon Melting
or boiling require the breaking of very strong covalent bonds. Giant covalent substances therefore have very high
melting and boiling points, which are often higher than that of most metals.
Page
334
Reference point: Chemical bonding
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 334
Important note
It is often said that metals have high melting points. Note that this is not true for the soft Group (I)
metals. K and Na have low boiling points, even lower than some covalent substances such as Sulphur.
1.1.3 Atomic radius
The atoms become smaller going across the period (Table 9.3). Nuclear charge increases across the period,
and this implies that the outer shell is pulled more strongly towards the nucleus. The atoms thus shrink in
size. Fig 9.2 below shows that the decrease in atomic radius is sharp at the beginning of the period, but
towards the end it becomes only gradual. This is because the increase in number of electrons in the outer
shell offsets to some extent the effect of increasing proton number.
Atomic
radius/nm
Na
Mg
Al
Si
P
S
Cl
Ar
0.156
0.136
0.125
0.117
0.110
0.104
0.099
-
Table 9.1.4 Atomic radii of Period 3 elements
Ionic radii
The variation in ionic radii of the Period 3 elements is shown in table 9.1.5 and
Fig 9.1.3.
Fig 9.1.2 Trend in atomic
radii across Period 3


Na
Mg
Al
Si
P
S
Cl
Ar
Ion
Na+
Mg2+
Al3+
-
P3-
S2-
Cl-
-
Atomic
radius/nm
0.095
0.065
0.050
-
0.212
0.184
0.181
-
Table 9.1.5 Trend in ionic radii across Period 3
The Period 3 cations have the same electronic configuration, 1s22s22p6. This corresponds to loss of
all valence electrons. The cations are therefore smaller than the neutral metal atoms because of the
loss of a shell. Moreover, the ions have fewer electrons which therefore feel a stronger attractive
pull from the nucleus.
The ionic radii of the ions Na+, Mg2+, Al3+ decrease in that order. This is because the number of
electrons remains constant, and yet number of protons in the nucleus increases. This increase in
nuclear charge implies that the nucleus has a greater pull on the outer most shell, causing the ions
to shrink in size.
The anions of Period 3 are larger than the cations, and they all have the same electronic
configuration, 1s22s22p63s23p6. The anions are larger than the cations because they have one more
shell than the cations.
Page
335

Element
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 335

The ionic radii of the anions P3-, S2-, Cl- decrease in that
order due to increasing nuclear charge. Meanwhile, the
number of electrons remains constant.
Fig 9.1.3 Trend in ionic radii of
the Period 3 elements
1.1.4 Variation in first ionization energy
In general, ionization energy increases across a period (Fig 9.3).
The increase in ionization energy is due to the increase in nuclear charge. The number of protons in the
nucleus increases, and yet electrons enter the same shell where they cannot effectively shield each other
from the attractive influence of the nucleus. We say effective nuclear charge increases across the period.
The effective nuclear charge refers to the strength of the attraction by the nucleus which an electron feels.
As the effective nuclear charge increases, the outer most electrons are bound more firmly to the atom and
more energy would be required to remove an electron from this shell.
A direct result of the increase in nuclear charge is that shells
are drawn more strongly towards the nucleus, resulting in a
decrease in atomic radii across the period. This shrinkage in
atomic size also contributes to the increase in IE across the
period. As the atoms become smaller, the outer most
electrons become closer to the nucleus, so they are held more
strongly to the atom.
Anomalies are found at S and Al. These two elements have
lower ionization energies than would have been expected
from their positions in Period 3. The electronic
configurations of P and S are shown below to help explain
this anomaly:
2 9.1.3
Fig
Trend in ionization
S: [Ne] 3s
energy3p
across
Period 3p
3 z
3p
x
y
3px
3py
3pz
The electronic configurations of Al and Mg are shown in the following diagrams to help explain why the
first ionization energy of Al is lower than that of Mg.
Page
336
P: [Ne] 3s2
In sulphur, the electron to be removed is paired with another
in a 3p orbital. This electron is already experiencing repulsion
and so it is more easily removed than in phosphorous. In P,
The 3p sub shell is also exactly half - filled, so its electrons are
relatively stable. More energy is required to remove an
electron from P than in S.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 336
For Al, the electron to be removed is in a p sub shell. In Mg, the electron is in
an s sub shell. Since the p sub shell is higher in energy(that is, it is further from
the nucleus), the electron in Al is more easily removed than the one in Mg, that
is, less energy is required to knock out an electron from the outer most shell of
Al.
Mg: [Mg] 3s22
Mg: [Mg] 3s
Al: [Ne] 3s22
Al: [Ne] 3s
3p
3p
1.1.5 Electronegativities
Electronegativity increase across a period.
Electronegativity measures the ability of an atom to attract electrons of a bond towards itself. Going across
the period, nuclear charge increases and atomic size decreases (an atom with a large number of protons
pulls shells strongly towards itself). The atom is then able to attract electron density of a bond more
effectively because the electron density is closer to the nucleus.
1.1.6 Electropositivity
This is the tendency of an atom to give away valence electrons during chemical reactions. Electropositivity
decreases across the period. The metals have the higher electropositivity.




They have large atoms whose valence electrons are a large distance from the nucleus. The electrons
are therefore weakly held to the atom and they are easily lost.
The number of protons in the metal atom is relatively small. The nucleus therefore has a weak
attraction for the valence shell electrons.
There are only a few electrons in the valence shell. The energy expenditure on removing the
electrons, as measured by ionization energies, is therefore small.
By losing the few valence electrons, a stable octet of electrons is exposed. The resulting cation has
a high stability and this more than compensates for the energy used to remove the electrons.
The electropositivity for Na, Mg and Al decreases in that order in response to an increase in nuclear charge,
a decrease in atomic size and an increase in the number of valence electrons to be removed.
1.1.7 Electrical conductivity
The electrical conductivities of the period (III) elements are shown in Table 9.4.
Element
Conductivity
x 10-8 / Ω
Na
Mg
0.218
0.224
Al
0.382
Si
2 x 10-7
P
S
10-17
10-23
Cl
-
Ar
-
Page
337
Table 9.4
The metals are good electrical conductors because of the presence of a sea of delocalized electrons. These
mobile electrons are responsible for conducting electric charge. The non-metals are very poor conductors,
or do not conduct at all. Electrons in these covalent substances are localized in covalent bonds. They cannot
move, so electric charge cannot be conducted through the substance.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 337
9.1.2
Chemical periodicity
The change from metallic to non metallic character results in a noticeable change in chemical properties
across the periodic table.
1.2.1. Redox properties
The metals act as reducing agents. This is expected. Because of the large size of their atoms and the
relatively small number of electrons in the nucleus, metals readily lose the valence electrons. By so doing
they are oxidized (oxidation is a loss of electrons) and in the process they act as reducing agents (reducing
agents are electron donors). Standard Reduction Potentials are given below for Na, Mg and Al.
The electrode potentials are negative, showing
that the metals are good reducing agents. The
value of E⁰ becomes less negative from Na to Al,
-2.71
Na+ + e- Ý Na
indicating decreasing reducing power. This is
because nuclear charge increases and outer
-2.38
Mg2+ + 2e- Ý Mg
electrons become more firmly held to the
atom. It becomes harder to remove these
-1.66
Al3+ + 3e- Ý Al
electrons during reactions.
reactions in which the metals act as reducing agents include reaction with water, oxygen and
Reduction equation
Na
Mg
Al
Typical
acids.
Eθ/V
1.2.2 Reaction with oxygen
Na and Mg burns rapidly in air, forming their respective oxides:
4Na(s) + O₂ (g) →2Na₂O(s)
2Mg(s) + O₂(s) → 2MgO(s)
These reactions are very exothermic, showing that the products being formed are very stable. Freshly
prepared Al reacts rapidly with oxygen on its surface to form a thin layer of aluminium oxide.
2Al(s) +3/2 O₂ (g) → Al₂O₃(s)
Any further reaction of the metal is prevented by the thin layer of oxide formed, since it is tenacious (sticks
firmly to the metal) and impervious to oxygen and water (the oxide layer is insoluble in water). It is the
presence of this layer of oxide which gives Aluminium its kinetic inertness to oxygen, water and dilutes
acids.
Silicon resists direct combination with oxygen because of the large number of very strong covalent bonds
present.
White phosphorous burns readily in air with a white flame, forming a white solid, P₄O₆ (limited oxygen) or
P₄O₁0 (excess oxygen).
Page
338
2P₄(s) + 5O₂ (g) → P₄O₁0(s)
When writing this equation, always remember that white phosphorous exists as P₄. In this
reaction, phosphorous attains the highest possible oxidation state (+5) in the oxide. In P4O6, the oxidation
number of P is +3, showing incomplete oxidation. P4O6 is therefore easily oxidized by oxygen to form P₄O₁0.
Note
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 338


P₄O₁0 is often written as P2O5. Note that in both formulae, the oxidation number of P is +5.
In the reaction with oxygen, P is oxidized, that is, it acts as reducing agent.
Sulphur burns steadily in oxygen with a blue flame, forming a colourless and choking gas.
S(g) + O₂(g) → SO₂(g)
In this reaction, sulphur does not attain the highest possible oxidation state of +6, showing that it is a
relatively weak reducing agent. Further oxidation of SO₂ to sulphur trioxide, SO₃, is possible
under industrial conditions, e.g. in the contact process for the manufacture of sulphuric acid. The reaction
is reversible and requires heating and the use of a catalyst. Again, this shows that S is relatively weak as a
reducing agent.
Chlorine is too weak a reducing agent to combine directly with oxygen.
Formula of the Oxides of Period 3 elements
Element
Na
Mg
Al
Si
P
S
Cl
Ar
Oxide
Na2O
MgO
Al2O3
SiO2
P4O10
P 4 O6
SO2
SO3
Unstable,
includes
Cl2O,
Cl2O7 and
ClO2
-
Oxidation
Number
+1
+2
+3
+4
+5
+3
+4
+6
+1
+7
+4
Table 9.5
The oxides of Period 3 elements
NB The highlighted oxides cannot be formed by direct combination of the element with oxygen under
ordinary conditions.
Note that the oxidation numbers of the elements in the oxides increases across the periodic table, in
response to an increase in the number of valence electrons. This is because the valence electrons are the
ones that participate in chemical reactions.
1.2.3 Reaction of the elements with water
Na and Mg are powerful enough as reducing agents to reduce water to hydrogen. Na reacts violently in cold
water in a very exothermic reaction, forming a strongly alkaline
solution of NaOH, and liberating hydrogen gas.
Page
339
Na(s) + H₂O (l) → NaOH (aq) + H₂ (g)
Because of its low density, a piece of sodium metal floats in water, so that a reaction takes place on the
surface of water. The increase in the kinetic energy of water molecules on the surface due to an increase in
temperature causes the metal to spin around on the surface. The part of the sodium metal exposed to the
air can also catch fire.
Magnesium is a weaker reducing agent than sodium. It has no appreciable reaction with cold water.
However, it reacts rapidly with steam, forming a white or yellow ash of magnesium oxide and hydrogen
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 339
Mg(s) + H₂O (g) → MgO(s) + H₂ (g)
Aluminium has no reaction with water. From an energetic point of view, we would have expected a reaction
to take place. The inertness of Al is kinetic; the metal is protected by a tough and impervious layer of Al 2O3,
as already explained.
The rest of the elements in Period 3 are rather weak as reducing agents. However, chlorine reacts slowly
and reversibly with water, to form a weakly acidic solution of HClO.
Cl2(g) + H2O (l) H+(aq) + ClO-(aq) + Cl-(aq)
This is in fact a disproportionation reaction in which chlorine is both oxidized to ClO - (oxidation number
increases from 0 to +1) and reduced to the -1 state in Cl-.
1.2.4. Reaction of the elements with chlorine
All period 3 elements except argon, combine directly with chlorine to form chlorides. The metals Na, Mg
and Al combine readily with chlorine to form ionic chlorides, for example, sodium metal burns vigorously
in chlorine.
2Na(s) + Cl2 (g)  2NaCl(s)
The reactions of Na, Mg and Al become less vigorous in that order, indicating a decrease in reducing
power of the metal. Fig 9.4 shows how AlCl3 could be prepared in the lab by the reaction
Al(s) + Cl2 (g)  Al2Cl6(s)
Heating is required to overcome activation energy associated with the tough impermeable oxide layer on
the surface of Al. The non metals, except Ar, also combine with chlorine, forming simple covalent
chlorides, for example,
P4(s) + 10Cl2 (g)  4PCl5(s)
Page
340
The reaction of the Period 3 elements with chlorine involves oxidation of the element, and reduction of
chlorine. The vigour of the reaction and chemical stability of the chloride decreases across the period due
to
 decreasing reducing power of the element
 decreasing difference in electronegativity between the element and chlorine.
The oxidation number of the element in its chloride increases across the periodic table, in response to increasing number of valence electrons. The total number of bonds that can be formed also increases (Table
9.6).
Element
Na
Mg
Al
Si
P
S
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Cl
Page 340
Chloride
NaCl
Oxidation
Number
+1
+2
Ionic
Ionic with
some covalent
character
Bonding
Behaviour
in water
No
reaction.
Dissolves to
give
solution of
pH 7
State at rtp
MgCl2
White solid
Dissolves
with slight
hydrolysis.
pH of
solution is
about 6
White solid
Lattice forces
I O N I C
Table 9.1.6
Al2Cl6
SiCl4
+3
+4
Covalent
with
significant
ionic
character
PCl5
PCl3
S2Cl2
unstable
+5
+3
+2
-
-
S i m p l e
C
o
v
U n d e r g o e s
a
l e
n
t
h y d r o l y s is
pH ≈ 3
pH ≈ 2
pH ≈ 2
pH ≈ 2
White solid
Volatile
colourless
liquid
PCl3 is a
volatile
colourless
liquid.PCl5 is
a pale yellow
solid
Orange
volatile
liquid
Mixed vdw
and ionic
V
D
W
Chlorides of Period 3 elements
The exact method of preparation of the chloride depends on its nature. Fig 9.1.4 shows how AlCl3 can be
prepared in the lab.
Page
341
Fig 9.1.4 Lab preparation of AlCl3.
1. Chlorine gas is produced by a suitable reaction. Can
you suggest such a reaction?
2. Aluminium metal is heated strongly with chlorine gas
passing over it. At the high temperature of the
reaction, gaseous AlCl3 is formed.
3. AlCl3 cools down and dimers to form Al2Cl6.
4. Unreacted chlorine is vented out.
9.2.3
Compounds of Period 3 elements
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 341
2.3.1
The chlorides
The formula and nature of the chlorides of Period 3 elements are given in Table 9.6 .
There is gradation in the bonding of the chlorides from giant ionic to simple covalent. This
corresponds to a change in the elements from metallic to non- metallic. The change in bonding from ionic
to simple covalent can be explained in terms of the increase in the electronegativity and the decrease in
electropositivity of the elements across the period. The tendency of the element to lose valence electrons
(electropositivity) and form an ionic compound decreases, and at the same time the tendency of the element
to stabilize a shared pair of electrons (covalent bond) by attracting it increases. The change in type of
bonding affects physical properties such as state at room temperature and melting points. NaCl and MgCl 2
have giant ionic bonding, and so at room temperature they are solids with high melting points. The rest of
the Period 3 chlorides are either volatile liquids or are solids with low melting points. In these covalently
bonded substances, there are weak Van der Waals operating between molecules. A small amount of heat
energy is required break these forces and separate the molecules from each other.
Shapes of the simple covalent chlorides
AlCl3
PCl5
900
Cl
Cl
P
Cl
0
120
Cl
Cl
P
PCl3
1200
Cl
trigonal planar
SiCl4
Cl
P
Cl
Cl
trigonal
bipyramidal
Cl
Cl
trigonal bipyramidal
1090
tetrahedral
Si
Cl
Cl
Cl
In PCl₅, phosphorous has more than an octet of electrons around it. This is because it has accessible d
orbitals which it can use to accommodate some of the excess electron density. We say P has the ability to
expand its octet. This ability is shown by the non- metallic elements from P (period III) onwards.
Reactions of the chlorides with water
Going across Period 3 there is an increasing tendency for the chlorides to hydrolyze in water. This coincides
with increasing covalent character across the period.
NaCl is ionic and has no reaction with water. It simply dissolves, forming a neutral solution.
Page
342
NaCl(s) → Na⁺ (aq) + Cl⁻ (aq)
The enthalpy change of solution for this reaction is +3Kj/mol. The dissolution of NaCl in water is
endothermic, that is, more energy is used to break the strong electrostatic attractions between the Na⁺ and
Cl⁻ ions than is released when these ions form electrostatic attractions with water. This observation can be
tracked down to the large size of the Na⁺ ion. It can form only weak attractions with water molecules, so
weak that the energy released is not enough to compensate for the energy used in breaking down the lattice
of NaCl.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 342
MgCl₂ is ionic, so it dissolves in water to form Mg² ⁺ and Cl⁻ ions. However, the solution formed is not
neutral. Rather, it is slightly acidic (pH= 6.5). This shows that there is a degree of covalent character in
MgCl₂. Strictly speaking, MgCl₂ dissolves in water with slight hydrolysis.
MgCl₂(s) → Mg²⁺ (aq) + 2Cl⁻ (aq)
Hydrolysis is a reaction in which a water molecule is split, forming the OH⁻ and H⁺ ions. It is the presence
of H⁺ ions from water that lowers the pH from 7. The Mg²⁺ ion is quite small compared to its charge. The
effect of small ionic size and large charge is to increase the charge density of the cation. A cation with a high
charge density has a large charge confined in a small volume. Such an ion is good at attracting a lone pair
in water, leading to hydrolysis.
H
Mg
2+
O
Mg2+ + OH- + H+
H
The OH⁻ ions produced form a co-ordination complex with Mg²⁺ and water, so they cannot recombine
with H⁺ ions.
Mg2+ (aq) + OH-(aq) + 5H2O (l)  [Mg(H2O)5OH]+(aq)
However, a much larger amount of Mg²⁺ ions will remain as hydrated ions, that is, Mg 2+ ions surrounded
by shells of water molecules.
Aluminium chloride, being covalent, hydrolyses in water more readily than does MgCl ₂. The solution
formed is also more acidic (pH ≈ 3).
AlCl3(s) + 6H2O (aq)  [Al (H2O)5OH]2+(aq) + H+(aq) + 3Cl-(aq)
Page
343
When a small amount of cold water is added to a sample of solid aluminium chloride (which exists as Al2Cl3
at low temperatures), a vigorous reaction takes place, producing copious fumes of HCl. The heat produced
during the reaction drives out the volatile HCl from solution. If a large amount of cold water is used, little
or no fumes are produced. This is because there is not enough heat to drive out the HCl.
The behaviour of AlCl₃ in water can be explained in two ways.
 Al in AlCl₃ has in incomplete octet around it (it has a sextet of electrons). AlCl₃ is therefore electron deficient at the Al atom. The Al atom can therefore attract a water molecule through a lone
pair of electrons on the oxygen. This is the initial step in the hydrolysis reaction.
 Since aluminium chloride has a high degree of ionic character, we can imagine some of the molecules dissociating in water to form Al³⁺ and Cl⁻ ions. The Al³⁺ ion has a very small radius and a
large charge. Because of its very high charge density, it easily hydrolyses water molecules, forming
an acidic solution
A clear distinction should be made between dissolution and hydrolysis. Dissolution is a physical process,
which results in the formation of a mixture (solution). Components of the mixture can easily be separated
e.g by evaporation (At junior levels it is often erroneously taught that physical changes such as dissolution
are not accompanied by heat energy changes. This is clearly wrong. Most dissolution processes absorb or
give out heat energy, known as the enthalpy of solution),
Hydrolysis is a chemical reaction which involves splitting of a water molecule. New products are formed
and this process is not as easy to reverse as is the case with dissolution.
SiCl₄ and PCl₃/PCl₅ hydrolyze in water much more readily than AlCl₃. A larger number of H⁺ ions are
produced in solution, so the solution formed is even more acidic.SiCl 4 and PCl3 are colourless liquids
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 343
which fume in moist air. This is a result of the reaction between the chloride and water in the atmosphere,
to produce fumes of HCl. PCl5 is a pale yellow solid which hydrolyses rapidly in water.
The behavior of these chlorides in water is typical of covalent chlorides with accessible d orbitals. These d
orbitals are used to accommodate a lone pair of electrons from water during the first step of hydrolysis.
Note that hydrolysis is not accompanied by a change in oxidation number, since it is not a redox reaction.
SiCl₄(l) + 2H₂O (l) → SiO₂(s) + 4H⁺ (aq) + 4Cl⁻ (aq) (= SiO₂ + 4HCl)
PCl₃(l) + 3H₂O (l) → H₃PO₃ (aq) + 3HCl (aq) (phosphonic acid + hydrochloric acid)
PCl₅ (l) + 4H₂O (l) → H₃PO₄ (aq) (aq) +5HCl (aq) (phosphoric acid + hydrochloric acid)
A note on phosphonic and phosphoric acids
Phosphonic acid, H3PO3 is the reduced form of phosphoric acid, H3PO4. It is produced from the hydrolysis of PCl3,
whereas H3PO4 is produced from the hydrolysis of PCl5. Note that PCl3 is also the reduced form of PCl5. (In the
presence of excess Cl2, PCl3 is oxidized to PCl5). A common error among students is to treat both H3PO3 and H3PO4
as tribasic acids. Only H3PO4 is tribasic. All its hydrogen atoms are bonded to oxygen, so they can be
dissociated in solution.
H3PO4  PO42- + 3H+
H3PO3 is dibasic. Two hydrogen atoms are bonded to oxygen. These can be dissociated in solution. The third
hydrogen atom is bonded to P and so cannot be dissociated in solution.
phosphoric acid
O
P
OH
OH
OH
phosphonic acid
O
P
H
OH
OH
More information on PCl5
Written as PCl5, the formula of this compound is misleading. At room temperature it is actually made up of
[PCl4]+ and [PCl6] - ions, that is, it exists as the ionic compound
[PCl4]+ [PCl6]-. What happens is that one PCl5 donates a chloride ion to another PCl5 molecule. The donating
molecule therefore becomes [PCl4]+ whilst the receiving molecule becomes[PCl6]-. The structures of these two ions
Disulphur
dichloride, S₂Cl₂ is an unstable orange liquid which readily reacts with water in a
are shown below.
disproportionation reaction.
+ + SO₂ + 4HCl)
Cl (aq) (= 3S
2S₂Cl₂ (l) + 2H₂O (l) → 3/8S₈(s) + SO₂ (g) + 4H⁺ (aq) + 4Cl⁻
Cl
Notice that the oxidation state of sulphur decreases from
0 in sulphur and
Cl+1 in S₂Cl₂ toCl
P
increases to +4 in SO₂.
This shows that the reaction involves
disproportionation.
This reaction is an
P
Cl
oversimplification. In practice,
the reaction is complex and produces several other sulphur containing
Cl
Cl
Cl
products. The observations
made include turbidity (cloudiness) dueClto the formation of solid sulphur and a
Cl
pungent smell due to the production of sulphur dioxide.
octahedral ( bond angle 900)
Page
344
tetrahedral (bond angle 1090)
More information on S2Cl2
Disulphur dichloride is an unstable orange -yellow liquid which fumes in air due to reaction with water vapour. The
white
fumes formed
are HCl. The
structure
of S2Cl2 is analogous
to that
hydrogen
peroxide,
H 2O 2.
Prepublication
manuscript
© L.
MWANAWENYU
2011. Please
doofnot
photocopy
or reproduce
Page 344
2.3.2 The oxides of Period 3 elements
Element
Oxide
Na
Mg
Al
Si
P
S
Cl
Na2O
MgO
Al2O3
SiO2
P4O10
P4O6
SO2
Cl2O7
(Cl2O)
State at 200C
Bonding
Behaviour
in water
Acid/base
nature
m.p/0C
S
o
G i a n t
i o n i c
Reacts,
forming
NaOH
Reacts,
forming
Mg(OH)2
Alkaline
pH≈13
Weakly
alkaline
pH≈9
Basic
1275°
sublimes
l
i
d
Giant
covalent
No reaction.
Insoluble
Basic
amphoteric
2 900
2 040
No
reaction.
Insoluble
Gas
Liquid
(Gas)
Simple molecular
Hydrolyzes, forming
strongly acidic solution.
pH≈2
A c i d i c
1 610
580
24
-10
Page
345
Table 9.1.7 The chemical and physical properties of the Period 3 elements. Note that
there is considerable discrepancy on the melting/sublimation point of Na 2O reported in
different sources.
Bonding and structure
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 345
The bonding in the oxides change from ionic to covalent across the period. This coincides with the change
in the elements from metallic to non- metallic. Na₂O and MgO have giant ionic structures. Al₂O₃ has a giant
ionic structure, but there is also a high degree of covalent character. SiO₂ has a giant covalent structure,
similar to that of diamond. The rest of the period three oxides are simple covalent.
The melting points of the elements depend on the bonding and structure. Na 2O and MgO have very high
melting points since they have giant ionic structures. The melting point of MgO is much higher than that of
Na2O. This is because of the doubly charged Mg2+ ion in MgO. This results in very strong ionic attractions
between the Mg2+ and O2- ions. A very large amount of heat energy is required to break down the lattice and
pull the ions from each other.
The melting point of Al2O3 is lower than that of MgO, something we might not have expected, given that the
Al3+ ion has the larger charge and so electrostatic attractions would be greater in Al 2O3 than in MgO. The
relatively low melting point of Al2O3 can be explained in terms of the high polarizing power of the Al3+ ion.
Polarization is when a cation withdraws electrons from, and therefore distorts the electron cloud of an anion.
In this case, Al3+, with its high charge density, withdraws some electron density from the oxide ion.
Consequently, there is some shared electron density (covalency) between the two atoms. The bonding in
Al2O3 is therefore not purely ionic; the presence of covalent character weakens the bonding present, and
consequently the melting point is relatively low. However, the melting point is not reduced by a large value.
This is because the oxide ion is relatively small and offers considerable resistance to being polarized.
Reactions of the oxides with water
Na₂O reacts readily with water in an exothermic reaction to form a strongly alkaline solution of NaOH
(pH is about 13)
Na₂O(s) + H₂O (l) → 2NaOH (aq)
MgO is sparingly soluble in water, hence the solution formed is weakly alkaline. The poor solubility of
MgO in water can be explained in terms of its high lattice energy. A large amount of energy is required to
break down the lattice of MgO since the electrostatic attractions between Mg 2+ and O2- ions are very
strong.
MgO(s) + H₂O (l) → Mg²⁺ (aq) + 2OH⁻ (aq)
Al₂O₃ is insoluble in water. The lattice energy of the oxide is high and water molecules are not able to
break it down. Also, Al2O3 has a complex structure which is not purely ionic. This structure is insoluble in
water.
An important note
Page
346
The chemistry of aluminum compounds is heavily influenced by the very high charge density of the Al³⁺ ion. For instance,
one would expect AlCl₃ to be ionic, not covalent, since it is formed between a reactive metal and a reactive non-metal. AlCl₃
is in fact mainly covalent, though it has a high degree of ionic character. Because of its high charge density, the Al³⁺ in AlCl₃
withdraws electrons from the chloride ion, resulting in a shared electron density between the two atoms. This shared
density of electrons is significantly stabilized through mutual attraction by the nuclei of Al and Cl. Electron density shared
between two atoms is, of course, a covalent bond.
It is also observed that some compounds of Al are very unstable, for example, aluminum carbonate, Al2 (CO3) 3. With its
high charge density, Al³⁺ readily polarizes the carbonate ion, and in the presence of water
or water vapour, carbon dioxide and aluminium (III) hydroxide are produced.
Al2 (CO3)3 + 3H2O → 2 Al (OH) 3 + 3CO2
SiO₂ is a giant covalent solid with a diamond like structure. It is insoluble in water. Water molecules are
not able to break the very strong and numerous Si –O bonds.
The oxides of phosphorous are simple covalent solids, soluble in water through hydrolysis.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 346
It should be noted that whereas metallic oxides react with water to form an alkaline solution, nonmetallic oxides, if they do dissolve, will hydrolyze to form acidic solutions.
P₄O₆(s) + 6H₂O (l) → 4H₃PO₃ (aq)
P₄O₁0 (s) + 6H₂O (l) → 12H⁺ (aq) +4PO₄²⁻ (aq) (= 4H₃PO₄)
P₄O₆ forms phosphonic acid, H3PO3 whilst P₄O₁0 forms phosphoric acid.
SO₂ and SO₃ are simple covalent gases. SO₂ is weakly soluble in water, forming a solution of sulphrous
acid. Notice that the oxidation state of S is +4 in both sulphrous acid (H₂SO₃) and SO₂. In fact, sulphrous
acid is sometimes written as SO₂ (aq).
SO₂ (g) + H₂O (l) ⇋ 2H⁺ (aq) + SO₃²⁻ (aq)
Sulphrous acid (H₂SO₃) is weakly acidic and is easily decomposed back to SO₂ and water. The acid has no
independent existence and it has a tendency to be oxidized in air, to form the more stable sulphuric acid.
H₂SO₃ (aq) + 1/2O₂ (g) → H₂SO₄ (aq)
In this reaction, the oxidation number of S changes from +4 to +6. This is the oxidation of aqueous SO₂ to
sulphuric acid. As such, aqueous SO₂ is a reducing agent, able, for instance, to reduce Cr2O72- (aq) to
Cr3+ (aq) . SO₃ is not naturally occurring. It is formed industrially by reacting SO₂ with oxygen in the
presence of a vanadium (V) oxide catalyst (Contact process).
SO₃ hydrolyzes readily and violently in water to form a mist of sulphuric acid.
SO₃ (g) + H₂O (l) → 2H⁺ (aq) + SO₄²⁻ (aq)
The oxidation state of S in both sulphuric acid and sulphur trioxide is +6. This is the highest possible
oxidation state for S.
Acid base nature of the oxides
The oxides of Period 3 change from basic through amphoteric to acidic.
Na₂O and MgO are ionic and basic. They are neutralized readily by cold dilute acids, forming their
respective salts and water, for example
Na₂O(s) + 2HCl (aq) → NaCl (aq) + H₂O (l) ...
pH = 7
MgO(s) + H₂SO₄ (aq) → MgSO₄ (aq) + H₂O (l) (pH is slightly less than 7 because of
hydrolysis of Mg2+ ions ).
These reactions can be reduced to the ionic form
O²⁻ + 2H⁺ → H₂O
In the ionic oxides, neutralization occurs in this way because of the presence of the oxide ion in the
lattice. The reaction basically involves the combination of two unstable ions, forming a neutral and stable
product, H₂O.
Al₂O₃ is amphoteric, that is, it behaves both as an acid and as a base. It reacts with dilute acids in the
same way as MgO and Na₂O.
Page
347
Al₂O₃(s) + 6H⁺ (aq) → 2Al³⁺ (aq) + 3H₂O (l)
pH ≈ 3
The solution formed is acidic because the Al³⁺ ions hydrolyze in water:
Al³⁺(aq) + 6H₂O(l) → [Al(H₂O)₆]³⁺(aq) →[ Al(H₂O)₅OH]²⁺(aq) + H⁺(aq)
Al₂O₃ reacts with a strong base such as NaOH, to form an aluminate
Al₂O₃ (s) + 2NaOH (aq) + 3H₂O (l) → 2Na [Al (OH)]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 347
The product of this reaction, sodium aluminate, Na [Al (OH)], contains the aluminate ion, which is a
complex ion. The formation of complex ions is usually associated with transition elements. The ability of
Al3+ to form a complex ion can once more be explained in terms of the high charge density of the ion. It is
able to strongly attract four OH- ions, and form dative bonds with the ions by using a lone pair of electrons
on the O atom of each OH- ion.
OH
Tetrahedral geometry in the aluminate ion. All of
the bonds (in bold) are dative. This is an example of
a complex ion.
Al3+
A complex ion is a covalent ion in which a central
metal cation is bonded datively to one or more
electron rich groups, known as ligands.
OH
HO
HO
bond angle = 1090
In the reaction of Al₂O₃ with OH- ions , Al₂O₃ acts as an acidic oxide, forming an aluminate salt. The
amphoteric nature of Al₂O₃ shows that though the oxide is predominantly ionic, there is some degree of
covalent character. Once more, this is caused by the high charge density of the Al³⁺ ion.
The oxides of the non- metals are acidic
SiO₂ is acidic, but it will not react with alkalis under normal conditions because of the very strong Si-O
bonds that must be broken in the reaction. SiO₂ does react with bases in a non- aqueous reaction and at
very high temperatures to form salts known as silicates. In the blast furnace for the extraction of iron,
SiO₂, present in sandy impurities, is neutralized by CaO to form calcium silicate, which is tapped out as a
slag.
SiO₂(s) + CaO(s) → CaSiO₃(l)
SiO2 also reacts with fused NaOH to form sodium silicate.
SiO₂(s) + 2NaOH  Na2SiO3
P₄O₆ and P₄O₁₀ are acidic oxides that readily react with alkalis, forming strongly acidic solutions, e.g.
P₄O₆ (s) +8NaOH (aq) →4Na₂HPO₃ (aq) + 2H₂O (l)
To balance this equation, it is necessary to know that the salt being formed contains the hydrogen
phosphorous ion, HPO₃²⁻, and not PO₃³⁻ (The hydrogen atom in the hydrogen phosphorous ion is bonded
to the P atom and so it is not dissociated).
A hint on balancing the equation
Page
348
Students often find this equation difficult to balance. It can be balanced in two steps. First, the oxide reacts with water
to form an acidic solution of phosphorous acid. The H⁺ ions from the diprotic phosphorous acid are then neutralized
by NaOH (aq). The two reactions are then added to obtain the net equation.
1. P₄O₆ (s) + 6H₂O (l) → 8H⁺ (aq) + 4HPO₃²⁻ (aq)
2. 8H⁺ (aq) + 8NaOH (aq) → 8H₂O (l) + 8Na⁺(aq)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 348
Overall
P₄O₆ (s) +8NaOH (aq)  4Na₂HPO₃ (aq) + 2H₂O (l)
Similar reactions can be balanced in the same way (that is, in reactions in which the oxide can react with both water
and a base such as sodium hydroxide)
P₄O₁₀ reacts with an alkali to form phosphate.
P₄O₁₀(s) + 12NaOH (aq) → 12Na⁺ (aq) + 4PO₄³⁻(aq) + 6H₂O(l) ( = 4Na₃PO₄ + 6H₂O)
Sulphur has two stable oxides, SO₂ and SO₃ (only SO2 exists naturally). Both react with NaOH (aq) to
form sodium sulphite and sodium sulphate respectively.
The equations can also be balanced by the 2 step method used in the previous example. It should be noted
that SO₂ produces a sulphite (SO₃²⁻) and SO3 produces a sulphate (SO₄²⁻).
SO₂ (g) + 2NaOH (aq) → 2Na⁺ (aq) + SO₃²⁻ (aq) + H₂O (l) (= Na₂SO₃ + H₂O)
The products of the reaction are the same as those produced when sulphrous acid, H 2SO3, reacts with
NaOH. This is because an aqueous solution of SO2 is essentially a solution of sulphrous acid. The equation
above can therefore be obtained by combining two equations,
SO2 + H2O  H2SO3
H2SO3 + 2NaOH  Na2SO3 + 2H2O
Net equation
SO2 + 2NaOH  Na2SO3 + H2O
Note that H2SO3 is dibasic, so it reacts with NaOH in the ratio 1:2, to produce two moles of water.
The reaction of SO₃ with NaOH (aq) is violent.
SO₃(l) + 2NaOH (aq) → 2Na⁺ (aq) + SO₄²⁻ (aq) + H₂O (l)
(= Na₂SO₄ + H₂O)
The products of the reaction are the same as those produced when sulphuric acid reacts with NaOH. This
is because in water, SO3 is converted to H2SO4.
Chlorine forms unstable oxides, Cl₂O₇ and Cl₂O. Both are acidic and react readily with NaOH (aq) to
form chlorate (VII) and chlorate (I) acid respectively.
Cl₂O₇ (l) +2NaOH (aq) → 2NaClO₄ (aq) +H₂O (l)
The product formed is an aqueous solution of sodium chlorate (VII), which is also known as sodium
perchlorate and is the conjugate base of chloric (VII) acid, HClO₄, also known as perchloric acid. The
products of the reaction above are therefore the same as those formed when HClO₄ reacts with NaOH
(aq). The reaction of dichlorine oxide with NaOH is
Cl₂O (g) + 2NaOH (aq) → 2NaClO (aq) + H₂O (l)
Page
349
This reaction forms sodium chlorate (I), which is the conjugate of chloric (I) acid, HClO.
Notes for use in qualitative analysis
The reader should be familiar with the tests and chemistry of the following Period 3 ions.
Mg2+
Al3+
Cl-
Prepublication
manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Here we discuss only Al3+. The other ions will be discussed later under the Groups (II) and (VII) ions.
A summary of the reactions of Al3+ (aq)
Page 349
350
Page
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 350
Questions, solutions and discussions
Q1
The elements phosphorus, sulphur, and chlorine are regarded as having simple molecular
structures.
(a)
What are the molecular formulae of each of these three elements?
(b)
(i) Place the three elements in order of their melting points with the highest first.
(ii) Suggest an explanation for the order you have given in (i).
(c)
(d)
Sulphur and chlorine can be reacted together to form disulphur dichloride, S2Cl2.
Disulphur dichloride, S2Cl2, is decomposed by water forming sulphur and a mixture of
hydrochloric acid and sulphurous acid. When 2.7 g of S2Cl2 is reacted with an excess of water,
0.96 g of sulphur, S, is produced.
(i)
What is the amount, in moles, of S2Cl2 present in 2.7 g?
(ii)
What is the amount, in moles, of S produced from 1.0 mol of S 2Cl2?
(iii)
Construct a balanced equation for the reaction of S 2Cl2 with water.
The reaction between S2Cl2 and water is a redox reaction.
Which product has been formed by oxidation and which by reduction?
9701/02/O/N/07
Solutions
3 (a)
P4 , S8 , Cl2
(b) (i) highest is S8 , Cl2 is lowest
(ii) The number of electrons in the molecules decreases in the order S8, P4, Cl2. Therefore Van der
Waals forces become weaker.
(c)
(iii)
2S2Cl2 + 3H2O → 3S + H2SO3 + 4HCl
Oxidation product is H2SO3 and reduction product is S
Page
351
(d)
(i) and (ii) is for the reader
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 351
Q2
The Periodic Table we currently use is derived directly from that proposed by Mendeleev in 1869
after he had noticed patterns in the chemical properties of the elements he had studied.
The diagram below shows the first ionization energies of the first 18 elements of the Periodic
Table as we know it today.
(a)
Give the equation, including state symbols, for the first ionization energy of fluorine.
(b)
Explain why there is a general increase in first ionization energies from sodium to
argon.
(c)
(i) Explain why the first ionization energy of aluminium is less than that of magnesium.
(ii) Explain why the first ionization energy of sulphur is less than that of phosphorus.
The table below refers to the elements sodium to sulphur and is incomplete.
(d)
(i)
Complete the ‘melting point’ row by using only the words ‘high’ or ‘low’.
(e)
When Mendeleev published his Periodic Table, the elements helium, neon and argon were not
included. Suggest a reason for this.
9701/02/M/J/08
Page
352
(ii) Complete the ‘conductivity’ row by using only the words ‘high’, ‘moderate’ or ‘low’.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 352
(a)
F(g) → F+(g) + e–
(b)
From Na to Ar, electrons are added to the same shell where they shield each other only slightly.
At the same time, nuclear charge increases and so the valence electrons feel a stronger attraction
(c) (i)
(ii)
Outermost electron in Al is in 3p which is higher in energy/farther from the nucleus than the 3s
outer shell in Mg.
Phosphorous and sulphur :
For phosphorous the valence 3p sub-shell is singly filled/exactly half-filled. For S one 3p orbital
has paired electrons which repel each other.
(d) (i) and (ii)
Element
(e)
Q3
Na
Mg
Al
Si
P
S
Melting point
low
…….
high
high
low
low
conductivity
high
…….
high
moderate
low
low
They had not been discovered yet
Magnesium chloride, MgCl2, and silicon tetrachloride, SiCl4, each dissolve in or react with water.
Suggest the approximate pH of the solution formed in each case. Explain, with the aid of an
equation, the difference between the two values.
9701/02/O/N/2008
Solutions
MgCl2 : 6 to 7
SiCl4: 0 to 3
MgCl2 has no reaction with water, it simply dissolves
SiCl4 reacts with water/undergoes hydrolysis in water
SiCl4 + 2H2O → SiO2 + 4HCl or
SiCl4 + 4H2O → Si(OH)4 + 4HCl or
NB MgCl2 actually undergoes slight hydrolysis in water
Page
353
SiCl4 + 4H2O → SiO2.2H2O + 4HCl
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 353
Q4
Copper and titanium are each used with aluminium to make alloys which are light, strong
and resistant to corrosion.
Aluminium, Al, is in the third period of the Periodic Table; copper and titanium are both
transition elements.
(a)
Write the electronic configuration of aluminium and of titanium, proton number 22.
Aluminium reacts with chlorine.
(b) (i) Outline how, starting from aluminium powder, this reaction could be carried out
in a school or college laboratory to give a small sample of aluminium chloride. A diagram
is not necessary.
(ii)
Describe what you would see during this reaction.
(iii)
At low temperatures, aluminium chloride vapour has the formula Al2Cl6.
Draw a ‘dot-and-cross’ diagram to show the bonding in Al2Cl6. Show outer electrons only.
Copper forms two chlorides, CuCl and CuCl2.
(c)
When copper is reacted directly with chlorine, only CuCl2 is formed.
Suggest an explanation for this observation.
Titanium also reacts with chlorine.
(d)
When an excess of chlorine was reacted with 0.72 g of titanium, 2.85 g of a chloride A
was formed.
(i) Calculate the amount, in moles, of titanium used.
(ii) Calculate the amount, in moles, of chlorine atoms that reacted.
(iii)
Hence, determine the empirical formula of A.
(iv)
Construct a balanced equation for the reaction between titanium and chlorine.
(e)
At room temperature, the chloride of titanium, A, is a liquid which does not conduct
electricity. What does this information suggest about the bonding and structure in A?
9701/02/M/J/2009
Solution
(b) (i) Pass chlorine gas over heated aluminium
(ii) Aluminium glows and a white solid is formed . Colour of chlorine fades.
Page
354
1 (a) Left for the reader
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 354
(iii) Left for the reader
(c) Chlorine is a powerful oxidizing agent
(d) (i) Moles of Ti =
0.72
47.9
= 0.015
(ii) Moles of chlorine =
(2.85−0.72)
35.5
= 0.06
(iii) Ti : Cl2 = 0.015 : 0.06 = 1:4
Therefore empirical formula of A is TiCl4
(e) Simple covalent.
Q5
Magnesium will react on heating with chlorine, or oxygen, or nitrogen to give the chloride, or
oxide, or nitride respectively. Each of these compounds is ionic and in them magnesium has
the same +2 oxidation state.
(a) (i) Write an equation, with state symbols, for the second ionization energy of
magnesium.
(ii) Use the Data Booklet to calculate the enthalpy change that occurs when one mole
of gaseous magnesium ions, Mg2+, is formed from one mole of gaseous magnesium
atoms.
Include a sign in your answer.
(b)
Separate samples of magnesium chloride and magnesium oxide are shaken with water.
In each case, describe what you would see when this is done, and state the approximate
pH of the water after the solid has been shaken with it.
(c)
Magnesium burns in nitrogen to give magnesium nitride, a yellow solid which has the
formula Mg3N2.
Magnesium nitride reacts with water to give ammonia and magnesium hydroxide.
Construct an equation for the reaction of magnesium nitride with water.
(ii)
Does a redox reaction occur when magnesium nitride reacts with water?
Use the oxidation numbers of nitrogen to explain your answer.
9701/21/M/J/2009
Page
355
(i)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 355
Solutions
2 (a) (i) Mg+(g) → Mg2+ (g) + e–
(ii) 736 + 1450 = +2186 kJ mol–1 (sum of first and second ionization energy)
(b) (i) MgCl2 dissolves , giving a solution of pH 6 – 7
(ii) MgO is slightly soluble in water, giving a pH of 8-11
(c) (i)
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3
(ii)
In Mg3N2, N has oxidation state of –3. In NH3, N has oxidation state of –3. Therefore reaction is
not redox since N does not change its oxidation state.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise
9.1
text box.]
1. The elements of the third period of the Periodic Table form chlorides of general formula EClx.
where E represents the element. These chlorides show a variation in oxidation number from
sodium to sulfur.
(a) (i) Use the information given to complete the table below.
Formula of chloride
NaCl
MgCl2
AlCl3
SiCl4
PCl3
SCl2
Oxidation number of
element in the
chloride
Page
356
(ii)
By considering the electron configurations of the elements, explain the variation in
oxidation number in the chlorides from Na to Al and from Si to S.
Sodium hydride, NaH, is a colourless crystalline solid which melts at 800 °C and has the
same crystal structure as sodium chloride which has a melting point of 808 °C. When molten
sodium chloride is electrolyzed using graphite electrodes, a shiny deposit, D, forms on the
cathode and a greenish-yellow gas is evolved from the anode. When molten sodium hydride
is electrolysed, under suitable conditions using graphite electrodes, the same shiny deposit
D is formed on the cathode and a colourless gas, G, is evolved from the anode.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 356
(b)
(i) Describe with the aid of a diagram the bonding in a sodium chloride crystal.
(ii) Suggest the type of bonding that is present in sodium hydride.
(iii) What is the oxidation number of hydrogen in sodium hydride?
(iv)
Draw a ‘dot-and-cross’ diagram for sodium hydride. Show outer electrons only.
(v)
The metals magnesium and aluminium form hydrides with formulae MgH 2 and
AlH3. The non-metals phosphorus and sulfur form hydrides with formulae PH 3 and
H2S.
By considering their positions in the Periodic Table, suggest oxidation numbers for
these four elements in their hydrides.
At room temperature, the chlorides of sodium, magnesium and aluminium are all solids which
dissolve in water.
The hydrides of sodium, magnesium and aluminium are also solids which react with water
with the rapid evolution of the same colourless gas G in each case.
(c)
(i) What is the pH of the solutions formed when separate samples of sodium chloride,
magnesium chloride, and aluminium chloride are dissolved in water?
(ii)
Suggest an equation for the reaction between sodium hydride and water.
(iii)
Suggest a value for the pH of the solution formed in (ii).
At room temperature, the chlorides of silicon, phosphorus and sulfur are all low melting point
solids or low boiling point liquids that can be seen to react with water.
(d)
(i) Suggest what type of bonding is present in sulfur dichloride, SCl2.
(ii) Write a balanced equation for the reaction between the chloride of silicon, SiCl4,
and water.
9701/22/O/N/2009
2. The formulae and melting points of the fluorides of the elements in Period 3, Na to Cl, are
Page
357
given in the table.
(c)
Formual of
fluoride
NaF
MgF2
AlF3
SiF4
PF5
SF6
ClF5
m.p. /K
1268
990
1017
183
189
223
170
(i) Suggest the formulae of two fluorides that could possibly be ionic.
(ii)
What is the shape of the SF6 molecule?
(iii)
In the sequence of fluorides in the table, the oxidation number of the elements increases
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 357
from NaF to SF6 and then falls at ClF5.
Attempts to make ClF7 have failed but IF7 has been prepared.
Suggest an explanation for the existence of IF7 and for the non-existence of ClF7
9701/22/M/J/2010
3 (a) State the formulae, and describe the reactions with water, of the oxides of the elements in the
third period from sodium to sulphur.
9701/22/M/J/1992
4. It has often been stated that the elements of the Periodic Table show a general trend across a period
from metallic to non-metallic behaviour, coupled with an increase of oxidation number in their
compounds.
Illustrate this statement by describing and explaining the formula, the properties and the reactions
with water of the chlorides of the elements sodium to phosphorous. 9701/22/M/J/1992
5 (a) Describe the variations in melting points and conductivities of the elements sodium to argon, and
explain these variations in terms of their structures and bonding.
(b) Alloys of aluminium and magnesium are often used in aircraft manufacture because of their high
strength and low density.
A 1.00 g sample of one such alloy was reacted with an excess of aqueous sodium hydroxide and
the volume of hydrogen given off was measured : 1.00 dm3 of gas was produced at 250C and a
pressure of 1.01 x 105 Pa.
The aluminium containing product of this reaction is the same as that from the reaction between
aluminum oxide and sodium hydroxide. From your knowledge of the latter reaction, construct an
equation for the reaction between aluminium and sodium hydroxide, and hence calculate the
percentage of aluminium in the sample of alloy.
9701/22/M/J/1992
Page
358
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 358
9.2 GROUP (II) ELEMENTS: Mg To Ba
Introduction
Some generalizations can be made about elements in the same group.
They have the same number of valence electrons, and so they are expected to have similar chemical
properties. This is observed most of the times, but exceptions do exist, and reactivities tend to change
down the group. Also, the similarity in chemical properties is well defined in some groups, and not so
well in others. It is also noticed that the first member of a group usually has properties which are
atypical of the group, for example, Beryllium in Group (II) forms a covalent chloride, BeCl₂, but the rest
of the elements form ionic chlorides.
9.2.1
Electronic structure of Group (II) elements
Valence shell configuration: ns²
The Group(II) elements therefore belong to the S block of the Periodic table (just as the Group(I) metals,
because only the s-sub shell is occupied in the valence shell. The value of n is equal to period number, for
instance, in Mg n=3 (Period 3) and in Sr, n=5 (Period 5).
From the configuration of the valence shell, we learn the following


The Group (II) metals react by losing two valence electrons. They therefore act as reducing agents.
They form +2 ions in their reactions. All of their compounds only have the +2 oxidation state.
A question which immediately arises is why the +1 oxidation state, corresponding to loss of one of the
valence 2s electrons, is not formed. This can be answered in two ways.


9.2.2
Electrons usually stabilize each other by pairing. A single electron in the 2s orbital is very unstable,
and so it is easily removed.
If only one valence electron is lost, the resulting +1 ion would have an outer shell with only one
electron, which is very unstable. By losing both electrons, a full octet of electrons is exposed. The
+2 state is therefore more stable, and its formation is favoured to that of the +1 state.
Structure and bonding
Page
359
The elements of Group (II) are all metallic. However, metallic character increases down the group.
Beryllium, the first member of the group, is the least metallic. It has a high degree of non- metallic character,
for instance, it forms a covalent chloride, BeCl2. Going down the group, more shells are opened,
and atoms become bigger in size. Two factors come into play:

The distance factor. The distance of the valence electrons from the nucleus increases. The
attraction they feel from the nucleus becomes weaker and weaker. Down the group, the outer most
electrons are easily detached and contributed into a common pool of delocalized electrons, which
characterizes metals.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 359

The screening (shielding) effect increases. Each additional shell opened screens (shields) the
valence electrons from feeling the full attractive influence of the nucleus.
One might expect the effect of shielding to be compensated for by the increasing nuclear charge. That is, as
the number of protons increases down the group, the attraction of the nucleus for the valence electrons
would increase and shells would be pulled inwards, keeping the size of atoms small. This does happen to
some extent, but the shielding effect has a greater impact than the increase in nuclear charge.
9.2.3
Physical properties
2.3.1 Atomic and ionic radius
Atomic
number
Metallic
radius/nm
Ionic radius of
M2+ ion/nm
Mg
12
0.160
0.065
Ca
20
0.197
0.094
Sr
38
0.215
0.110
Ba
56
0.211
0.134
Fig 9.2.1
Trend in atomic
and ionic radii
down Group (II)
Table 9.2.1 Trend in atomic and ionic radii
down Group (II)


Atomic and ionic radii increase down the group. This is because additional shells are being opened.
The more shells an atom has, the larger it is. However, the increase in atomic radii from one element
to the next is not large because the increase in nuclear charge has a counteractive effect: as proton
number increases, the ability of the nucleus to pull shells towards itself increases, keeping the atoms
relatively small.
The ions are smaller than the atoms from which they are formed. This is because each ion has one
less shell than the neutral atom.
2.3.2 Ionization energies
Page
360
Ionization energy decreases down the group.
The outer most electrons become more and more loosely held to the atom for two reasons:
 Increasing size of the atoms (due to opening of new shells) implies that the distance of the valence
electrons from the nucleus increases.
 The opening of new shells implies that the valence electrons become more and more shielded from
feeling the full attractive influence of the nucleus.
These two factors work together so that valence electrons become loosely held to the atom down the
group, and less energy would be required to remove them
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 360
2.3.3 Density
Table 9.2.2
Densities of Group
(II) elements
Density gives a measure of how much mass is packed in a given volume. A large mass packed in a small
volume results in high density. The trend in density is not so straightforward for two reasons
 Density is made up of two terms, mass and atomic volume, which has opposite effects to each
other. The larger the mass, the greater the density, but the larger the volume, the smaller the
density. Down the group, volume of the atoms increase, so density is expected to decrease (D is
inversely proportional to V). This observation is made only for the first three elements. Density
decreases from Be to Ca, but begins to increase from Sr to Ba.
The increase in density from Sr to Ba is a result of the sharp increase in mass. In Sr and Ba the
increase in mass is much greater than the increase in volume, so density increases.
 It should also be noted that the elements of Group (II) have different crystal structures and this
has an effect on the trend on densities.
2.3.4
Melting points and boiling points
The Group (II) elements generally have high melting points because of the presence of strong metallic
bonding. In the metallic lattice, metal atoms have a charge of +2, and so they form strong electrostatic
attractions with the sea of delocalized electrons. In general, melting points decrease down the group. As
the size of atoms increase, the charge density on the positive atomic cores decrease, and the electrostatic
attractions between the atoms and the sea of delocalized electrons become weaker, that, is less energy would
be required to separate the atoms from each other and break down the metal lattice during melting. The
increase in atomic size weakens attraction in another way. When the atoms are large, then the sea of
delocalized electrons is a large distance from the nuclei of the metal atoms. The sea of delocalized electrons
is therefore weakly attracted to the positive nuclei of the atoms.
Mg has an abnormally low melting point. This is a result of its specific crystal structure. The metals in group
(II) have different crystal structures, so that the trend in melting points is not as regular as would have been
expected from a consideration of atomic radii alone.
Page
361
9.2.4 Trends in chemical properties
The Group (II) elements are metals and so they show chemical properties usually associated with metals.
 They react by losing their valence electrons. Thus they are reducing agents as they tend to give
away the valence electrons to oxidizing agents such as oxygen, chlorine and dilute acids. The
Group (II) elements have a fixed oxidation number of + 2 in their compounds, corresponding to
loss of the two valence electrons.
 In general, reactivity increases down the group and reactions become more violent. As the size of
atoms increases, the tendency for the valence electrons to be lost during chemical reactions
increases as the electrons become weakly held to the atoms. This is supported by electrode potential
data for M2+(aq)/M(s) for the Group (II) metals. The values given below are for the forward
reduction half equation M2+(aq) + 2e- ∏ M(s)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 361
Table 9.2.3 Reduction potentials
(M2+ (aq) + 2e ⇋ M(s)) of the group (II) metals
In general, the electrode potentials are large and
positive in the reverse (oxidation) direction. This
shows that the Group (II) metals are very reactive
and are easily oxidized to the +2 state. The
electrode potentials become more positive in the
reverse direction, showing that the metals are
more easily oxidized on going down the group
(they become more powerful reducing
agents), that is, the tendency to lose the valence electrons increases.
The increase in the reactivity of the metals is illustrated by their reaction with oxygen, water and dilutes
acids.
2.4.1 Reaction of the metals with oxygen
The metals reduce oxygen to O²⁻. Meanwhile the metal is oxidized to its +2 state. The vigour of reaction
increases down the group, as already explained. The reactions produce a basic metal oxide MO. The
reaction of the metal with oxygen can be generalized as
2M(s) + O2(g)  2MO(s)
Magnesium burns rapidly with a bright white flame to form a white ash of MgO, which may be tainted
gray by the formation of magnesium nitride.
2Mg(s) + O₂ (g) → 2MgO(s)
3Mg(s) + ½ N₂ (g) → Mg₃N(s)
Formation of Mg₃N does not occur readily because the reaction has a very high activation energy due to
the very strong triple bonds that must be broken in nitrogen molecules for the reaction to take place. If Mg
is burnt in a closed vessel, Mg₃N might be formed toward the end of the reaction when oxygen is almost
used up.
Mg might not burn easily because of the heat resistant layer of magnesium oxide that forms on the surface
of the metal. In that case, prolonged heating with a very hot Bunsen flame will be necessary. Alternatively,
the magnesium metal can be scraped first to remove the oxide layer.
MgO forms a weakly alkaline solution in water because of its poor solubility
Calcium burns faster and more readily than Mg. This particularly applies to the freshly cut metal which
is not coated with a surface layer of calcium oxide.
CaO dissolves in water more readily than MgO, forming an alkaline solution of Ca(OH)₂.
Sr and Ba burns explosively in air, forming their respective oxides, which dissolve appreciably in water
to form strongly alkaline solutions.
Ba also forms a peroxide with oxygen, in which the oxidation state of oxygen is -1.
Ba(s) + O2 (g) BaO2(s)
Page
362
The anion present in this solid is the peroxide ion, O₂²⁻. This anion is large enough to be polarized by the
smaller metal cations at the top of the group. However, BaO₂ is stable. Because of its large size and low
charge density, Ba²⁺ is not able to polarize and destroy the peroxide ion.
The oxides of group (II) metals
The Group (II) oxides have giant ionic structures in which the lattice particles are the M²⁺ and the O²⁻
ions. The first member of Group (II), Be, forms an amphoteric oxide. This shows that although BeO is
ionic, it has a high degree of covalent character.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 362
The rest of the oxides are basic, and they react readily with dilute acids, forming their respective salts, for
example,
MgO(s) + H2SO4 (aq)  MgSO4(aq) + H2O(l)
The lattice energies of the Group (II) metal oxides become smaller (less negative) down the group.
This is a result of the increase in the ionic radius of the M²⁺ ion. Going down the group, additional shells
are opened and this results in an increase in the size of the ions. This weakens the electrostatic attractions
between the metal ion and the oxide ion in two ways.
 Charge density of the M2+ ion decreases down the group. An ion with a small charge density has a
relatively weak attraction for an anion.
 The distance between the nucleus of the metal ion and the anion increases. This weakens the
attraction between the protons in the cation and the negative charge on the anion.
MgO, at the top of the group, has a very high lattice energy, and for this reason it is used as a refractory
material for lining the walls of furnaces.
Solubility of the oxides
Solubility increases down the group.
BeO is insoluble in water, though it is dissolved by both dilute acids and alkalis. MgO is poorly soluble in
water, so it forms a weak alkaline solution of Mg (OH)₂
MgO(s) + H₂O(s) → Mg²⁺ (aq) + OH⁻(aq)
CaO dissolves much better than MgO. SrO and BaO dissolve well in water.
To dissolve in water, the electrostatic attractions between the metal cation and the oxide ion must be
broken. The stronger the forces, the harder it is for water to break down the solid lattice. The strength of
these attractions is measured in terms of the lattice energy of the oxide. The higher the lattice energy (that
is, the more negative it is), the stronger the forces of attraction. Lattice energies of the Group (II) oxides is
high for the members at the top, and it decreases down the group (Table 9.2.4). It is therefore harder to
break down the lattice structure of MgO, but going down the group, the forces holding the ions of the lattice
together become weaker, and it becomes easier for water molecules to separate the ions.
Cation
present
Ionic
radius
of cation
Lattice energy of
oxide/Kjmol-1 x -1
MgO
Mg2+
0.065
3 929
CaO
Ca2+
0.099
3 477
SrO
Sr2+
0.113
3 205
BaO
Ba2+
0.135
3 042
2.4.2
Table 9.24 The lattice enthalpies
(theoretical) of the Period 2 metal
oxides. Note the inverse relationship
with ionic radii of the M2+ ions. As the
ionic size increases down the group,
the lattice enthalpy decreases, that is ,
it becomes less exothermic.
Reaction of the Group (II) metals with water
The metals become more reactive down the group for reasons that have already been explained.
Page
363
M(s) + 2H₂O → M(OH)(s) + H₂ (g)
In these reactions, the metals act as reducing agents, reducing water to OH⁻. Water acts as an oxidizing
agent, oxidizing the metal to its +2 state. The readiness and vigour of reaction increases down the group.
Beryllium has no reaction with water. Mg metal has no appreciable reaction with cold water, but it
burns rapidly in steam, forming its oxide and liberating hydrogen.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 363
Mg(s) + H₂O (g) → MgO(s) + H₂(s)
The initial product of the reaction is actually Mg(OH)₂, but it is dehydrated because of the high
temperature, forming MgO.
Calcium reacts vigorously with cold water in a very exothermic reaction which may cause the water to
boil. The water is immediately clouded by calcium hydroxide.
Ca(s) + 2H₂O (l) → Ca(OH)(s) + H₂ (g)
Sr and Ba react explosively with cold water. The reaction of Ba is more violent. The suspension formed is
not as cloudy as that for Ca. This is because the hydroxides of Sr and Ba are more soluble in water.
The resistance of Beryllium to reaction with water can be explained in two ways
 The atoms are very small and therefore do not easily lose their outer shell electrons
during the reaction with water.

The metal is protected by a thin, tough and impermeable layer of BeO, which protects the metal
underneath. This renders the metal kinetically stable with respect to its oxide.
Solubility of the group (II) hydroxides
The hydroxides dissolve reversibly in water to form a suspension. Solubility increases down the group.
M(OH)₂(s) ⇋ M²⁺(aq) + 2OH⁻(aq) … (i)
Going down the group, the equilibrium lies more and more to the right and the suspension becomes less
cloudy as the amount of undissolved metal hydroxide decreases.
Mg(OH)₂ is poorly soluble in water. It therefore forms a weakly alkaline solution. For this reason, it is frequently used as an antacid, for example in toothpaste and in milk of magnesia (for treating acid indigestion).
If it were very soluble in water, it would not be suitable as an antacid because the high concentration of OH ions would lead to poisoning.
The solubilities given in Table 9.25 show the increasing tendency of the hydroxides to dissolve in water
down the group.
Solubility/moles per 100g of
water
Mg(OH)2
0.2 x 10-4
Ca(OH)2
16 x 10-4
Sr(OH)2
330 x 10-4
Ba(OH)2
240 x 10-4
M(OH)₂(s) → M²⁺(aq) + OH⁻(aq) …. (i)
First the ionic crystal is broken down into its constituent gaseous ions. This is the reverse of lattice
energy, and it is an energy absorbing process (endothermic). The gaseous ions are then hydrated in water.
This process is exothermic. Whether the enthalpy of reaction (i) is relatively positive or negative therefore
depends on which of the two terms, lattice enthalpy and hydration enthalpy, has the greater numerical
value.
Page
364
Fig 9.25 Solubilities of the Group (II) metal
hydroxides.
The enthalpy change of this reaction (∆Hθsolution)
becomes more exothermic down the group, that is,
the dissolution becomes more energetically
favourable. Reaction (i) might be thought of as
happening in two steps, as shown in the energy
cycle in Fig 9.2.2 below.
Solubilities are best explained in terms of
energetics. The dissolution process involves
dissociation of the solid lattice to form aqueous
ions.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 364
M(OH)2(s)
HOsol
L.E
Mg 2+(aq) + 2OH-(aq)
HOhydr
Fig 9.2.2 An energy cycle for the
determination of the enthalpy change of
solution for the dissolution of a Group (II)
metal hydroxide.
Mg2+(g) + 2OH-(g)
By Hess’ law,
∆Hθ sol = - L.E + ∆Hθhydr ... (ii)
Here we consider only the hydration of the M2+ ions, since the hydration enthalpy of the OH - ion remains
constant. Going down the group, both LE of the metal oxides and the hydration enthalpy of the metal ion
decrease because of the increasing size of the M 2+ ion. However, the decrease in LE (the endothermic
component in equation (ii)) is faster than the decrease in hydration enthalpy (the exothermic component
in equation (ii)), that is, going down the group, ∆Hθsol becomes more exothermic. The dissolution process
therefore becomes more energetically favourable.
An important note
The size of the anion is important in determining whether or not the fall in LE is faster that the fall in enthalpy of
hydration. When the size of the anion is small, such as the hydroxide ion, then the fall in LE is faster and the enthalpy
of solution becomes more negative, that is, solubility increases down the group. When the anion is a large species such
as the sulphate and the carbonate ions, the fall in LE is very slow, much slower than the fall in hydration enthalpy. In
that case, enthalpy change of solution becomes more positive down the group and solubility decreases.
2.4.3
The Group (II) sulphates
The elements of Group (II) form stable sulphates, M(SO₄).
The solubilities of these sulphates decrease down the group, as shown in the Table 9.2.6 . For example,
from this table you can easily verify that MgSO4 is about 111 times more soluble than BaSO4.
Solubility/moles per 100g of
water
10-4
365
M(SO)4(s)  M2+(aq) + SO42-
MgSO4
3 600 x
CaSO4
11 x 10-4
The enthalpy change of solution, ∆Hθsol for this process
can be calculated from Hess’ Law and a suitable energy
cycle diagram (see Fig 9.2.2).
SrSO4
0.62 x 10-4
∆Hθsol = -L.E + ∆Hθhydr ... (i)
BaSO4
0.009 x 10-4
We consider only the enthalpy of hydration of the M2+
ion, since the hydration enthalpy of SO42- remains
constant.
Table 9.2.6 Solubilities of the
Group (II) sulphates
Page
This trend in solubilities can be explained in terms of
chemical energetics. The dissolution process is
Whether the solid sulphate dissolves or not depends on whether the dissolving process is favourable
(relatively negative enthalpy change of solution = exothermic) or unfavourable (relatively positive
enthalpy change of solution = endothermic).
Both L.E and ∆Hθhydr decrease down the group (they become less exothermic due to the decrease in size of
the M2+ ion). However, the change in LE of the metal sulphate down the group is only slight, so that
enthalpy change of solution is to a large extend influenced by the change in the hydration enthalpies of the
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 365
M2+ ions. In other words, the decrease in ∆Hθhydr is much faster than the decrease in L.E. Going down the
group, the radius of the metal cation increases, and hydration of the ions become more endothermic, since
weaker attractions are formed with water. Enthalpy of solution therefore becomes more endothermic down
the group, and more exothermic up the group.
We can rewrite equation (i) as
∆Hθsol = constant + ∆Hθhydr ... (ii)
since the change in L.E is very slight . Now, ∆Hθhydr becomes more positive down the group. Substituting
these more positive values in equation (ii) causes ∆Hθsol to become more positive as well. In other words,
∆Hθsol becomes more endothermic and the tendency of the metal sulphate to dissolve in water becomes
less energetically feasible. MgSO4 is soluble. CaSO4 is less soluble, but it does dissolve appreciably. SrSO 4
is sparingly soluble and BaSO4 can be considered as being insoluble.
Explain why lattice energy changes only slightly for the Group (II) sulphates?
Q
A
The change in lattice energy is slight because of the large size of the sulphate ion.
L.E α
q₊q₋
r+ + r₋
where q+ and q- are the charges on the positive and negative ion respectively and r+ and
r- are the radius of the positive and the negative ions respectively. Lattice energy varies
inversely as the sum of radii (r+ + r-) of the cation and anion. When (r+ + r-) increases,
lattice energy decreases. Now, since r-(radius of sulphate ion) is large, the change in the
sum (r+ + r-) remains fairly constant when the radius of the M2+ ion increases down the
group. Since (r+ + r-) remains almost constant, the change in L.E of the Group (II)
sulphates is small.
Note
The hydroxides of Group (II) elements show the reverse trend, that is, they become more soluble down the group. This
time lattice energy decreases faster than enthalpy of hydration. This is because the OH - ion is relatively small. The
increase in the radius of the M2+ ion down the Group results in a significant increase in the sum (r+ + r-), leading to a
decrease in lattice energy down the group.
2.4.4
The carbonates and nitrates of the Group (II) elements
Page
366
The elements from Mg to Ba form stable carbonates and nitrates.
All of the Group (II) nitrates are soluble. The carbonates are rather insoluble, as shown in
Table 9.2.7. In general, solubility of the carbonates decreases down the group just as is observed for the
sulphates, and for the same reason.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 366
Solubility/moles per
100g of water
MgCO3
1.3 x 10-4
CaCO3
0.13 x 10-4
SrCO3
0.07 x 10-4
BaCO3
10-4
Thermal stability of the group (II) nitrates and
carbonates
he nitrates and carbonates of the group (II) nitrates are
decomposed by heat. The decomposition of the nitrates can be
generalized by the equation
𝟏
M(NO₃)₂(s)  MO(s) +2NO₂(g) + O₂(g)
𝟐
0.09 x
f the
MCO3(s)  MO(s) + CO2 (g)
When the decomposition takes place, red brown fumes of NO2
are produced, together with a gas that relights a glowing splint
(oxygen). A white residue of metal oxide is left behind.
The carbonates decompose according to the equation
When the decomposition is complete, a white residue is left which dissolves in a strong acid such as HCl
without producing effervescence. This shows that the product is a metal oxide, and not a carbonate.
The ease of decomposition of the nitrates and carbonates by heat decreases down the group, that is, it
becomes more difficult to decompose them, and higher temperatures would be required. Another way of
putting it is that the stability of the carbonates and nitrates towards heat increases down the group. As an
example, consider the decomposition of magnesium and Barium nitrate. Mg(NO 3)2 decomposes fairly easily
when strongly heated in a Bunsen flame but Ba(NO3)2 resists decomposition and only starts to decompose
at much higher temperatures.
Explaining the trend in thermal stabilities of the Group (II) nitrates and
carbonates
The nitrate ion, being large, can be easily polarized by a cation. That is, some electron density from the
anion is attracted by the cation, resulting in the distortion of the electron cloud of the anion. This creates a
region of shared electron density between the metal cation and the N atom of the nitrate ion. This region of
electron density is shared between the cation and the anion. The result is that some covalent character is
introduced, and some ionic character is removed from the metal nitrate. This has the effect of destabilizing
the metal nitrate so that it can be decomposed fairly easily by heat.
At the top of the group, the metal ion, M²⁺, because of its small size, has the greatest charge density, and
hence is the most effective at polarizing the nitrate ion. Mg (NO₃) is therefore the least stable, and it is
easily decomposed by heat. Going down the group, the size of the cation increases and its charge density
decreases. The cation becomes less and less able to effectively polarize the anion. The metal nitrate therefore
becomes more thermally stable. Higher temperatures would be required to decompose the nitrate. Ba
(NO)₃, at the bottom of the group, has the largest ionic radius, and hence the smallest charge density.
Ba(NO₃)₂ is therefore thermally stable. A very high temperature is required to decompose it. The same
explanation applies for the metal carbonates.
Page
367
The alternative explanation
Decomposition of the nitrates and carbonates result in the formation of metal oxides. The more stable the
oxide, the more favourable is its formation, and the easier it is to decompose the metal nitrate or carbonate.
Stability of the oxide is best measured in terms of its lattice energy. The larger the magnitude of the lattice
energy, the more stables the oxide. Lattice energy of the metal oxide depends on the charge density of the
cation. The higher the charge density, the stronger the attraction between the cation and the oxide ion. This
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 367
Page
368
time a large charge density does not result in polarization and instability, since the oxide ion is small and
not easily polarized. The Mg²⁺ ion, being the smallest, has the highest charge density, so MgO has the largest
magnitude (most negative) lattice energy. The formation of MgO is therefore energetically favourable and
happens fairly easily at relatively low temperatures. Going down the group, the size of the metal cation
increases, its charge density decreases and its ability to attract the oxide ion decreases. The magnitude of
lattice energy of the oxide correspondingly decreases and the metal oxide becomes less stable. The
formation of the metal oxide becomes less favourable down the group, that is, decomposition of the nitrate
becomes less likely.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 368
Group (II) elements: quick facts
Magnesium
Named from the ancient region Magnesia in Greece, from which the metal was discovered. This
silvery white metal is relatively strong and less dense than Al. Its alloy with Al, sometimes known as magnalium, is
prized for its relative lightness and it is accordingly used in applications where lightness and strength are important,
such as in aircraft bodies.
Magnesium is the 11th most abundant element by mass in the human body. It is required for the proper functioning
of many enzymes. Magnesium (II) is found at the centre of chlorophyll molecules. This explains the addition of
magnesium salts to fertilizers.
Magnesium compounds have a number of medical uses.

As common laxatives (substances that induce bowel movement. They are also known as purgatives). A
well known example is Epsom salts in which the active ingredient is MgSO₄.7H₂O.

Milk of magnesia is a suspension of magnesium hydroxide in water. This suspension is slightly alkaline
because of the poor solubility of Magnesium hydroxide in water. Milk of magnesia is used as an antacid
to treat acid indigestion.
Magnesium hydroxide is also the antacid found in many types of toothpastes. It neutralizes acids formed
by bacteria in the mouth, thus preventing corrosion of teeth.
 MgO is used as a refractory (heat resistant) lining in furnaces because it has a very high melting point.
Calcium
From Latin, calx, meaning ‘lime’.
Calcium was known as early as the first century when the ancient Romans prepared lime as calcium oxide. Ca is a
soft gray alkaline metal which can be cut with a knife. It is the fifth most abundant metal by mass in the human body.
It is required for teeth and bone formation (bone is essentially calcium phosphate). Ca (II) is also important in the
transmission of nerve impulses.
Limestone is a rock in which the major component is CaCO₃. This explains the chemical erosion of limestone
landscapes by acid rain. Limestone is mainly used as a source of CaO (quicklime) and slaked lime (calcium
hydroxide). CaO is made by heating CaCO₃ at high temperature:
CaCO₃(s) → CaO(s) + CO₂ (g)
CaO, formed insitu in the blast furnace for the extraction of iron, is used to remove sandy impurities by neutralizing
silicon dioxide (silica), which is the major component of sand.
CaO(s) + SiO₂(s) → CaSiO₃(s)
The calcium silicate formed is then tapped out as a slag which is used for road surfacing.
CaO, as lime, is used in agriculture to neutralize acidic soils. Slaked lime (Ca(OH)2) is also used for the same purpose.
CaO and Ca(OH)2 are poorly soluble in water, producing a weakly alkaline solution. The advantages are that

They work for a long time, since they dissolve only gradually.

They do not cause a sudden increase in soil pH (excessive alkalinity), since they have a poor solubility in
water.
CaO, obtained from the decomposition of limestone or marble, is also used in the manufacture of cement.
Plaster of Paris is another important compound derived from calcium. The chief ingredient present is
calcium sulphate hemihydrate, CaSO₄.1/2H₂O. It is made by heating gypsum (CaSO₄.2H₂O) to about 150⁰C.
Plaster of Paris is used, among other things, in plaster casts for broken limbs. This is because it hardens
when it absorbs water, due to the reaction
Page
369

Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 369
1
CaSO₄. H₂O + 3/2 H2O  CaSO₄.2H₂O
2

Gypsum, CaSO4.2H2O, is used in a wide variety of applications:
1.
2.
3.
4.
As a fertilizer and soil conditioner
In Plaster of Paris, as already explained.
A component of Portland cement used to prevent flash setting of concrete.
In foot creams, shampoos and many other hair products.
A large deposit of gypsum at Montmartre in Paris is the source of the name ‘Plaster of Paris’.
Barium
From Greek ‘barys’ meaning ‘heavy’ (It is a heavy metal).
The uses of metallic Barium are limited and include as an ‘oxygen scavenger’. Being very reactive, it removes oxygen
from sealed glass tubes (oxygen can interfere, for example, with the operation of a vacuum tube). Barium sulphate
is used in X-ray studies of the gastrointestinal (GI) system. The GI system includes the stomach, intestines, and
associated organs. Prior to the examination, the patient swallows a barium ‘meal’ containing a slurry of barium
sulphate in water. The barium sulphate then coats the GI tract, for example, the lining of the small intestines. A
fluoroscope is then held over the patient’s abdomen. The fluoroscope produces x-rays which can easily pass through
tissues, but is blocked (absorbed) by BaSO4. The absorption pattern, as seen, for example, on an X-ray film, can reveal
some problems of the GI lining by highlighting abnormalities.

The ability of BaSO4 to absorb X-rays is a result of its very low solubility in water.

BaSO4 is insoluble so it does not release ions that would be absorbed into the blood stream, leading to
poisoning (Ba2+ ions are toxic).
Questions, solutions and discussions
Q1
Magnesium is the eighth most common element in the Earth’s crust. The metal is widely used in
alloys which are light and strong. Some reactions of magnesium and its compounds are shown in
the following reaction scheme .
H2(g) + A (aq)
Mg (s)
Na 2CO 3(aq)
C (s)
heat
dil. HCl
B(aq)
heat in air
D (s)
heat
E (s)
ev aporate
F (aq)
Page
370
NaOH (aq)
dil. HNO3
heat
F (s)
+ H 2 (g)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 370
(a)
Identify, by name or formula, compounds A to F.
(b) (i) Construct balanced equations for the following reaction
magnesium to compound A
compound C to compound D
compound F to compound D
(ii)
Suggest a balanced equation for the effect of heat on compound E.
9701/02/M/J/2005
Solution
(a) A MgSO4
B MgCl2
C MgCO3
D MgO
E Mg(OH)2
F Mg(NO3)2
(b) (i) Mg + H2SO4 → MgSO4 + H2
MgCO3 → MgO + CO2
2Mg(NO3)2 → 2MgO + 4NO2 + O2
(ii) Mg(OH)2 → MgO + H2O
Q2
This question is about the elements in Group II of the Periodic Table, magnesium to barium.
(a) Complete the following table to show the electronic configuration of calcium atoms and of
strontium ions.
1s
2s
2p
Ca
2
2
6
Sr2+
2
2
6
3s
3p
3d
4s
4p
4d
(b) Explain the following observations.
(i) The atomic radii of Group II elements increase down the Group.
(ii)
The strontium ion is smaller than the strontium atom.
Page
371
(iii) The first ionization energies of the elements of Group II decrease with increasing
proton number.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 371
(c)
Samples of magnesium and calcium are placed separately in cold water and left for some time.
In each case, describe what you would see and write a balanced equation for each reaction.
(d)
Strontium nitrate, Sr(NO3)2 undergoes thermal decomposition.
(i) State one observation you would make during this reaction.
(ii)
Write a balanced equation for this reaction.
9701/02/M/J/07
Solutions
(a) Left for the reader
(b) (i) Number of shells increases down the group
(ii) outermost shell has been removed
(iii) Number of shells increases, so outermost electrons are further from nucleus/shielding of outer
electrons increases .
(c) (i) Mg reacts very slowly, bubbles of gas are slowly produced
Mg + 2H2O → Mg(OH)2 + H2
(ii) Ca reacts more vigorously reaction, forming a white suspension and rapidly evolving a gas/Calcium
is rapidly used up.
Ca + 2H2O → Ca(OH)2 + H2
(d) (i) Brown fumes are gas evolved
(ii) 2Sr(NO3)2 → 2SrO + 4NO2 + O2
Q3
Agricultural lime is manufactured from limestone (calcium carbonate) by first heating the
rock to a high temperature in a lime kiln. The product is allowed to cool and a calculated
amount of water is added. A highly exothermic reaction takes place and a white powder
called ‘slaked lime’ is produced.
(a) Write balanced equations for these two reactions.
(c) How does the temperature required to decompose the carbonates of Group II elements
vary down the group, and why is this so?
Page
372
(b) Give two reasons why lime is used in agriculture.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 372
(d) The mineral dolomite is a double carbonate of magnesium and calcium, with the
formula CaMg(CO3)2. When 1.000 g of an impure sample of dolomite was completely
dissolved in an excess of hydrochloric acid, 0.450 g of carbon dioxide was given off.
(i) Write a balanced equation for the reaction.
(ii) Calculate the percentage purity of the dolomite.
9701/04/SP2007
Solutions
(a) CaCO3 → CaO + CO2
CaO + H2O → Ca(OH)2
(b) To neutralize acidic soils
To precipitate clays , thereby improving soil quality.
(c)
Increases
Going down the Group , the metal cation increases in size . The ability of the cation to polarize the
anion therefore decreases. This increases stability of the carbonate.
(d) (i) CaMg(CO3)2 + 4HCl → CaCl2 + MgCl2 + 2CO2 + 2H2O
(ii) Mr of dolomite = 40 + 24 + (2 x 60) = 184
184 g of dolomite ( 1 mole) should produce 2 moles of CO2 = 2 x 44 g of CO2
Hence 1 g of dolomite should give
88
184
= 0.478 g of CO2
% purity of the dolomite is
0.450
0.478
x 100
= 94.1%
Calcium is the fifth most common element in the Earth’s crust. Calcium compounds occur in bones
and teeth and also in many minerals.
Some reactions of calcium and its compounds are shown in the following reaction scheme .
Page
373
Q4
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 373
U (aq)
dilute HCl
Ca (s)
roast in air
V (s)
dilute HNO3
H2O (l)
H2O (l)
Na2CO3 (aq)
X (s)
W (aq)
dilute
H2SO4
Y (s)
(a)
reaction 1
Z (s)
State the formula of each of the calcium compounds U to Y.
(b) Compound Y may be converted into compound V.
Outline how this reaction would be carried out in a school or college laboratory using a
small sample of Y.
(c) (i) Construct balanced equations for the following reactions.
calcium to compound U
compound V to compound W
compound U to compound Y
(ii) Construct a balanced equation for the effect of heat on solid compound W.
(d)
Suggest the formula of an aqueous reagent, other than an acid, for reaction 1.
(e)
What would be observed when each of the following reactions is carried out in a test tube?
the formation of X from Ca(s)
the formation of X from V
9701/21/M/J/11
(a)
U CaCl2
(b)
Heat strongly in a boiling tube
(c) (i)
V CaO
W Ca(NO3)2
X Ca(OH)2
Y CaCO3
Ca + 2HCl → CaCl2 + H2
CaO + 2HNO3 → Ca(NO3)2 + H2O
Page
374
Solutions
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 374
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
(ii) 2Ca(NO3)2 → 2CaO + 4NO2 + O2
(d)
Any soluble salt, for example, Na2SO4(aq)/K2SO4(aq)
(e) (i) effervescence/bubbles /Ca dissolves/ white suspension formed
(ii) Heat given out/vigorous reaction occurs/steam is formed
Q5
The element magnesium is used in fireworks and warning flares.
(a) (i) Describe, with the aid of an equation, the reaction of magnesium which makes it suitable
for use in fireworks.
(ii) Draw the lattice structure of the product formed in part (a)(i) and explain why it is used in
furnace linings.
(b) The reaction between magnesium oxide and water is very slow and is reversible to form
magnesium hydroxide which is used in indigestion remedies and in toothpastes.
(i) Suggest two reasons for the partial solubility of magnesium oxide in water.
(ii) Comment on the use of magnesium hydroxide in toothpastes.
(c) (i) Explain why the solubilities of the Group (II) sulphates in water decrease down the group.
(ii) Describe a simple test for the sulphate ion
9189/01/O/N/2010
solutions
(a)(i) Magnesium burns in air with a bright white flame
Mg(s) + O2 (s)  MgO (s)
(ii)
2+
Mg
O
2-
(b) (i) Lattice energy of MgO is very high/ forces of attraction between Mg2+ and O2- are very strong water does not easily break down the crystal OR
Page
375
MgO is used in furnace linings because it has a very high melting point and so it is resistant to
heat.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 375
Sum of hydration enthalpies of Mg2+ and O2- is less exothermic than the lattice energy of MgO.
(ii) Being partially soluble, Mg(OH)2 makes toothpaste slightly alkaline, thus neutralizing acids
produced by bacteria in the mouth. These acids are responsible for decay of teeth.
(c)(i) ΔHsolution = - L.E + ΔHhydration
Going down the group, ΔHhydration of the metal cations decrease rapidly (becomes more
endothermic) due to increase in ionic radii. However, L.E remains fairly constant (it decreases only
slightly) due to the large size of the SO42- ion. ΔHsolution therefore becomes more endothermic down
the group, that is, the dissolution of the Group (II) sulphate becomes less energetically feasible.
(ii) White precipitate formed with Ba 2+ (aq) or Pb2+ (aq)
Q5
A 5.00 g sample of an anhydrous Group (II) metal nitrate loses 3.29 g in mass on strong heating.
Which metal is present in the nitrate?
A. Mg
B. Ca
C. Sr
D. Ba
Solution
2M(NO3)2  2MO + 4NO2 + O2 … (i)
moles of M(NO3)2 =
mass
Mr
=
5.00
(Ar of M)+ 124
( 124 is the ‘Mr’ of (NO3)2 )
Mass of oxide MO formed = 5.00 - 3.29 = 1.71 g
Moles of MO =
1.71
Ar of M
From equation (i), moles of M(NO3)2 = moles of MO, that is
5.00
1.71
=
(Ar of M)+ 124
Ar of M
Let Ar of M be x, then
5.00
𝐱 + 124
=
1.71
𝐱+M
Page
376
Solving for x gives 37, so the metal M is Ca (Ar = 40)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 376
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
9.2in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
1(a) Describe the reactions, if any, of magnesium and calcium with cold water and write a balanced
equation for any reaction.
(b) Explain qualitatively why magnesium sulphate and calcium sulphate have different solubilities in
water.
(c) Give a commercial use of either magnesium sulphate or calcium sulphate (specify which).
(d) Magnesium reacts with nitrogen only at very high temperature, producing a yellow solid D. D
contains 72 % of magnesium , by mass.
(i) Suggest why a high temperature is required for this reaction.
(ii) Calculate the empirical formula of D.
(iii) When water is added to the yellow solid D, a white suspension forms and ammonia is
liberated. Identify the white suspension and write an equation for the action of water on D.
9701/02/O/N/1993
2(a)(i) Describe how the ease of thermal decomposition of the hydroxides and nitrates of the Group (II)
metals varies down the group. Write generalized balanced equations for the reactions.
(ii) Suggest an explanation for this trend.
(b) A Group (II) hydroxide exists as a hydrate M(OH)2. nH2O. On heating, 1.ooo g of this hydrate lost
0.542 g in the form of steam, as it was converted into the anhydrous M(OH)2. Subsequent heating to
constant mass produced a further mass loss of 0.068 g.
Calculate the value of n, and the Ar of the metal M.
9701/01/O/N/1996
3(a) When water is added to the solid remaining after barium nitrate has been heated, an alkaline
solution is produced. The addition of sulphuric acid to this solution produces a white precipitate.
Identify the products of these two reactions and write equations for them.
Describe clearly what these differences are , and explain why they occur.
Page
377
(b) When the procedure described in (a) is repeated using the solid left after magnesium nitrate has
been heated, several differences are observed.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 377
(c) A 4.50 g sample of a carbonate of a Group (II) metal (known to be either calcium or strontium) lost
1.34 g in mass when heated strongly. Identify the metal, showing clearly your working.
9701/01/O/N/2000
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
9.3 GROUP (IV) ELEMENTS: CARBON TO
LEAD
Introduction
The Group (IV) elements are characterized by a change from metallic through metalloid, to metallic
structure. In fact, it is noticed in all groups that there is a tendency for metallic character to increase
down the group, but this observation is clearly noticeable in Group (IV). The similarity in the structure
and properties of elements which is obvious in Group (II) is not so apparent in Group (IV), as shown in
Table 9.3.1.
9.3.1
Brief overview of the elements
Carbon
There are two well known allotropes of carbon, namely, graphite and diamond.
Both graphite and diamond have crystalline giant covalent structures, characterized by a very large number
of C to C bonds extending throughout the structure. This explains the very high boiling and melting points
of these substances.
Silicon and germanium
These two elements have giant covalent structures, similar to that of diamond. However, in electronics, they
are referred to as metalloids because under some conditions they conduct electricity, thus behaving as if
they were metals.
Page
378
Tin and lead
Tin has two forms, white tin and gray tin. White tin, which is metallic, is the more stable allotrope at
temperatures equal to or greater than room temperature. Its metallic nature is shown by its behavior as a
reducing agent, although it is a mild reagent. Gray tin has a giant covalent structure, and it is formed when
white tin is cooled well below room temperature. The conversion from white tin to gray tin by cooling is a
very slow process.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 378
Lead has a giant metallic structure. Metallic lead has a bluish - whitish colour after being cut, but it soon
tarnishes on exposure to air. It is therefore clear that metallic character increases down the group. This is
caused by the increase in atomic radii down the group. As the atoms get larger, overlap of atomic orbitals
become poor, and covalent bond formation becomes less likely. At the same time, the tendency of the atoms
to lose their outer most electrons and contribute them into a common pool of delocalization increases,
which is what characterizes metals. This tendency to lose electrons is a result of the increasing distance of
outer shell electrons from the nucleus. Moreover, as the number of shells increase, these outer electrons
experience greater shielding from the nucleus.
C
Atomic
number
Outer shell
configuration
Atomic
radius/nm
Si
Ge
Sn
Pb
50
82
6
14
32
2s22p2
3s23p6
4s24p2
5s25p2
6s26p2
0.077
0.117
0.122
0.141
0.154
1 410
937
232
327
2 680
2 830
2 270
1 730
1 x 106
Giant
molecular
2 x 106
Giant
molecular
3 730(d)
m.p/oC
(g)
4 830(d)
b.p/oC
(g)
Conductivity
Type of
structure
I.E1
Giant
molecular
1 090
Table 9.3.1
786
762
8 x 106
Giant metallic
5 x 106
Giant metallic
707
833
The properties of group (IV) elements ( d = diamond, g = graphite)
Electronic structure
The valence shell configuration of the Group (IV) elements is
... ns²np²
Where n is equal to period number, for example, for C in period 2, n=2 and for Si in period 3, n=3.
9.3.2
Physical properties of the Group (IV) elements
Page
379
Physical properties of the elements are influenced by the bonding and structure present. As the nature of
bonding changes down the group, so does the physical properties such as melting points and electrical
conductivities.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 379
3.2.1 Trend in melting points
In general, melting points decrease down the group. Notice
the unexpected melting point of lead.
Fig 9.3.1 Trend in melting
points of the Group (IV)
elements
Melting involves breaking forces which hold lattice particles
together. The stronger the forces, the higher the melting
point. Strength of these forces depends on the type of bonding
present, and on the specific crystal structure of the solid.
Carbon, silicon and germanium have giant covalent
structures. This explains their high melting points. A large
amount of energy is required to break a large number of
strong covalent bonds. However, the melting points of these
three elements decrease down the group, showing that the
covalent bonds become weaker with increasing atomic size.
As atomic size increases, overlap of atomic orbitals to form covalent bonds becomes less efficient.
Consequently the bonds become longer and weaker.
Tin and lead have giant metallic structures, so they have lower boiling points. In general, metallic bonds
are weaker than the covalent bonds in giant covalent structures, and less energy is required to break them.
The melting points of tin and lead depends on the strength of the metallic bonds. Lead has a higher melting
point, showing that it has the stronger metallic bonds. The nuclear charge of lead is much higher than that
of tin. As a result, the nucleus has a stronger attraction for electrons in the delocalized sea, despite the fact
that in Pb there are more shells (the change in nuclear charge from Sn to Pb is very large, and so has the
greater effect than that of increase in number of shells). The atoms of lead are therefore more firmly held
to the lattice and more energy is required to pull them from each other during melting.
3.2.2 Trend in atomic radii
The size of atoms increase down the group (Table 9.3.1)
Atomic radii increases down the group as new shells are opened. The increase in atomic size down a group
is not sharp because the increase in nuclear charge has a counterbalancing effect. As number of protons
increase, there is a greater pull of shells towards the nucleus and this prevents atomic size from increasing
rapidly.
3.2.3 Trend in ionization energies
Ionization energies follow the same trend as melting points (Table 9.3.1 and Fig 9.3.2).
Ionization energy is a measure of how firmly valence electrons are held to the atom. The more strongly held
the atoms, the higher the energy required to remove them.
In general, ionization energies fall down the group, showing that the valence shell electrons become more
weakly held. Down the group, atomic size increases due to the opening of new shells. The valence electrons
become weakly held to the atoms because of shielding by underlying shells and because of the increasing
distance of the valence electrons from the nucleus.
Page
380
Explaining the trend
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 380
Lead has a higher first ionization energy than
expected. Once more, the explanation lies in the
nuclear charge of lead atoms. The largest jump in
proton number from one element to the next is found
at Sn and Pb. This sharp increase in proton number
implies that the valence shell electrons in lead are
more firmly held to the atom than in tin.
Fig 9.3.2 Trend in ionization
energy of the Group (IV)
elements
3.2.4 Electrical conductivity
In general, electrical conductivity increases down the group (Table 9.3.1)
Carbon’s two allotropes differ in their conductivities. Diamond is an insulator but graphite is a good conductor of electricity along its layers.
Si and Ge are semi-conductors. The conductivity of Ge is twice that of Si. Sn and lead are better conductors
than Ge and Si but note that the conductivity of Pb is slightly less than that of Sn.
Explaining the trend
The increase in conductivity down the group coincides with the increase in metallic character. Down the
group, atomic size increases and valence electrons become loosely held to the atoms. It becomes easier for
these valence electrons to be detached and contributed into a common sea. It is this sea of delocalized
electrons which gives metals their ability to conduct electricity. Pb has a slightly smaller conductivity than
tin. Once more, this is a result of the very high nuclear charge of lead. The large number of protons in the
nucleus has a greater attraction for electrons in the 4s sub shell than those in the 4p (this is because the 4s
is closer to the nucleus). As a result, for lead, it is easy to lose the 4p electrons, but the 4s remain tightly
held to the atom. For tin, the nuclear charge is relatively small and both the 4s and the 4p electrons are
delocalized in the metal lattice. The increased number of delocalized electrons results in greater conductivity.
Diamond is an insulator because there are no free delocalized electrons to carry charge. Instead, the
valence electrons in the atoms are localized in covalent bonds. The ability of graphite to conduct electricity
has nothing to do with size of atoms (the size of atoms is the same as in diamond), but with the bonding
present. Each carbon atom uses only three of its valence electrons to bond to other carbon atoms. The fourth
electron of each atom is then delocalized in the structure. This delocalized system of electrons is responsible
for the conductivity of graphite along the layers.
Page
381
9.3.3
Trends in chemical properties
The following observations will be important in explaining trends in chemical properties
 Atomic size increases down the group and this coincides with increasing metallic character (valence
electrons become more loosely held to the atom).
 Nuclear charge increases down the group (nuclear charge has an indirect effect on chemical
properties).
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 381
3.3.1
Oxidation states
The valence shell of the Group (IV) elements has the configuration.
ns²np²
The +4 oxidation state is therefore common to all group four elements. The +2 state is also important in all
of the elements.
The group (IV) elements exhibit two stable oxidation states in their compounds, the +2 and the +4. This
differs from the Groups (I) and (II) elements which exhibit only one stable oxidation state, the +1 and the
+2 respectively.
Why two oxidation states?
The valence shell is made up of two sub-shells, the ns and the np. The +4 state corresponds to loss of all
the four valence electrons. However, it is also possible to lose only the two electrons in the np sub-shell,
which is higher in energy (less stable) than the ns sub-shell. When this happens, two electrons are left in
the 2s sub-shell. Since these remaining electrons are paired, they are quite stable, so formation of the +2
state is also energetically favourable.
Relative stabilities of the +4 and the +2 oxidation states
The +4 oxidation state becomes more stable up the group. The +2 state becomes more stable down the
group. CO₂, in which C has the +4 state, is more stable than CO, in which carbon has the oxidation state of
+2. Similarly, SiO₂ is the stable oxidation state for Si. SiO does not exist.
It is difficult to reduce carbon dioxide to carbon monoxide. The reduction can take place, but the reaction
is endothermic and strong heating is required. On the other hand, carbon monoxide is easily oxidized to
carbon dioxide and the reaction is exothermic, that is, the product is more stable than the reactant.
The +4 state becomes less stable down the group, with respect to the +2 state. However, the +4 state
remains more stable than the +2 state in all elements up to tin. In lead, the +4 state becomes less stable
than the +2 state. Thus all of the group (IV) elements, except lead, react with oxygen to form the dioxide.
Lead forms the monoxide, PbO.This trend in the relative stabilities of the +4 and the + 2 states can be
explained in terms of the inert pair effect.
Redox data can be used to show the trend in the relative stabilities of the +4 and the +2 states.
Reduction equation
Eθ/V
Ge
Ge4+ + 2e- Ý Ge2+
-1.60
Sn
Sn4+ + 2e- Ý Sn2+
+0.15
Pb
Pb4+ + 2e- Ý Pb2+
+1.80
Table 9.3.2 The reduction potentials
of Group (IV) elements
The reduction potentials become more positive, showing that the formation of the X 2+ ion becomes more
feasible. In other words, stability of the +2 state increases whilst that of the +4 state decreases, as illustrated
in Fig 9.3.3.
Page
382
Element
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 382
The inert pair effect
+4
Relative
stability
+2
C
Si
Ge
Sn
Pb
Fig 9.3.2 Change in the relative stabilities of
the +4 and +2 oxidation states in the Group
(IV) elements. Note that in lead, the +2 state
is more stable than the +4 state.
The valence shell of the Group (IV) elements has the
configuration ns²np². Down the group, the two
electrons in the ns sub-shell become more firmly
held to the atom, that is, they become more stable
and their tendency to be used in chemical reactions
decreases. This observation has been termed the
‘inert pair effect’.
The +2 state, corresponding to loss of the np
electrons, (which are higher in energy and therefore
less stable) becomes more important down the
group. The increase in nuclear charge down the
group is a key factor in explaining the inert-pair
effect.
As the number of protons increase, the attraction for the valence electrons in the ns and np sub shells
increase. However, the ns electrons, being closer to the nucleus, feel a stronger attraction. It therefore
becomes harder to remove or unpair these electrons during reactions. Loss of the np electrons also help to
stabilize the ns electrons. Once the two electrons in the np sub-shell have been lost, the remaining electrons
feel a greater attraction by the nucleus since the attractive influence of the nucleus is now being shared by
fewer electrons. This stabilization of ns electrons by loss of np electrons becomes more significant as nuclear
charge increases down the group.
3.3. 2
Reactions of the group four elements
Reactions with oxygen
All of the Group (IV) elements combine directly with oxygen. In all cases, except lead, the dioxide is
formed.
X(s) + O₂ (g) → XO₂(s)
Lead reacts with oxygen to form the monoxide
Pb(s) +1/2 O₂ (g) → PbO(s)
This shows that in lead the +2 state is more stable than the +4 state. It is however possible to form the
unstable lead dioxide by other means than direct combination with oxygen.
Carbon
Graphite burns in oxygen to form carbon dioxide.
Page
383
C(s) + O₂ (g) → CO₂ (g)
The reaction is very exothermic. The enthalpy change of formation of carbon dioxide, -394Kj/mol, shows
that carbon dioxide is more stable relative to its elements.
The enthalpy change for the formation of carbon monoxide is also negative (-111Kj/mol). This also shows
that carbon monoxide is relatively stable with respect to its elements, but the more negative value for carbon
dioxide shows that carbon dioxide is more stable, hence in an excess supply of oxygen, carbon
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 383
dioxide will be formed instead of carbon monoxide.
Diamond does not react with oxygen at temperatures that can easily be achieved. In fact, for practical
considerations, diamond can be considered as having no reaction with oxygen. However, it has been
possible to burn diamonds at extremely high temperatures, producing carbon dioxide as the only product.
This shows that diamond is indeed carbon. The inertness of diamond is due to kinetic factors. Diamond
contains a very large number of strong covalent bonds which must be broken to allow reaction with oxygen.
The reaction has a very high activation energy.
Silicon
Silicon does not react with oxygen at ordinary temperatures. Even at very high temperatures, the reaction
is very slow. The resistance to reaction with oxygen is, once more, kinetic. Silicon has the same crystal
structure as diamond.
The +4 state of silicon, as in silicon dioxide, is more stable than the +2 state. In fact the +2 state is too
unstable to exist. Silicon dioxide is abundant in nature in the form of quartz (sand is small fragments of
quartz).
Germanium
Germanium reacts with oxygen to form germanium dioxide. The fact that the +2 state is not formed shows
that the +4 state is more stable. GeO does exist under certain conditions, but it is unstable and is easily
oxidized to GeO₂. It cannot be formed by direct combination with oxygen but by reducing GeO₂ at 1000⁰C.
Tin
Tin reacts with oxygen at high temperatures to form tin (IV) oxide. Lead (II) oxide is not formed by this
direct combination with oxygen, since the +4 state is more stable. At normal temperatures, the reaction of
tin with oxygen is very slow. It is this inertness which makes tin useful in the manufacture of tin cans used
in the canning industry. Tin cans are not made of tin. They are made of steel and then coated with a very
thin layer of tin to protect the steel from rusting.
Lead
Lead combines directly with oxygen. This time, the +2 state is more stable so it is formed in favour of the
+4 state. Lead dioxide can be formed, but not by direct combination with oxygen. At normal temperatures,
the reaction of lead with oxygen occurs readily. However, formation of a tough layer of PbO protects the
metal from further attack, so that the reaction with air is only superficial (occurs on the surface). This kinetic
inertness explains why lead was at one time used to make water pipes. Lead is no longer used for this
purpose because it leads to poisoning by Pb2+ ions formed by the slow dissolution of lead in water.
3.3.3 The compounds of Group (IV) elements
3.3.1 The oxides
Page
384
Table 9.3.2 shows the formulae and properties of the +4 oxides of Group (IV) elements.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 384
The dioxides (MO2) of Group (IV) elements : Oxidation state of the element M is +4
Oxide
CO2
Preparation by
direct
combination
with O2
Structure
B.p /OC
Acid/base
nature
Acid base
reactions
SiO2
Y
Simple
covalent
-78
GeO2
E
Giant covalent,
diamond
like
structure
2 590
A c i d i c
Reacts with strong alkali, giving
MO32- salts.
SnO2
PbO2
No. Prepared by
oxidation of
Pb2+ by the
electrolysis of a
Pb2+ solution.
S
Mixed giant covalent and ionic
1 200
1 900
Decomposes
when heated
A m p h o t e r i c
React with both acids and alkalis. With fused alkali,
MO32- salts are formed.
CO2 + 2OH-  CO32- + H2O
Carbonate
e.g. SnO2 + 2OH-  SnO32- + H2O
stannate(IV)
SiO2 + 2OH-  SiO32- + H2O
silicate
Similarly, PbO2 forms PbO32- (plumbate (IV)) and GeO2
forms GeO32-. Notice that this is analogous to the
reaction of CO2 and SiO2 with alkali.
With concentrated acids, +4 salts are formed.
e.g. PbO2 + 4HCl  PbCl4 + 2H2O.
(ionic: PbO2 + 4H+  Pb4+ + 2H2O)
Thermal
stability
Table 9.3.2
Page
385


Resists decomposition even at very high temperatures
Decomposes on
warming, to
produce PbO
PbO2  PbO +
½ O2
The oxides of Group (IV) elements in the +4 oxidation state
Carbon dioxide has a very low boiling point and is a gas at room temperature. This is because the
bonding present is simple covalent, and so weak Van der Waals forces exist between the CO 2
molecules. Compare with the other oxides, for example, SiO 2, in which there is giant covalent
bonding.
SiO2, GeO2 and SnO2 are solids at room temperature, with vey high melting and boiling points. This
is a result of the nature of bonding present, which is giant covalent or giant covalent mixed with
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 385

ionic character. A large amount of heat is required to break the large number of very strong covalent
bonds and decompose the structure during melting or boiling.
PbO2 is a solid at room temperature but it is very sensitive to heat. It decomposes on warming to
form the +2 oxide, PbO, which is more stable. PbO2 is therefore a good oxidizing agent because
1. It has the ability to donate oxygen atoms during reactions.
2. Whilst it is being reduced to PbO, it gains electrons from another substance (reduction is gain
of electrons). This other substance therefore loses electrons, and is thus oxidized (oxidation is
loss of electrons).
PbO2 is powerful enough to oxidize an acidified solution of iodide ions to iodine.
EO/V
PbO2 +
4H+
+
2e-
Ý
Pb2+
+ 2H2O
I2 + 2e- Ý 2IPbO2 + 4H+ + 2I- Ý Pb2+ + 2H2O + I2
+1.47
+0.54
+ 0.93
Note that the second reduction equation and its electrode potential have been reversed
before adding up the two equations.
The reaction is energetically feasible because it has a positive Eθcell.
SnO2 is not able to oxidize iodide ions to iodine because this would reduce tin from a stable
oxidation state (+4) to a less stable oxidation state (+2). The very small electrode potential for
Sn4+/ Sn2+ shows that the conversion of Sn (IV) to Sn (II) is unlikely.
Sn4+ + 2e- Ý Sn2+
Eθ = +0.15V
Sn(II) is easily oxidized to the +4 state, which is the more stable state. In other words, Sn(II)
compounds are good reducing agents, for example, they can reduce Fe 3+ to Fe2+.
Q
Construct a balanced equation for the reaction between SnCl2 (aq) and FeCl3 (aq)
and show that it is energetically feasible. Write down two observations that are
made in this reaction.

SiO2 and CO2 are acidic, so they react with alkalis such as NaOH. Notice that the two reactions are
analogous. They both form a salt containing the ion MO32-. CO2 forms sodium carbonate (Na2CO3)
whilst SiO2 forms sodium silicate (Na2SiO3). The behaviour of SiO2 and CO2 towards alkalis is
expected. In general, covalent oxides are acidic whereas ionic oxides are basic.

The rest of the Group (IV) oxides are amphoteric. This is because their structures are intermediate
between giant covalent and ionic. The reaction with bases is analogous to that of SiO2 and CO2.
The reaction with concentrated acid produces the +4 chloride, MCl 4. In dilute solution, presence
of a large amount of water causes the tetrachloride to decompose by hydrolysis.
Page
386
Table 9.3.3 summarizes the main properties of the +2 oxides of the group (IV) elements.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 386
The monoxides (MO) of Group (IV) elements : Oxidation state of the element M is +2
Oxide
CO
Preparation by
direct
combination of
the element
with O2
SiO
No. Can be
produced by the
reduction of
CO2.
No. Exists only
at very high
temperatures
GeO
No. Can be
produced by the
reduction of
GeO2 at 1000OC
SnO
PbO
No. Can be prepared by heating the
corresponding hydroxide or nitrate.
e.g. Pb(NO3)2  PbO + 2NO2 + ½
O2
Pb(OH)2  PbO + H2O
Structure
Acid/base
nature
Acid base
reactions
Simple covalent
Neutral
No reaction with acids or bases
Predominantly ionic, but have some
covalent character.
A m p h o t e r i c
React with both acids and alkalis to form +2 salts. With
alkalis, salts of the formula M(OH)3- are formed.
e.g. SnO + OH- + H2O  Sn(OH)3- + H2O
With acids, +2 salts are also formed.
e.g. PbO + 2HCl  PbCl2 + H2O.
Thermal
stability
When warmed or heated in air, they are easily oxidized to the +4 state, MO2.
Table 9.3.3


Key properties of the Group (IV) oxides with the +2 oxidation state
All of the monoxides are unstable, except PbO. These monoxides are therefore easily oxidized to
the +4 oxidation states, which are more stable. For lead, the +2 state is more stable, so it resists
being oxidized to the +4 state. This observation has already been explained in terms of the inert
pair effect.
CO is a simple covalent substance that has weak Van der Waals between particles. It is therefore a
gas at room temperature.
The chlorides of Group (IV) elements
All Group (IV) elements form +4 chlorides of the general formula XCl4. They are all oily liquids with simple
molecular structures. The shape of the tetrachlorides is tetrahedral.
Page
387
3.3.2
Stable
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 387
Cl
1090
X
Cl
Cl
Cl
The tetrachlorides have simple covalent bonding with weak Van der Waals forces between
the molecules. They therefore have low melting and boiling points. However, the melting
and boiling points increase down the group as the strength of Van der Waals forces increases
in response to increasing number of electrons.
Preparation of the +4 chlorides
SiCl4, GeCl4 and SnCl4 are easily prepared by direct combination of the element with chlorine gas, for
example,
Si + 2Cl2  SiCl4
Preparation of CCl4 by direct combination with chlorine is difficult because the reaction has a very high
activation energy due to the large C-C bond energy. These bonds must break first to allow combination with
chlorine. As bond energies in the elements decrease down the group, (Table 9.24) activation energy
decreases, and so it becomes possible for Si, Ge and Sn to combine directly with chlorine to form the
tetrachloride.
This decrease in bond enthalpy is expected. Going down
Bond
Average bond
the group, atomic size
increases and-1 overlap of atomic
Energy/Kjmol
orbitals to form the X-X bond becomes poor. As a result,
PbCl
not be prepared by the direct combination of
the distance of the bond pair of electrons from the
nuclei
4 canof
PbCl2 is formed instead, since the +2
both
atoms increases. The
bond pair therefore lead
feelswith
only chlorine.
a
C-C
348
state
lead is more stable than the +4 state. PbCl4 can be
weak attraction from the nuclei of both atoms of
theof
bond.
by the reaction of PbO2 with cold concentrated
TheSi-Si
result is that the X-X226
bond becomes longer prepared
and weaker
HCl. The reaction mixture should be cold to prevent
down the group.
Ge-Ge
188
Sn-Sn
151
n the
Thermal stability of the tetrachlorides
The tetrachlorides become less stable down the group as a result of the increasing size of the atom of the
element, X. As the size of X increases, overlap of its valence orbitals with that of chlorine becomes less
efficient. As a result, the X-Cl bond becomes longer and weaker.
CCl4, SiCl4 , GeCl4 and SnCl4 are thermally stable and resist decomposition by heat.
PbCl4 decomposes readily on warming to produce PbCl 2 and Cl2. Even without warming, PbCl4 undergoes
gradual decomposition.
Page
388
PbCl4
 PbCl2 + Cl2(g)
Yellow liquid
white solid
This reaction occurs for two reasons.
 The bonds in PbCl4 are weak and so easily break, as already explained.
 The product formed is more stable since the +2 oxidation state of lead is more stable relative to the
+4 state.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 388
Hydrolysis of the tetrachlorides
All Group (IV) tetrachlorides undergo hydrolysis in water, except CCl 4, for example
SiCl4 + 4H2O  SiO2.2H2O + 4HCl
A strongly acidic solution of pH ≈2 is formed. The product SiO 2.2H2O is sometimes written as Si(OH)4.
GeCl4 also hydrolyzes rapidly in water, forming GeO 2. The hydrolysis of SnCl4 and PbCl4 to SnO2 and PbO2
is only partial. CCl4 resists hydrolysis in water because of the absence of d-orbitals in C to accept the electron
rich water molecules during the first step of the reaction. The rest of the chlorides hydrolyze in water
because there are energetically accessible (low energy) d-orbitals to form bonds with water, as illustrated
for silicon in Fig 9.3.3.
vacant and energetically
accessible d - orbitals
3s
3d
3p
Si in SiCl 4
dative bond
O
H
H
Fig 9.3.3 The use of d-orbitals in the initial step of the
hydrolysis of SiCl4
The +2 chlorides, XCl2
Only tin and lead form stable +2 chlorides, either by direct combination with chlorine or by decomposition
of XCl4. In tin, the +4 chloride is more stable than the +2. SnCl₄ is therefore not decomposed to SnCl₂ by
warming, but by strong heating. For lead, the +2 chloride is more stable. PbCl₄ is therefore easily
decomposed by warming to form lead (II) chloride and chlorine gas.
C, Si and Ge do not form the +2 chlorides for two reasons
 the +2 state is unstable compared to the +4 oxidation state.
 Excitation of an electron from the ns sub shell to the np sub shell of the element occurs before
bonding, accompanied by sp3 hybridization. This creates four hybrid orbitals that are equivalent in
terms of shape and energy. Thus the element can form four bonds, not two, by sigma overlap of
the four hybrid orbitals with four chlorine atoms.
The bonding in PbCl2 and SnCl2 is ionic. These compounds are therefore crystalline solids.
Page
389
The tendency of Sn and Pb to form their respective +2 chlorides by direct combination with chlorine can be
explained in this way
 The inert pair effect: As already explained, going down Group (IV), the two ns electrons become
inert due to the strong attraction they feel from the nucleus. The formation of PbCl 2 and SnCl2
involves the loss of the two np electrons, which are relatively loosely held in Pb and Sn.
 Pb cannot form PbCl4 when it reacts with chlorine. This is because in Pb, the +2 state is more stable
than the +4 state.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 389
Some uses of Group (IV) elements and their compounds
The allotropes of carbon have a wide variety of uses
Graphite


CARBON


Inert electrodes in electrolytic processes
Dry lubricant in machines which are likely to generate large
amounts of heat (oil based lubricants decompose)
Mixed with clay for use in ‘lead’ pencils
Moderator in nuclear reactors. It slows neutrons by absorbing
Them.
Diamond


Tips in drilling tools (It is very hard)
jewelry (It has a high refractive index, which gives it an attractive
sparkle)
Buckminsterfullerene (C60)
 Trapping and protecting reactive atoms
 plugs in nano test tubes
Organic and inorganic compounds of C also have many uses. Some of the
useful inorganic compounds include carbon dioxide, which is used

to produce effervescence in fizzy drinks

in fire extinguishers

as a dry ice refrigerant. Its advantage in this sense is that it leaves
no residues when it sublimes, and it is relatively environmentally
friendly.

as a solvent. The use of supercritical carbon dioxide in
extractions is fairly modern. Substances which can be extracted
in this way include caffeine and limonene which is found in the
peels of citrus fruits
The use of CO2 in fizzy drinks
Carbon dioxide reacts reversibly with water
Equilibrium (i) lies well to the left, that is, the dissolution of
carbon dioxide in water is only slight. To increase the amount of carbon
dioxide that dissolves the gas is pumped under high pressure into the
liquid. Increasing pressure forces the equilibrium to the right, thus
producing more hydrogen ions and hydrogen carbonate ions (Le
Chatelier’s principle). The hydrogen ions produced are mainly
responsible for the acidic taste of Coca-cola.
Some American consumer lobby groups are known to have filed against
the Coca-cola company on claims that the level of acidity of the beverage is
toxic. These claims were rejected in American courts (home of the Cocacola company) as baseless.
Page
390
CO2(g) + H2O(l) Ý H2CO3(aq) Ý HCO3-(aq) + H+(aq) ...(i)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 390
In fact, the pH of coca-cola, which is approximately 5, is about 1000 times
less acidic than the contents of the stomach, whose pH is about 2 (due to
hydrochloric acid which is secreted by the walls of the stomach).
Furthermore, carbonic acid, which is the chief acid in coke, is immediately
destroyed when a bottle or can of Coke is opened (the fizzing sound is caused
by CO₂ escaping under pressure). The production of bubbles and fizzing that
occurs when a bottle of Coca-Cola is opened can be explained using Le
Chatelier’s principle. Before the bottle is opened, pressure inside the bottle
is high due to undissolved carbon dioxide. When the bottle is opened, the
undissolved gas escapes, thus lowering pressure. By Le Chatelier’s principle,
equilibrium (i) above shifts to the left, thus producing more carbon dioxide
to replace the gas which is escaping. This results in the decomposition of
carbonic acid, producing bubbles of carbon dioxide. This explains why an
opened bottle of Coca–Cola soon goes ‘flat’. The acidic taste is lost as
carbonic acid is converted to carbon dioxide and water. Once it gets in the
stomach, this loss of carbonic acid continues so that the pH of the stomach
is not altered to any significant extent. Some people have also reported that
Coke is acidic enough to corrode an iron nail. This is not correct. To dissolve
an iron nail, the reaction below should
take place:
H₂CO₃(aq) + Fe(s) → FeCO₃(s) + H₂(g)
This reaction does not take place. Carbonic acid is too weak an acid to
corrode metals.
SILICON AND
GERMANIUM
Tin is used to coat steel cans (by electroplating) which are used for packing
foods. The tin metal reacts with oxygen to form a thin, tough, impermeable
and insoluble layer of tin (II) oxide. This oxide layer protects the steel from
rusting. Other uses of tin include

making useful alloys such as bronze (tin and copper)

the chloride (SnCl2.H2O) is used as a reducing agent and as a
mordant in calico printing. A mordant is a substance that is used
to set dyes on fabrics by forming a co-ordination complex with the
dye.
Page
391
TIN
Silicon is the most abundant element on Earth, after oxygen. It is the major
component of sand, sandstone and silicates found in clay.
Ultrapure silicon and germanium are used in the manufacture of
transistors. These are electronic components found, for example, in
computer chips. Silicon dioxide is found in a large number of ceramics. This
is because silicon dioxide has a very high melting point and is therefore heat
resistant.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 391
LEAD




Lead metal and PbO2 are used as electrodes in storage batteries
such as the car battery.
Tetraethylead (Pb(C2H5)4 is used as an antiknock in petrol
engines therefore allowing smooth running of the engine. Petrol
that contains this compound is known as leaded petrol. Health
concerns related to the toxic nature of Pb2+ has resulted in a
decrease in the use of leaded petrol.
Lead metal is used as a shield to absorb harmful radiation
around X-ray equipment and nuclear reactors.
PbCrO4, a bright yellow substance of very low solubility , is used
to make road marks.
Notes for use in qualitative analysis
Reactions involving Pb2+ and lead compounds
Three oxides of lead are known

Oxide
colour
Oxidation state
of lead
PbO
yellow
+2
PbO2
black
+4
Pb3O4
red
+2 and +4
Heating PbO2 on a Bunsen flame results in a change of colour from black to yellow. A colourless gas is
given off which relights a glowing splint.
PbO2(s)  PbO(s) + 1/2O2
This decomposition is accompanied by reduction of Pb (IV) in PbO2 to Pb(II) in PbO. This
proves that the +2 state of lead is more stable than the +4 state.
Red lead oxide (Pb3O4) is in fact a double oxide, containing PbO and PbO2. The formula of Pb3O4 can
be written as 2PbO.PbO2. PbO, which is more stable, has the greater proportion. Red lead oxide
behaves in its chemical reactions as if it were a mixture of PbO 2 and PbO.
Action of heat on read lead oxide
When Pb3O4 is heated, the PbO2 in it decomposes to PbO and oxygen is evolved. The PbO component is not
altered.
Page
392

Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 392
2PbO.PbO2 (s)  3PbO(s) + ½ O2
There is a change of colour from red to yellow as PbO is formed.
Action of dilute HNO3 on red lead oxide
PbO is ionic and is therefore basic (In fact, it is amphoteric). PbO2 is less basic because it contains a degree of covalent
character. When dilute HNO3 is added to red lead oxide, the PbO in it is neutralized to form Lead (II) nitrate. PbO2 is not
affected and remains as a black residue.
Pb3O4(s) + 4HNO3 (aq)  2Pb (NO3)2(aq) + 2H2O (l) + PbO2(s)
red solid
colourless solution
black solid
Q Predict the effect of adding the following to red lead oxide
a) concentrated HCl followed by heating
b) NaOH(aq)
Reactions of and tests for Pb2+
Reactions 1, 2 and 6
Page
393
Pb2+ is precipitated by both NaOH (aq) and ammonia as Pb (OH)2. Now, this product is amphoteric, so it reacts further
with strong alkali, in this case, NaOH, producing a soluble complex, Pb (OH)42-. The white precipitate of Pb(OH)2 dissolves
in the process (reaction 2). However, Pb(OH)2 does not react with a weak base such as ammonia. The precipitate formed in
reaction 6 is therefore insoluble in excess ammonia.
Reactions 1 and 2 can be reversed by carefully adding a dilute acid. In this case, HNO3 acid is used because using H2SO4
and HCl would result in the formation of a precipitate
.
Reactions 3 and 4
Pb(OH)42- formed in reaction 2 has lead in the +2 state. This is the stable state of Pb. Oxidation of Pb2+ to Pb(IV) would
require a good oxidizing agent such as H2O2. This is what happens in reaction 3. The Pb(OH)2 formed is unstable. Heating
it results in the formation of PbO2, which then decomposes to PbO, releasing, O2. Lead (IV) is thus converted back to lead
(IV) (reaction 4.)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 393
Note that PbO is also obtained by heating Pb (OH)2 (reaction 5)
Reactions 7, 8, 9 and 10
Pb2+ ions form precipitates with Cl- ions (reaction 9), iodide ions (reaction 8), sulphate ions (reaction 7) and chromate
ions (reaction 10). Note that PbI2 and PbCrO4 are yellow solids whereas PbCl2 and PbSO4 are white. Also note that
PbCl2 and PbI2 are insoluble in cold water but dissolves when heated due to complex ion formation.
Questions, solutions and discussions
Q1
(a) (i) Sketch the variation in melting points of the elements in Group (IV).
(ii) Explain how this variation in melting point is related to the structure and bonding of
the elements.
(b)
CCl4 and SiCl4 behave differently with water.
(i) Describe the reaction (if any) of CCl4 with water.
(ii) Describe the reaction (if any) of SiCl4 with water.
(iii) Write equations for any reactions that occur
(iv) Explain why these two chlorides differ in their behaviour with water.
9701/4/O/N/2002
Solutions
(a)(i)
m.p
X
X
X
X
Si
Ge
Sn
Pb
Page
394
C
X
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 394
(ii) C, Si and Ge have high melting points due to the presence of giant covalent bonding. A large amount
of energy is required to break a large number of strong covalent bonds. Melting points decrease in
the order C, Si and Ge because of increasing bond length. The bonds therefore become weaker and
less energy is required to break them.
Sn and Pb are metals. The metallic bonds are weaker than the covalent bonds in giant covalent
structures.
(b)(i) CCl4 has no reaction with water.
(ii) SiCl4 undergoes hydrolysis in water, liberating white fumes (of HCl) and forming a white solid
(SiO2).
(iii) SiCl4 + 2H2O  SiO2 + 4 HCl
(iv) SiCl4 has accessible d - orbitals which makes initial attack by water molecules possible.
Q2
All the Group IV elements form chlorides with the formula MCl4.
(a) Describe the bonding in, and the shape of, these chlorides.
The boiling point of lead (IV) chloride cannot be measured directly because it decomposes
on heating. The following table lists the boiling points of three Group IV chlorides.
chloride
b.p. /0C
SiCl4
58
GeCl4
83
SnCl4
114
Page
395
(b) (i) Plot these data on the following axes and extrapolate your graph to predict what the
boiling point of PbCl4 would be if it did not decompose.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 395
(ii) Suggest why the boiling points vary in this way.
(c)
SiCl4 reacts vigorously with water whereas CCl4 is inert.
(i) Suggest a reason for this difference in reactivity.
(ii) Write an equation for the reaction between SiCl4 and water.
(iii) Suggest, with a reason, whether you would expect GeCl4 to react with water.
(d)
SiCl4 is used to make high-purity silicon for the semiconductor industry. After it has been purified
by several fractional distillations, it is reduced to silicon by heating with pure zinc.
(i) Suggest an equation for the reduction of SiCl4 by zinc.
(ii) Use your equation to calculate what mass of zinc is needed to produce 250 g of
pure silicon by this method.
9701/04/M/J/2004
Solutions
(a) Simple covalent, tetrahedral geometry, bond angle 109.5 0
(b) (i) From a plotted graph, melting point of PbCl4 is 138 – 151°C
(ii) Van der Waals forces become stronger due to increasing number of electrons in the metal as the
group is descended.
(c)(i) and (ii) See Question 1
(iii) GeCl4 reacts with water since Ge has accessible/ low lying d - orbitals to accept water molecules.
(d) (i) SiCl4 + 2Zn  Si + 2ZnCl2
(ii) mass = 250 x 2 x 65.4/28.1
= 1163.7 g
(a) (i) Use the following sets of axes to sketch graphs of the variations in the melting
points and the electrical conductivities of the Group IV elements.
(ii) Explain how the variation in conductivity is related to the structure and bonding in
the elements.
Page
396
Q3
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 396
melting point
electrical conductivity
High
High
medium
medium
low
low
C
Si
Ge
Sn
Pb
C
Si
Ge
Sn
Pb
(b) Going down Group (IV) there is a variation in the relative stabilities of the higher and
lower oxidation states of the elements in their oxides.
Illustrating your answers with balanced chemical equations, in each of the following
cases suggest one piece of chemical evidence to show that
(i) CO is less stable than CO2,
(ii) PbO is more stable than PbO2.
(c)
Name one ceramic based on silicon (IV) oxide, and explain what properties of the
oxide make it suitable for this use.
(d)
Tin(II) oxide reacts with both acids and alkalis.
(i) What name is given to this property of an oxide?
(ii) Write suitable equations to show these two reactions of tin(II) oxide. 9701/04/O/N/2005
Solution
(a) (i) melting point: graph should show
low : silicon
medium : germanium
High : carbon
Sn and Pb: lower than Si/Ge
conductivity: graph should show
Low : carbon (diamond)
High : carbon (graphite)
Page
397
Medium : silicon and germanium
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 397
Sn and Pb: higher than Si/Ge
(ii) Sn, Pb are metals and so have delocalized electrons to conduct electricity. Graphite has a giant
layered structure, with electrons delocalized along layers.
Si, Ge and diamond have electrons localized in covalent bonds.
(b) (i) e.g CO burns to give CO2 , 2CO + O2 → 2CO2
OR CO is converted to CO2 by Fe2O3 , 3CO + Fe2O3 → 3CO2 + 2Fe
(ii) e.g. PbO2 is decomposed by heat 2PbO2 → 2PbO + O2
(c)
pottery - high melting point, hard
(d) (i) amphoteric
(ii) SnO + 2HCl → SnCl2 + H2O
SnO + 2NaOH → Na2SnO2 + H2O
Q4
Carbon forms two stable oxides, CO and CO2. Lead forms three oxides: yellow PbO, black
PbO2 and red Pb3O4.
(a) Carbon monoxide burns readily in air. Heating black lead oxide produces oxygen gas,
leaving a yellow residue.
(i) Suggest a balanced equation for each reaction.
(ii) Explain how these two reactions illustrate the relative stabilities of the +2 and +4
oxidation states down Group IV.
(b) Red lead oxide contains lead atoms in two different oxidation states.
(i) Suggest what these oxidation states are, and calculate the ratio in which they occur
in red lead oxide.
(ii)
Predict the equation for the action of heat on red lead oxide.
(iii)
Suggest an equation for this reaction.
(iv)
Explain how this reaction illustrates the relative basicities of the two oxidation states
of lead.
Page
398
When red lead oxide is heated with dilute nitric acid, HNO 3, a solution of lead (II) nitrate
is formed and a black solid is left.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 398
(c) Both tin(II) oxide and tin(IV) oxide are amphoteric.
Write a balanced equation for the reaction between tin (II) oxide and aqueous sodium
hydroxide.
9701/04/M/J/07
Solutions
(a) (i) 2CO + O2 → 2CO2
2PbO2 → 2PbO + O2
(ii) +4 state becomes less stable / +2 state becomes more stable down the group
(b) (i) Pb (II) : Pb (IV) = 2:1
(ii) Pb3O4 → 3PbO + ½O2
(iii) Pb3O4 + 4HNO3 → 2Pb(NO3)2 + PbO2 + 2H2O
(iv) PbO is more basic than PbO2 since PbO2 does not react with HNO3
(c)
SnO + 2NaOH → Na2SnO2 + H2O (or Na2Sn(OH)4
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
9.2in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
1 (a) Describe how the behaviour of the oxides of tin and lead in their +4 oxidation states
differ on heating.
(b)
Explain the following by using data from the Data Booklet where appropriate, and writing
equations for all reactions.
Page
399
(i) A sample of liquid PbCl4 is placed in a flask and the flask is gently warmed. A gas is evolved
and a white solid is produced. When the gas is bubbled through KI(aq), purple fumes are
produced.
(ii) Repeating the same experiment using liquid SnCl4 instead of PbCl4 results in no evolution of
gas, and no reaction with KI(aq).
(c)
The molecule dichlorocarbene, CCl2, can be produced under certain conditions. It is highly
unstable, reacting with water to produce carbon monoxide and a strongly acidic solution.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 399
(i) Suggest the electron arrangement in CCl2 and draw a dot-and-cross diagram showing this.
Predict the shape of the molecule.
(ii) Construct an equation for the reaction of CCl2 with water.
2
9701/41/M/J/10
The most typical oxides of tin and lead are SnO, SnO2, PbO and PbO2.
The following two generalizations can be made about the oxides of the elements in Group (IV).
• As the metallic character of the elements increases down the Group, the oxides become
more basic.
• The oxides of the elements in their higher oxidation states are more acidic than the oxides of the
elements in their lower oxidation states.
(a) Use these generalizations to suggest which of the above oxides of tin or lead is most
likely to react with each of the following reagents. In each case write a balanced
equation for the reaction.
(i) with NaOH(aq)
(ii) with HCl(aq)
(b) ‘Red lead’ is used as a pigment, and as a metal primer paint to prevent the corrosion of steel. It
is an oxide of lead that contains 9.30% oxygen by mass.
Calculate to 3 significant figures the number of moles of oxygen and lead contained in a
100.0 g sample of red lead. Hence calculate its empirical formula. 9701/41/O/N/10
Page
400
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 400
9.4
GROUP (VII) ELEMENTS
Introduction
The Group (VII) elements are known as the halogens, which literally means salt formers. Table 9.4.1
below shows key facts about the Group (VII) elements, from chlorine to iodine. The halogens are
diatomic and have a simple molecular structure. This explains why they have low melting points and
boiling points. Weak Van der Waals forces exist between the molecules and a small amount of energy
is required to break these forces and bring about melting or boiling.
The Group (VII) elements illustrate some important points, including
 the effect of number of electrons (Mr) on physical properties
 the effect of atomic size on chemical reactivity
Once more, as you go through this section, try to link up with concepts which you studied in physical
chemistry, for example, the use of reduction potentials in predicting the outcome of a reaction.
Group (VII) elements: Facts at a glance
Element
Atomic number
Outer shell
configuration
Cl2
Br2
I2
17
35
53
3s23p5
4s24p5
State at 200C
gas
Colour
Greenish yellow
m.p0C
Solubility in water
-101
moderate
liquid
Red-brown
-7
slight
5s25p5
solid
black
113
insoluble
Table 9.4.2
Properties
of the Group (VII)
elements, from chlorine
to iodine
Page
401
9.4.1.
Physical properties
The elements show a clear trend in physical properties. Chlorine is a greenish-yellow gas, bromine is a redbrown liquid and iodine is a shiny black solid at room temperature
 There is a clear gradation from gaseous state, through liquid state, to solid state.
This change is a result of increasing strength of VDW forces down the group. All of the elements
are simple molecular, so the forces of attraction present between molecules are VDW. These forces
are a result of instantaneous distortions of the symmetric electron cloud of the molecules. Down
the group, the number of electrons (Mr) in the molecules increases, so the extent of distortion also
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 401



increases. Thus strength of VDW forces increases down the group. Molecules are held more tightly
and more closely, and this results in a change of a state.
Intensity of colour increases down the group. This shows that particles hold to each other more
tightly down the group. If particles of a substance are held close, its colour becomes intense. Solid
iodine appears black, but when its particles are separated by heating to form a gas, the colour of
the vapour becomes less intense (it is purple)
The melting and boiling points of the elements increase down the group. VDW forces become
stronger down the group, so more energy is required to separate molecules from each other.
Solubility of the elements in water decrease down the group. The dissolution of the halogens
involves slight reaction with water, to form the acid HXO -. Reactivity of the halogen with water
decreases down the group and so does the solubility.
9.4.2
Oxidation numbers in compounds
The halogens are able to expand the octet by using d-orbitals in bonding. Thus they exhibit variable
oxidation numbers in their compounds, ranging from -1 to +7. The lowest oxidation number, -1, refers to
the gain of one electron in the outer shell of the halogen atom, for example, during the formation of ionic
compounds such as NaCl. The halogen atom has 7 electrons in the outer most shell. Any number of these
electrons can be used in covalent bond formation, for example, the use of one valence electron leads to the
formation of compounds such as HCl. Examples of compounds in which the halogen has exceeded an octet
in its outer shell are ICl6 and IF7 molecules . Table 9.4.3 gives representative compounds for the various
oxidation states of the halogens. The electronic structures of CF 3, ClO3- and CF7 are also shown. Note that
in all three compounds, Cl has more than an octet of electrons in the outer shell. The halogens, except for
fluorine which has no d orbitals, are able to expand the octet .
Oxidation
number
Examples
-1
Ionic: NaCl, MgBr2
Covalent: ICl
+1
ClO- e.g. in KClO
+3
ClF3
+5
ClO3- e.g. in NaClO3
+7
ClO4-
F
x
x
..Cl. . x F
.. .
x
x
402
Some halogen
ClO3-
CF3
Page
Table 9.4.3
compounds
xx
x
O
x
charge
.
.. Cl ..xx O
..
x x
F
Cl uses 3 electrons out of 7 for
covalent bonding
CF
O
Cl uses 5 electrons for bonding
Prepublication manuscript © L. MWANAWENYU7 2011. Please do not photocopy or reproduce
F
Fx . x. .x F
. .x
F x .Cl
. F
x x
F
Page 402
Cl uses all 7 electrons for bonding
. Cl. x F
.. .
x
.. Cl ..xx O
..
x x
F
O
Cl uses 5 electrons for bonding
Cl uses 3 electrons out of 7 for
covalent bonding
CF7
F
Fx . x. .x F
. .x
F x .Cl
. F
x x
F F
9.4.3
Cl uses all 7 electrons for bonding
Reactions as oxidizing agents
The Group (VII) elements usually participate as oxidizing agents in reactions.
Table 9.4.2 shows the Eθ values for the reaction X2 + 2e Ý 2X- where X2 is a Group (VII) element. Redox
data in this table shows that the Group (VII) elements become weaker oxidizing agents down the group.
Chlorine has the highest oxidizing power of the three elements being discussed, and iodine has the least.
Element
Reduction equation
Eθ/V
Cl2
Cl2 + 2e- Ý 2Cl-
1.36
Br2
Br2 + 2e- Ý 2Br-
1.07
I2
I2 + 2e- Ý 2I-
0.54
Table 9.4.2 Redox data for the
Group (VII)
Make sure you can interpret redox data such as
that given in this table. Chlorine has the most
positive Eθ/V, showing that the forward reaction
which forms the halide ion (by reduction of the
halogen) is the most energetically feasible. In this
conversion, the halogen gains electrons (it is
reduced) from another substance. This other
substance therefore loses electrons, that is, it is oxidized by the halogen. Chlorine, with the most
positive Eθ value, accepts electrons most easily. It is
the most powerful oxidizing agent of the three
halogens.
Explanation of the trend in oxidizing power
Page
403
To start with, all three halogens are oxidizing agents, that is, they tend to react by gaining electrons. Note
that all the electrode potentials are positive, meaning that the halogens can oxidize a large number of substances.
 Each halogen has a large number of protons in the nucleus, compared with the elements in the
same period at the left hand side of the Periodic Table. The halogens therefore gain electrons during
chemical reactions, since the incoming electrons feel a relatively strong attraction from the nucleus.
In addition, the halogen atom is much smaller than atoms on the left hand side of the Periodic
Table. Small non-metallic atoms are good at attracting an electron because the incoming electron
would be close to the nucleus from which it experiences a relatively strong attraction.
 The valence shell of the halogen has 7 electrons. Each halogen atom only has to accept a single
electron to complete the octet in the outer most shell, which is energetically feasible.
The oxidizing power of the halogens is affected by atomic size and shielding (Fig 9.4.1).
An oxidizing agent accepts an electron or electrons from a reducing agent. The outer shell of the atom must
have the ability to accept and stabilize the electron. Chlorine has the smallest atoms. The outer shell is
therefore close to the nucleus, and so it is able to strongly attract the incoming electron. Once incorporated
into the outer shell, the electron is strongly held to the atom (stabilized) for two reasons.
 It is close to the nucleus from which it feels a strong attraction.
 There are a few inner shells to shield the electron from the attractive influence of the nucleus.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 403
Down the group, the size of atoms increases as new shells are opened. The distance of the outer shell from
the nucleus increases, so it does not easily attract an electron. If the electron is accepted into the outer shell,
it is also shielded by inner shells from the nucleus. The electron would therefore be weakly held to the atom.
Fig 9.4.1 Oxidizing power of the halogens decrease down the group
in response to increasing atomic size.
Electron affinities of the Group (VII) elements give evidence that going down the group, the elements
become more reluctant to accept an electron in the outer most shell (the electron affinities become less
negative down the group).
Element
First electron affinity
/Kjmol-1
Cl
-364
Br
-342
I
-314
Table 9.4.3 Electron affinities of the
Group (VII) elements
4.3.1
Reactions with metals
The halogens, being oxidizing agents, react with metals, which are reducing agents, for example
2Na(s) + X₂(g) → 2NaX(s)
Page
404
This is a redox reaction which forms a salt. In fact, the term halogen means ‘salt former’. The metal acts
as the reducing agent and the halogen as the oxidizing agent. The speed and vigour of reaction decreases
down the group. For instance, fluorine reacts rapidly and explosively with Na.
2Na(s) + F₂ (g) → 2NaF(s)
The reaction of Na with chlorine is rapid and vigorous but less violent. The reaction of iodine with Na is
relatively slow.
Explanation
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 404
There are two ways of explaining this trend in reactivity.


In terms of the reactivity of the halogen. It has already been noted that the reactivity of the
halogens decreases down the group, that is, the affinity of the halogen for electrons decreases. In
this reaction, Na metal readily donates its valence electron to the halogen. The vigour and speed of
reaction depends on how readily the halogen atoms can accept an electron in the outer shell.
In terms of stability of the resulting salt
If the product of a reaction has high stability, then its formation is favourable and the reaction may
occur readily. The stability of ionic compounds can be measured in terms of their lattice energies.
NaI, with the smallest lattice energy (least stable) has the least tendency to be formed, that is, the
reaction between sodium and iodine is the slowest.
Salt
Lattice energy/
Kjmol-1
NaCl
771
NaBr
733
NaI
684
Table 9.4.4 Lattice energies
of the sodium salts of the
Group (VII) elements
4.3.2
The lattice energies of the Group (VII) metal salts, NaX decreases
down the group due to the increasing size of the halide ion (Table
9.4.4). As the ion gets bigger:
(i) Interionic distance in the salt increases, and this results in
weaker electrostatic attraction between the metal ion and the
halide ion.
(ii) Distance between the positive charge on the metal ion and the
nucleus of the halide ion increases. Attractions between the
nucleus of the halide and the positive charge on the metal ion
therefore become weaker. The tendency of the salt to be
formed therefore decreases down the group.
The halogen - halide redox displacement reactions
Consider the reaction:
Cl₂(g) + 2KI (aq) → I₂ (aq) + 2KCl (aq)
In this reaction, chlorine displaces iodine from its salt. A halogen will displace the halogen which is below
it from a solution of its salt. Once more, this shows that the oxidizing power of the halogens decreases down
the group.
The reaction essentially involves oxidation of the iodide ion by chlorine, to form iodine. Note that the iodide
ion loses an electron from the outer shell to form neutral iodide atoms, which immediately bond to form I 2
molecules. By losing an electron to chlorine, iodide is oxidized, that is, it acts as a reducing agent.
We can use electrode potentials to deduce the feasibility of such reactions.
Eθ/V
Page
405
Q
A
Cl₂ + e⁻ → Cl⁻
+1.36 ... (i)
Br₂ + e⁻ → Br⁻
+1.07 ... (ii)
I₂ + e⁻ → I⁻
+0.54 ... (iii)
Is the reaction between chlorine and KI (aq) energetically feasible?
Using half equations (i) and (iii) above, and reversing (iii), which is the less
positive, gives Cl2(g) + 2KI(aq)  I2(s) + 2KCl(aq).
Eθcell = 1.36 + (-0.54) = +0. 82V
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 405
The reaction will take place since the Eθ cell is large and positive.
The reaction will take place, since the Eθ cell is large and positive.
4.3.3 Reactions of the halogens with transition metal ions in low
oxidation states
Chlorine gas and liquid bromine can oxidize Fe (II) in solution, to iron (III).
Fe³⁺ (aq) + e⁻ → Fe²⁺ (aq)
Eθ = +0.77
(IV)
Reversing equation (iv) and combining with equation (i)
Cl₂ (g) + 2Fe²⁺ (aq) → 2Fe³⁺ (aq) + 2Cl⁻ (aq)
Eθcell = 1, 36 + (- 0.77) = +0.59V
The reaction is feasible. The light green colour of Fe (II) changes to the yellowish brown of Fe (III).
4.3.4
Reaction of the halogens with hydrogen
X₂ + H₂ (g) → 2HX
The halogens oxidize hydrogen to form hydrogen halides, HX, where X is Cl -, Br – and I- . In this reaction,
the oxidation number of hydrogen increases from 0 to +1, showing that it has been oxidized. The oxidation
number of the halogen decreases from 0 to -1, showing that it has been reduced.
The vigour, speed and extent of reaction decreases down the group. The enthalpy of reaction also becomes
less negative, that is, it becomes more endothermic.
Fluorine reacts explosively with hydrogen under all conditions, even at freezing temperatures.
F₂(g) + H₂(g) → 2HF(g)
Hydrogen burns in chlorine. No heating is required. The reaction becomes explosive if the mixture is exposed to light.
A mixture of bromine and hydrogen gases will not react unless the mixture is heated to 200⁰C in the presence of Ni catalyst.
A mixture of iodine and hydrogen gases will not react unless the mixture is strongly heated at about 400⁰C,
in the presence of Pt catalyst. The reaction is reversible, that is, once formed, some molecules of
HI will decompose back to its constituent elements.
I₂ (g) + H₂ (g) ⇋ 2HI (g)
Page
406
The trend observed in the vigour, speed, extent and enthalpy of reaction can be explained in two ways.

Oxidizing power of the halogens, which decreases down the group?

Stability of the hydride, which decreases down the group. The hydride becomes less stable due to
the increase in size of the halogen atom. Overlap between the valence s orbital of H and the p orbital
of the halogen atom becomes less efficient. The distance between the nuclei of the atoms of the HX bond and the electrons of the bond also increases. Consequently, the H-Hal bond becomes longer
and weaker. This explains why once formed, HI has a tendency to decompose back to its elements.
The enthalpy change of reaction becomes less exothermic because the bonds being formed become
weaker and the product is relatively high in energy (that is, unstable).
Hydride
Bond
length/nm
Bond energy
Kj/mol
∆Hθf
Kj/mol
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 406
HCl
0.128
431
-92.3
HBr
0.141
366
-36.2
HI
0.160
299
+25.9
Table 9.4.5 Some properties
of the Group (VII) hydrides
The hydrides of Group (VII) have the general formula HX. Important information about the Group (VII)
hydrides has been given in Table 9.4.5.
 The hydrides are strong acids. In water, they dissolve and dissociate completely, giving a strongly
acidic solution, for example,
HCl (aq)  H+ (aq) + Cl-(aq)
The dissolution of HX in water can also be regarded as Brønsted-Lowry neutralization in
which HX acts an acid and water as a base.
+
H
H
O
H
H-Cl
H
H
O
+ Cl-
Thermal stability of the hydrides
Stability of the hydrides with respect to decomposition by heat decreases down the group. This is because
the H-Hal bond becomes weaker down the group as the size of the halogen atom increases.
HCl resists decomposition by heat, even at very high temperatures. HBr will decompose slightly when
strongly heated. Plunging a red hot platinum rod in a jar containing HI will cause it to decompose. A
purple vapour containing I2 and H2 is formed.
2HI (g) Ý H2 (g) + I2 (g)
4.3.5
Reaction of the halogens with thiosulphate, S₂O₃²⁻
The halogens are powerful enough to oxidize sodium thiosulphate. The sulphur containing product depends
on the strength of the halogen as an oxidizing agent. Chlorine and bromine are powerful enough to oxidize
thiosulphate to SO42-. The oxidation state of sulphur increases from +2 in S2O32- to +6 in SO42-. In SO42- ,
sulphur attains the highest possible oxidation state. Let the halogen be X 2, and then the reaction
with thiosulphate can be written as
S₂O₃²⁻ + 4X₂ + 5H₂O → 2SO4²⁻ + 10H⁺ + 8X⁻ … (i)
Page
407
Since the oxidation state of sulphur changes from +2 to +4, each sulphur atom loses two electrons
(oxidation is loss of electrons; it is accompanied by an increase in oxidation number.) Since there are two S
atoms in thiosulphate, each S₂O₃²⁻ molecule loses four electrons. Thus four moles of the halogen are
required per mole of thiosulphate, since each halogen atom requires one electron.
When X2 = Cl2, the Eθ cell for reaction (i) is.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 407
Reduction equation
Eθ/V
Cl2 + 2e Ý 2Cl-
+1.36
Br2 + 2e- Ý 2Br-
+1.07
I2 + 2e- Ý 2I-
+0.54
Table 9.4.6 Some redox potentials
involving the
2- halogens
2-
S 4 O6
+ 2e Ý 2S2O3
+0.09
Iodine is not strong enough as an oxidizing agent to oxidize thiosulphate to sulphate. Instead, the
tetrathionate ion is produced, in which the sulphur has an average oxidation number of +2. The fact that
iodine is not able to bring sulphur to the +6 state shows that it is a weak oxidizing agent.
2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻ ... (ii)
Q
Calculate the cell potential for reaction (ii) above
4.3.6
Reactions of the halogens with water
Halogens react with water in a redox displacement reaction. For chlorine, bromine and iodine, the
reaction can be generalized as
X2(g) + H2O (l) Ý HOX (aq) + X- ,
E.g. for chlorine, the reaction forms chloric (I) acid, HOCl and Cl Cl2 + H2O Ý HOCl + HCl


This is a redox disproportionation reaction because the chlorine undergoes both oxidation and
reduction. Its oxidation number increases from 0 in chlorine to +1 in HOCl, and decreases to -1 in
HCl.
The HOCl formed is itself a weak acid, so it dissociates to yield the chlorate (I) ion and a H + ion.
HOCl (aq) Ý H+ (aq) + OCl-(aq) ... (i)
Page
408

The reaction of chlorine with water is important in water purification. The OCl - ion is an oxidizing
agent. It oxidizes organic matter, including bacteria and food materials that sustain them. The
presence of H+ ions also makes the water acidic, and this also kills the bacteria. HClO is more
effective as a bactericide than the chlorate ion, so it is important to ensure that its concentration in
the water being treated remains fairly high. This can be achieved by closely monitoring pH. If pH
is too high (that is, large [OH-]), H+ ions in equilibrium (i) are neutralized and removed as water.
By Le Chatelier’s principle, this causes the equilibrium to shift to the right, resulting in reduction
of [HOCl]. However, pH should not be made too low, because this causes chlorine to be lost from
water by shifting equilibrium (ii) below to the left. This also results in lowering of the
concentration of HOCl.
Cl2 + H2O Ý HOCl Ý H+ (aq) + OCl-(aq) + Cl- (aq) ... (ii)

Concentration of chlorine in water must be kept relatively low to prevent formation of poisonous
substances by the free radical reactions between chlorine and organic matter
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 408

Bromine and iodine are less reactive towards water. This is because they are weaker oxidizing
agents. Concentration of HBrO and HIO in water would be smaller. This explains why Br 2 and I2
are poorly soluble in water.
4.3.7
Reactions of the halogens with NaOH (aq)
Consider the reaction of chlorine with water
Cl2 (g) + H2O (l) Ý H+(aq) + ClO- + H+(aq) + Cl-(aq) ... (i)
[HClO (aq]
[HCl (aq)]
Now, suppose that cold NaOH (aq) is added to the resulting solution. It will neutralize the two H+ ions on
the right hand side of the equilibrium. This forces the equilibrium to the right, increasing the amount of
chlorine that reacts with the water.
2H+ (aq) + 2OH- (aq)  2H2O (l) ... (ii)
Combining equations (i) and (ii)
Cl2 (g) + H2O (l) Ý H+ (aq) + ClO- + H+ (aq) + Cl-(aq)
2H+ (aq) + 2OH- (aq)  2H2O (l)
Cl2 (g) + 2OH- (aq)  ClO- + Cl-(aq) + H2O (l)
... (i)
... (ii)
... (iii)
Equation (iii) is a summary of what happens when chlorine reacts with cold dilute NaOH (aq) (temperature
< 150C). A disproportionation reaction takes place, in which some chlorine molecules are converted to Cl and some to ClO- . The formation of water is expected since the reaction essentially involves neutralization
of an acidic solution of Cl2 (aq).
This reaction is used to make household bleach, which is a solution containing sodium chlorate (I) and
NaCl.
With hot concentrated NaOH (aq), further oxidation of chlorate (I) to chlorate (V), ClO 3-, occurs.
Again, this is a disproportionation reaction.
3Cl2(g) + 6NaOH (aq) → 5NaCl (aq) + NaClO3 (aq) + 3H2O (l)
Check that the oxidation state of chlorine in the chlorate (V) ion (ClO3- in NaClO3) is +5.
NaClO3 is used as a weed killer.
Cl2 undergoes disproportionation in NaOH (aq). The products formed depend on the conditions
employed. With dilute and cold NaOH, a mixture of NaCl and NaClO is formed. With hot and
concentrated NaOH (aq), a mixture of NaCl and NaClO 3 is formed.
Page
409
The reactions discussed above are also undergone by bromine and iodine. Bromine forms bromate (I) and
bromate (V) ions (BrO- and BrO3- respectively), depending on the conditions. Similarly iodine forms IO and IO3-. However, IO- is very unstable, even at O0C and any attempt to prepare it results in the formation
of IO3- as well, even under ice cold conditions.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 409
9.4.4
Reactions of the halide ions
Important note
Students often treat the terms ‘halogen’ and ‘halide’ as if they were interchangeable. For example, when asked to
describe a particular reaction of a halide ion, they tend to discuss in terms of the halogen. A halogen refers to a neutral
Group (VII) element or atom. A halide is the negative ion formed from a halogen. Reactions and chemical properties
of halogens are different from that of the halides, for instance, the halogens are oxidizing agents, but the halides are
reducing agents.
The halide ions are involved in three major types of reactions in which they behave as
 reducing agents
 precipitating agents
 ligands
The halide ion, being negatively charged, is electron rich, and so has the ability to act as an electron donor
in chemical reactions, that is, as a reducing agent. The trend in reducing power of the halide ions down the
group can be shown and explained using reduction potentials (Table 9.4.7).
When the halide ion acts as a reducing agent, it loses
θ/V
Element
Reduction
potential
its
charge (electron)
and becomes
a neutral E
atom.
Two
halogen atoms then combine to form a molecule. To
Chlorine the Cl
+1.36
2 + 2e Ý 2Cl
understand
behaviour
of - the halide
ions as
reducing agent, the electrode potentials in Table 9.4.7
Bromine
+1.07
Br2 + 2e
2Br
above
should therefore
be- Ý
read
in- reverse to
show the
halide ions losing electrons (a reducing agent is an
Iodine donor). I2 + 2e- Ý 2I+0.54
electron
After
reversing
the reduction
potentials,
the I- to I2
Table
9.4.7 Reduction
potentials
for the
θ
conversion
haselements
the most positive E value, showing
Group (VII)
that this is the most feasible. The Eθ value becomes
more negative up the group, showing that it becomes
more difficult for the halide to lose an electron,
that
Halide
Reduction potential Eθ/V
is, to act as a reducing agent. Cl- is therefore the
weakest
the most
- Ý Cl and
- iodide is -1.36
Chloridereducing
2Clagent,
2 + 2e
powerful.
Bromide
-1.07
2Br- Ý Br2 + 2eReducing power of the halide ions
increases
down the
group.
Iodide
-0.54
2I- Ý I2 + 2e-
Page
410
the
Explanation
It has already been explained that the halide ion acts as a reducing agent by losing the extra electron (negative charge) in the outer most shell. Reducing power of the halide ion therefore depends on how firmly or
loosely held this electron is to the atom. Going down the group, size of the halide ion increases as new shells
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 410
are opened. As a result, distance of the outer shell from the nucleus increases (Fig 9.4.2). Moreover, the
outer shell is shielded by inner shells from feeling a strong attractive influence from the nucleus. Going
down the group, it becomes easier to remove an electron from the outer shell of the halide ion, since this
electron becomes loosely held to the atom.
Important note
Students often confuse the trend in the reactivity of the halide
ions with that of the halogens.
The halide ions become more reactive (as reducing agents) down
the group, whereas the halogens become less reactive (as
oxidizing agents) down the group. This is because the halide ions
react by losing an electron, a process which becomes easier down
the group as ionic size increases. On the other hand, the halogens
react by attracting an electron into the outer shell, a process
which becomes harder down the group as the atoms become
larger. Distance of the outer shell from the nucleus increases,
and moreover, the outer shell becomes more shielded by inner
shells. It therefore becomes harder for the halogen atom to
accept an electron, that is, to act as an oxidizing agent.
Fig 9.4.2 Reactivity of the halide ions as
reducing agents increase down the group
4.4.1
Reaction of the halide ions with concentrated sulphuric
acid (see Table 9.4.8)
Concentrated sulphuric acid is an oxidizing agent. As it oxidizes a substance, it is itself reduced to lower
oxidation states such as sulphrous acid, (H2SO3 = SO2 + H2O), H2S and S.
The final sulphur containing product depends on the strength of the reducing agent.
When concentrated H2SO4 is added to NaCl (aq), white fumes of HCl are produced.
H2SO4 (aq) + Cl-(aq)  HSO4-(aq) + HCl (g)
Note that this is not a redox reaction. Cl- ions are too weak as reducing agent to reduce H2SO4. This is an
example of a displacement reaction in which a volatile acid (HCl) is displaced from a solution containing a
less volatile acid (H2SO4). What happens is that H2SO4 dissociates to produce a hydrogen ion.
H2SO4  H+ + HSO4- ... (I)
The hydrogen ions then combine with chloride ions to produce the less volatile acid, HCl.
Page
411
(i) If a dilute solution of H2SO4 is used, no white fumes are formed, because the large concentration
of water causes the HCl to dissolve, instead of escaping as a gas.
(ii) In reaction (I) above, H2SO4 does not dissociate fully to form 2H+ and SO42-, because
once formed, the negatively charged HSO4- ion has very little tendency to lose a positive
particle (H+).
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 411
Add concentrated sulphuric acid
Salt
HX produced(X= Cl-,
Br- , I-)
Other
products
formed
Type of
reaction
Observation
NaHSO4
displacement
White and dense
fumes (of HCl)
Yes , but a little
Br2 , SO2 , H2O,
NaHSO4
Displacement and
redox, displacement
is predominant
Orange vapour,
acrid smell of SO2,
white fumes of HBr.
Yes, but it is immediately
reduced to iodine
A mixture of
products , including
I2, SO2, H2O, H2S
Displacement and
redox, the product
of displacement
(HI) is immediately
oxidized to I2
Dark brown
precipitate of iodine
, acrid smell of SO2 ,
foul smell of H2S and
yellow deposit of
sulphur. Some of
these products,
especially H2S and S,
may not form,
depending on initial
concentration of NaI
and H2SO4
NaCl
Yes
NaBr
NaI
Table 9.4.8
Reactions of the halide ions with concentrated sulphuric acid
When concentrated sulphuric acid is added to a solution of NaBr (aq), white fumes of HBr are produced,
but the white colour is soon masked by an orange vapour of Br 2.
Two types of reactions occur here.
 HBr is displaced
H2SO4 (aq) + NaBr (aq)  HBr (g) + NaHSO4 (aq) ... (i)
Some bromide ions (in the form of HBr produced in reaction (i)) reduce H2SO4 to SO2 and H2O
(a mixture of SO2 and H2O can also be thought of sulphrous acid, H2SO3). Meanwhile, the
bromide ions are oxidized to bromine.
We can construct the equation of this reaction with the aid of a Data Booklet.
H2SO4 + 2H+ + 2e- Ý SO2 + 2H2O
(or SO42- + 4H+)
(or 2H2SO3)
Page
412
Br2 + 2e- Ý 2Br -
Eθ/V
+0.17
+1.07
We reverse the second reaction so that Br- ions go to the same reactant side as H2SO4 (the
electrode potential for the reaction then becomes -1.07V).
Adding the two equations gives the net reaction for the reduction of H 2SO4 by bromide
ions.
H2SO4 + 2H+ + 2Br -  SO2 + 2H2O + Br2
(or 2 HBr )
Eθ cell = 0.17 + (-1.07) = -0.90V ... (ii)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 412
The reaction has a negative cell potential and so will not occur when standard conditions
are employed. In this case, hot concentrated sulphuric acid is used (non-standard
conditions), so the reaction does happen, but only a small amount of Br - is converted to
Br2. The major reaction that takes place is displacement of HBr.
(i) If a dilute solution of H2SO4 is used, no bromine is produced, and no white fumes are
observed. With dilute solutions, bromide ions cannot reduce sulphuric acid, since the cell
potential of the reaction is negative. When the sulphuric acid is dilute, the concentration of
water is high, so the HBr which is formed simply dissolves in water, instead of escaping as
white fumes.
(ii) The fact that Br - reduces sulphuric acid to SO2 shows that Br - is a stronger reducing
agent than Cl-.
When hot concentrated sulphuric acid is added to NaI (aq), HI is produced by displacement, but it immediately reduces sulphuric acid to SO2, S and H2S. No white fumes of HI are observed because the HI reacts
with H2SO4 as soon as it is produced.
H2SO4 (aq) + NaI(s)  HSO4-(aq) + HI (aq) ... (i)
Some of the reactions in which HI acts as a reducing agent are shown below
H2SO4 (aq) + 2HI (aq)  I2(s) + SO2 (g) + 2H2O (l) ... (ii)
The net cell potential for this reaction is obtained using the following half equations
Eθ/V
H2SO4 + 2H+ + 2e- Ý SO2 + 2H2O
(or SO42- + 4H+)
(or 2H2SO3)
I2 + 2e- Ý 2I-
+0.17
+0.54
Reversing the second equation and its reduction potential, and adding the two equations together gives
reaction (ii) . The cell potential is then 0.17 + (-0.54) = -0.37 V.
The cell potential is still negative (because sulphuric acid is a weak oxidizing agent). Using dilute solutions
of sulphuric acid and NaI would therefore result in no reaction. However, the cell potential is more positive
than the one obtained for the reaction of NaBr with sulphuric acid. This shows that iodide ions are more
powerful than bromide ions as reducing agent.
The net equation for the reaction of concentrated sulphuric acid and NaI is sometimes obtained by adding
reactions (i) and (ii), giving.
2NaI(s) + 3H2SO4 (aq)  2NaHSO4 (aq) + I2(s) + SO2 (g) + 2H2O (l) ... (iii)
The observations would therefore be a brown precipitate of iodine, and the pungent smell of SO 2. However,
this reaction is misleading because it does not reveal that first HI is formed, which then reduces sulphuric
acid to SO2. Also, further reduction of SO2(aq) to H2S and S is possible.
Page
413
4.4.2 Redox displacement reactions involving the halide ions
A halide ion can reduce a halogen which is above it in the group. Consider what happens when chlorine
gas is bubbled into a solution containing KI (aq).
Cl2 (g) + 2I-(aq) → I2 (aq) + 2Cl- (aq)
It has already been shown that the reaction is energetically feasible since it has a positive E θ value. In this
reaction, I- acts as a reducing agent (electron donor), and Cl 2 as an oxidizing agent (electron acceptor).
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 413
Iodide, being relatively large, easily lets go of its charge (extra electron in the outer shell), and gives it to
chlorine (Fig 9.4.3). Iodide is thus oxidized, because it loses an electron. Meanwhile, chlorine, having
relatively small atoms, easily accepts an electron from iodine. Cl 2 is thus reduced, since it gains electrons
(each atom in chlorine gains one electron, therefore a total of two electrons is gained per molecule of
chlorine).
Fig 9.4.3 Diagram illustrating why the reaction between iodide ions and chlorine
is feasible. Verify that this reaction is indeed feasible by means of electrode
potentials.
Now consider what happens when iodine is added to a solution containing chloride ions.
I2 + 2Cl - → no reaction
Page
414
This time no reaction takes place. The chloride ion is small and so it has a firm hold on the extra electron
its outer shell. It will not easily give it away to iodine. Iodine atoms are large and so they will not easily
accept an electron in their outer shell.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 414
The Group (VII) elements and their compounds: uses and other useful information
Chlorine
Origin of name: Greek ‘chloros’ meaning ‘greenish yellow’ or ‘yellowish green’
Colour and state at room temperature: Greenish yellow gas, bleaches litmus paper to white.
Uses of chlorine and its compounds

Water purification (in water it forms HOCl and HCl, which kill bacteria)

Manufacture of PVC ( a polymer)

Manufacture of pesticides, solvents (such as CCl4) and aerosol propellants.

Used to oxidize Br- ions to Br2 during the extraction of bromine from sea water.

NaCl is used in food seasoning and as a food preservative.

Manufacture of mild antiseptics such as trichlorophenol, which is the active ingredient in dettolTM.
Lab preparation of chlorine
Chlorine can be produced on a small scale by the oxidation of chloride ions (from concentrated HCl) by MnO2 (Fig
9.4.4).
4
3
1
2
Fig 9.4.4
Lab preparation of chlorine gas
1. A mixture of concentrated HCl and solid MnO2 (oxidizing agent) are heated. Cl- is oxidized to Cl2 while MnO2 is
reduced to Mn2+.
Cl2 +
2e-
Ý
Eθ/V
+1.36
2Cl-
MnO2 (aq) + 4H+ (aq) + 2e– Ý Mn2+ + 2H2O(l)
net
2Cl-
+ MnO2 (aq) +
4H+
(aq) →
Mn2+
+1.23
+ 2H2O (l) + Cl2
Page
415
Eθcell for this reaction is (-1.36 + 1.23) = -0.13V
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 415
Note that the first reaction and its electrode potential have been reversed so that Cl - is on the same reactant side as
MnO2. Since the cell potential for the reaction between Cl- and MnO2 is negative, the reaction will not take place if the
concentration of Cl- ions is low. Since a high concentration of hot HCl is used, the reaction does take place.
2. Chlorine produced in flask 1 is mixed with HCl(g). When the mixture is passed through water, HCl is absorbed.
Chlorine is slightly soluble in water, but the presence of a strong acid, in this case HCl, makes chlorine almost completely
insoluble.
3. Chlorine leaving flask 2 is contaminated with water vapour. The gas is dried by passing it through concentrated
sulphuric acid, which has a high affinity for water (it is a good drying agent).
4. Dry chlorine is collected in a closed jar.
Industrial preparation of chlorine
The industrial preparation of chlorine by electrolysis of brine (concentrated NaCl) using a diaphragm cell was discussed
in Section 6.
Check to see if you still remember the details of the process
1. State the material used at the anode and the cathode.
2. Write down the equations for the reactions that occur at each electrode.
3. Why is a porous material (asbestos) used to divide the anode compartment from the cathode
compartment ?
4. Why is the level of brine kept higher in the anode compartment than in the cathode compartment?
4. What other two products are produced during this process, besides chlorine gas, and in what
ratio are they formed?
5. If the contents of the two compartments are mixed, another substance is formed. Name this
substance, and by means of a balanced equation, explain how it is formed.
Bromine
Origin of name: Greek ‘bromos’ meaning ‘stench’ due to its characteristic smell.
Colour and state at room temperature: red brown volatile liquid with a characteristic stench.
Solubility: sparingly soluble in water. Dissolves well in organic solvents such as chloroform, in which it forms a red
solution.
NB: Bromine is extremely poisonous.
Uses of bromine and its compounds
Some compounds of bromine are used as flame retardants, that is, they prevent the spreading of fires.
Manufacture of pesticides.
Page
416


Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 416
Lab preparation
The method involves oxidation of Br- ions by MnO2 in the presence of hot concentrated H2SO4 (aq), which is also an
oxidizing agent. The reaction is carried out in a distillation apparatus, so that bromine is collected as a liquid.
Industrial preparation
Bromine is obtained from sea water, which is rich in halide ions. Sea water is treated with chlorine, which oxidizes
the bromide ions to bromine in a redox displacement reaction. Bromine, which is volatile, is then driven out by
blowing air into the water.
Iodine
Origin of name: From Greek ‘iodes’ meaning ‘violet’ because of the colour of the gas.
Colour and state at room temperature: black solid, when heated, it sublimes to a purple vapour.
Solubility: poorly soluble in water, but dissolves well in organic solvents such as chloroform (CCl4) to form
beautiful purple solutions.
Uses of iodine and its compounds

A mixture of iodine and ethanol, commonly referred to as ‘tincture of iodine’, is used as a mild antiseptic
suitable for small wounds.

Its salts are useful in photography

The iodide ion is an important mineral nutrient. Deficiency of iodine in the diet leads to goiter.
Lab preparation of iodine
Same method as the preparation of bromine, that is, it involves oxidation of I- ions, for example, from NaI, by MnO2
in the presence of concentrated sulphuric acid. Solid iodine is produced, which sublimes to a gas due to the heat
applied during the process. The gas can be sublimed back to solid when it strikes a cold surface, for example, the
bottom of a flask containing cold water.
Commercial extraction of iodine
Iodine is found in the earth’s crust as sodium iodate (V), NaIO3, which is usually found mixed with Chile salpetre,
NaNO3, so named because large amounts of this compound are found in Chile.
In the production of iodine from NaIO3, IO3- ions are first reduced to iodide ions by sodium hydrogen sulphite
IO3- (aq) + 3HSO3-(aq)  I-(aq) + 3HSO4-(aq)
A reaction then occurs between IO3- and iodide ions, producing iodine
IO3- (aq) + 5I-(aq) + 6H+ (aq)  3I2(s) + 3H2O (l)
Notes for use in qualitative analysis
Tests for and properties of the halogens

Chlorine is a greenish yellow gas which bleaches litmus paper. This is in fact an oxidation process.
CAUTION: Chlorine is poisonous. Any experiments that result in the production of chlorine should be
carried out in a fume hood.
Bubbling chlorine gas into a solution containing Fe2+ results in a change of colour from light green to yellowish
brown. This is a result of the oxidation of Fe2+ (aq) to Fe3+ (aq) by chlorine
Page
417
2Fe2+ (aq) + Cl2 (g)  2Fe3+ (aq) + 2Cl-(aq)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 417



Passing chlorine gas into a solution containing KI (aq) results in formation of a brown solution, and a dark
brown deposit may also be formed. This is due to the oxidation of iodide ions to iodine by chlorine. When
formed in the presence of excess iodide ions, iodine dissolves significantly due to the formation of I3- ions. The
solution formed is brown. Any iodine which does not dissolve forms a dark brown deposit.
Liquid bromine is too poisonous for use in a school lab.
Iodine is a shiny black solid. It dissolves poorly in water but dissolves significantly in a solution that contains
iodide ions. When the solid is heated strongly, it sublimes to a purple vapour. Iodine solution can be tested for
using
(i) Starch solution. A blue-black coloration is formed due to the formation of a complex between iodine and
starch molecules.
(ii) Sodium thiosulphate solution. Iodine solution is decolourized because iodine is reduced to iodide ions.
I2 (aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
Iodine is a weak oxidizing agent, for instance, it has no effect on litmus paper, and it can not
convert Fe2+ (aq) to Fe3+ (aq).
Test for the halide ions
Halide ions usually behave in two ways in their reactions

as agents of precipitation

as complexing agents
In addition, iodide is a good reducing agent, and this often allows its detection in qualitative analysis.
Halide ions give precipitates with Pb2+ (due to the formation of the insoluble salt, PbX2, where X is the halide ion) and
with Ag+ due to the formation of insoluble AgX.
Precipitation reactions with Pb2+
A common source of Pb2+ ions in the lab is the white crystalline salt, Pb (NO3)2, which, like all nitrates, is soluble in
water.
Cl- ions form a white precipitate of PbCl2 with Pb2+. Br- forms a cream precipitate and I- a yellow precipitate. All three
precipitates dissolve when heated.
Testing for the halide ions using Ag+ ions followed by aqueous ammonia
Cl- forms a white precipitate of AgCl, which is soluble in dilute ammonia. Br - forms cream coloured AgBr, which is
sparingly soluble in dilute ammonia but dissolves in concentrated ammonia.
I- forms a yellow precipitate of AgI, which is insoluble in both dilute and concentrated ammonia.
The underlying principle
The halide ion, X- is precipitated out of solution by Ag+ ions.
Ag+ (aq) + X-(aq)  AgX(s)
The solubility of these precipitates in water is very small, that is, the equilibrium
AgX(s) Ý Ag+ (aq) + Cl-(aq) ... (i)
Salt
Ksp at 298K
AgCl
1.8 x 10-10
AgBr
7.7 x 10-13
AgI
8.3 x 10-17
Page
418
has a very small solubility product, Ksp , as shown below
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 418

Note that solubility of the silver halides decrease down the group. Compared to the three salts, AgCl is
the most soluble. In water, some AgCl dissolve, releasing free Ag+ ions. Upon addition of aqueous
ammonia, a complexation reaction takes place which removes the Ag+ ion solution.
Ag+ (aq) + 2NH3 (aq)  [Ag (NH3)2] + (aq)


This forces equilibrium (i) to the right, producing more Ag+ ions and causing more AgCl to dissolve. If
an excess of NH3 (aq) is added, all of the AgCl will dissolve, forming a colourless solution containing the
diamine (I) silver ion, [Ag (NH3)2] +.
In water, the solubility of AgBr is very small and consequently the concentration of Ag+ (aq) is minute.
Dilute ammonia will not result in any reaction with the Ag+ ions, because the chances of there being a
collision between NH3 molecules and Ag+ ions is close to zero. However, when concentrated ammonia is
used, the probability of collision between ammonia and silver ions increases. Removal of Ag+ ions from
solution will promote dissolution of AgBr.
The solubility of AgI in water is so small that no reaction occurs with dilute or concentrated ammonia.
Table 9.4.9 summarizes the common reactions and tests for the halide ions.
Ion
Test
Cl-
Observation
Forms white ppt
Add Pb2+(aq)
Br-
Forms cream ppt
I-
Forms yellow ppt
Cl-
Forms white ppt, which dissolves
upon addition of dilute NH3, to
form a colourless solution.
Br-
Forms cream ppt, insoluble in
dilute ammonia.
Forms yellow ppt, insoluble in
ammonia.
Insoluble precipitates PbX2 formed.
The precipitates dissolve when
heated.
Insoluble precipitates AgX formed.
AgCl, which is more soluble in water,
dissolves in dilute ammonia due to
formation of the complex
[Ag (NH3)2] +.
AgBr, which is less soluble in water,
does not dissolve in dilute ammonia.
However, it dissolves in
concentrated ammonia, to form a
colourless solution containing the
complex
[Ag (NH3)2] +.
AgI, whose solubility in water is
extremely low, does not dissolve in
dilute or concentrated ammonia.
Page
419
I-
Add Ag+(aq), followed
by dilute NH3
Notes
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 419
Ion
Test
Cl-
Observation
Forms white ppt
Add Pb2+(aq)
Br-
Forms cream ppt
I-
Forms yellow ppt
Cl-
Forms white ppt, which dissolves
upon addition of dilute NH3, to
form a colourless solution.
Br-
Add Ag+(aq), followed
by dilute NH3
Forms cream ppt, insoluble in
dilute ammonia.
Notes
Insoluble precipitates PbX2 formed.
The precipitates dissolve when
heated.
Insoluble precipitates AgX formed.
AgCl, which is more soluble in water,
dissolves in dilute ammonia due to
formation of the complex
[Ag (NH3)2] +.
AgBr, which is less soluble in water,
does not dissolve in dilute ammonia.
However, it dissolves in concentrated
ammonia, to form a colourless
solution containing the complex
[Ag (NH3)2] +.
AgI, whose solubility in water is
extremely low, does not dissolve in
dilute or concentrated ammonia.
I-
Forms yellow ppt, insoluble in
ammonia.
Cl-
White fumes produced.
HCl (g) displaced.
White fumes, which are soon
coloured by orange fumes. Acrid
smell detected.
HBr(g) gas displaced(white fumes)
Sulphuric acid reduced to SO2(acrid
smell)
Br- oxidized to Br2 (orange fumes)
Purple vapour produced, acrid
smell produced. Smell of rotten
eggs may also be detected, and a
yellow ppt may form.
Iodide oxidized to iodine (purple
vapour). Sulphuric acid reduced to
SO2 , H2S and S.
Br-
To concentrated
solutions of the anions,
add hot concentrated
sulphuric acid
I-
Cl-
No change.
Solution turns orange or red
brown, a greenish yellow gas is
evolved
Redox displacement occurs; Br- is
oxidized to Br2, whilst chlorine is
reduced to chloride.
2Br- + Cl2  2Cl- + Br2
Page
420
Br-
Pass chlorine gas into a
solution containing the
anion.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 420
Ion
I-
Test
Pass chlorine gas into a
solution containing the
anion.
ClBr-
Observation
Dark brown deposit or brown
solution formed
Notes
Redox displacement occurs; I- is
oxidized to I2, whilst chlorine is
reduced to chloride.
No change
Add bromine solution
No change
I-
Dark brown precipitate or brown
solution formed, bromine
decolourized
Cl-
No change
Redox displacement occurs; I- is
oxidized to I2, whilst bromine is
reduced to bromide.
Redox displacement not possible.
Br-
Add iodine solution
No change
I-
No change
Cl-
No change, yellow brown colour of
Fe3+ (aq) observed.
Br-
Fe3+(aq),
Add dilute
followed by starch
solution
I-
Cl-
Br-
No change, yellow brown colour of
Fe3+ (aq) observed.
Brown solution formed
I- is powerful enough to reduce Fe3+
to Fe2+. Meanwhile, the iodide ions
are oxidized to iodine (brown).
2I- + 2Fe3+  I2 + 2Fe2+
White ppt formed, turns purple
grey in sunlight.
White AgCl precipitated. AgCl
undergoes superficial(surface)
photochemical decomposition in light
AgCl(s)  Ag(s) + ½ Cl2(g)
Cream ppt, turns green yellow in
sunlight.
Photochemical decomposition of
AgBr * takes place, as for AgCl.
No effect
Page
421
I-
Add Ag+ (aq), filter any
solid formed. Leave the
solid in sunlight for a
few minutes
Cl- and Br- are too weak as reducing
agents to convert Fe3+ to Fe2+.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 421
Ion
Test
Observation
Notes
No change
Cl- and Br- are not powerful enough
to reduce hydrogen peroxide.
Solution changes from colourless
to brown
Iodide ions reduce hydrogen
peroxide. Iodine is formed.
ClBr-
I-
Add dilute hydrogen
peroxide
H2O2 (aq) + 2I- (aq) + 2H+ (aq) → I2
(aq) + 2H2O
Cl-
No change
Redox displacement not possible.
*The photochemical decomposition of AgBr is used in black and white photography. AgBr on photographic film is
Add iodine solution
Br
No
change
decomposed by light to Ag(s). Sodium thiosulphate, Na2S2O3 (referred to as ‘hypo’ in photography) is used during
development of the film. The S2O32- ions remove excess AgBr from the film by formation of a soluble complex, [Ag
I-3)2]3-. This leaves an opaque shadow
No change
(S2O
of silver metal on the film.
The halic ions, XO- and XO3Clare
No change,
yellow
brown
of (I) ion and ClO3 (Chloric (V) ion. These
-(chloric
You
now aware of the existence of halic
ions, for
example
ClOcolour
3+(aq), followed
3+ (aq) observed.
Add
Fe
Fe
Cl-substances.
and Br- are too weak as reducing
ions are good oxidizing agents. It whitens fabric by oxidizing dirt to colourless
3+ to Fe2+.
by
starch
solution
agents
to convert
Fehalogen
Br
Recall that the halic (I) ions such as ClO are prepared by the disproportionation reaction
of the
with cold
No
change,
yellow
brown
colour
of
dilute NaOH (aq). When hot concentrated NaOH (aq) is used, the halic (I) ion undergoes further oxidation, forming
Fe3+ (aq) observed.
the halic (V) ion, XO3-.
IO is unstable with respect to IO3 . It is therefore difficult to prepare the iodate ion, IO-(also known as the iodic (I)
3+
Brown
I- is powerful
enough towith
reduce
ion)I because it is immediately converted
to IOsolution
under ice cold conditions.
Of the oxyanions
theFe
formula
3-, even formed
2+. Meanwhile, the iodide ions
to
Fe
XO , ClO is the most stable. Of the oxyanions with the formula XO3 , IO3 is the most stable.
are oxidized and
to iodine
(brown).
In the presence of a suitable reducing agent and acid, the halic (V) ions are decomposed
reduced
to the halogen,
- + 2Fe3+  I2 + 2Fe2+
2I
for example
2ClO3-(aq) + 12H+ (aq) + 10e-  Cl2 (g) + 6H2O (l)
All three halic (V) ions are therefore powerful oxidizing agents, that is, they are easily reduced. This is the basis of
- ions, for example, IO3-(aq) oxidizes I-(aq) according to the equation
the
between
XO3-filter
and X
Cl-reactionAdd
Ag+ (aq),
any
White ppt formed, turns purple
White AgCl precipitated. AgCl
+ (aq)
formed.
Leave
the
grey
in sunlight.
undergoes superficial(surface)
IOsolid
+ 5I–(aq)
+ 6H
 3H
3–(aq)
2O (l) + 3I2 (aq)
solid
in
sunlight
for
a
photochemical
in light by
There is a change of colour from colourless to brown. This reaction (whose
equation youdecomposition
can easily construct
few
minutes
AgCl(s)

Ag(s)
+
½
Cl
2(g)
combining relevant half equations from the Data Booklet) can be used to determine the concentration of IO3– in a
given solution. The procedure involves back titration of the iodine
Brppt,
turns green yellow in
produced
with sodium thiosulphate of Cream
a known
concentration.
sunlight.
I2(s) + S2O32-(aq)  2I-(aq) + S4O62-(aq)
No effect
Page
422
I-
Photochemical decomposition AgBr *
takes place, as for AgCl.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 422
Questions, solutions and discussions
Q1
This question is about the elements of Group VII, the halogens.
(a) Complete the following table.
Halogen
colour
Physical state at room
temperature
chlorine
bromine
iodine
(b) Concentrated sulphuric acid is added to separate solid samples of magnesium chloride,
magnesium bromide, and magnesium iodide.
(i) Describe, in each case, one observation you would be able to make.
(ii) Give an equation for the reaction of concentrated sulphuric acid with magnesium
chloride.
(c) When dilute nitric acid and aqueous silver nitrate are added to a solution of a magnesium
halide, MgX2, a pale cream precipitate is formed. This precipitate is soluble in concentrated aqueous
ammonia but not soluble in dilute aqueous ammonia.
(i) What is the identity of the precipitate?
(ii) Give an equation, with state symbols, for the reaction of the precipitate with
concentrated aqueous ammonia.
(d) A hot glass rod is plunged into separate gas jars, one containing hydrogen chloride and
one containing hydrogen iodide.
(i) For each gas, state what you would observe, if anything, and write an equation for any reaction
that takes place.
(ii) Explain your answer to (i) in terms of enthalpy changes.
Page
423
(iii) What is the role of the hot glass rod in any reaction that occurs?
9701/02/M/J/06
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 423
Solutions
(a) Left for the reader
(b) (i) MgCl2 : white fumes (of HCl)
MgBr2 : red colour (of Br2) and white fumes (of HBr)
MgI2 : purple fumes (of I2 gas) / black solid (I2) / yellow solid (S) /stinking gas (H2S)
(ii) MgCl2 + H2SO4 → MgSO4 + 2HCl / MgCl2 + 2H2SO4 → Mg(HSO4)2 + 2HCl
(c) (i) AgBr
(ii) AgBr(s) + 2NH3(aq) → Ag(NH3)2Br(aq)
(d) (i) HCl: no reaction
HI : purple vapour, 2HI → H2 + I2
(ii) The H- Cl bond energy is high but the H-I bond is weaker and is more easily broken .
(iii) Provides activation energy
Q2
The elements in Group (VII) show trends in a number of their physical and chemical properties.
(a) Explain the trend in boiling points of the elements as shown.
element
Boiling point/0C
chlorine
- 35
bromine
+59
iodine
+184
(b) There are also clear trends in both the H-X bond energy and the Eθ values for the cell
X2(aq)/X -(aq), where X = Cl, Br, I.
Use relevant data from the Data Booklet to describe and explain each of the following.
Give an equation for each reaction that occurs.
(i) The reactions of the halide ions with concentrated sulphuric acid.
9701/04/SP2007
Page
424
(ii) The action of heat on the hydrogen halides
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 424
Solutions
(a) Strength of Van der Waals forces increases down the group with increasing number of electrons.
Stronger temporary dipoles are established and so the force of attraction between molecules
becomes stronger.
(b) (i) Reaction of sulphuric acid with the halide ions involves oxidation and displacement of a volatile
acid, HX, where X = Cl, Br, I. The following E 0/V are relevant in explaining the ability of the
halide ion to reduce sulphuric acid, that is, to be oxidized by sulphuric acid to form the halogen
X2 .
½ Cl2 + e-  Cl-
Eθ/V = +1.36
½ Br2 + e-  Br-
Eθ/V = +1.07
½ I2 + e-  I-
Eθ/V = +0.54
The Eθ values become less positive down the group, indicating that the halogens become stronger
reducing agents. Thus Cl- is too weak as a reducing agent to reduce sulphuric acid. The only
reaction that takes place is the displacement of HCl (white fumes).
H2SO4 (aq) + Cl-(aq)  HCl(g) + HSO4-(aq).
Br- and I- are powerful enough to reduce sulphuric acid to SO2, according to the equation.
2X- (aq) + 3H2SO4 (aq)  2HSO4- (aq) + X2(s) + SO2 (g) + 2H2O (l)
For Br- , which is a weaker reducing agent than I-, white fumes of HBr are also formed as a
result of displacement. I- is powerful enough to continue reducing sulphuric acid until low
oxidation states (H2S and S) are obtained. When this happens, a foul smell like that of rotting
eggs is detected due to H2S and a yellow deposit of sulphur is formed.
(ii) The H-X bond energy decreases down the group.
Bond
Bond energy/
Kjmol-1
H-Cl
431
H-Br
366
H-I
299
Thus HI has the weakest bond and it is easily broken by heat, resulting in decomposition
Copious violet fumes of iodine are produced when a red hot needle is plunged into a jar
containing HI.
The H-Br bond is stronger and brown fumes of Br2 are only produced upon very strong heating.
The H-Cl bond is the strongest. HCl therefore resists decomposition by heat.
Page
425
2HI (g)  H2 (g) + I2 (g)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 425
Q3
Chlorine is a yellow-green gas whilst iodine is a black solid.
With the aid of chemical equations compare and contrast the chemistry of chlorine and iodine by
describing and explaining the following:
(a) Reaction of the two elements with sodium thiosulphate
(b) Reactions of solid halides of the two elements with concentrated sulphuric acid
(c) Reactions of the two elements with hydrogen
9189/01/O/N/ 2010
Solution
(a) The halogens act as oxidizing agents. Oxidizing power decreases down the group. Chlorine is therefore
more powerful than iodine, as exemplified by the reaction of the two elements with sodium
thiosulphate, Na2S2O3. Chlorine is powerful enough to oxidize thiosulphate all the way to the +6 state
of sulphur in SO42-.
4Cl2 (aq) + S2O32-(aq) + 5H2O (l)  8Cl- (aq) + 2SO42- (aq) + 10H+ (aq)
Iodine is a much weaker oxidizing agent, and it can only oxidize S 2O3 to the tetrathionyl ion, S4O62- ,
in which the the oxidation state of sulphur is 2.5.
I2(aq) + S2O32- (aq)  2I- (aq) + S4O62- (aq)
(b) See question 2
(c) Both chlorine and hydrogen react with hydrogen in a redox reaction
X2 (g) + H2(g)  2HX.
However, chlorine, being a powerful oxidizing agent, reacts explosively and completely with hydrogen.
The reaction of hydrogen with iodine is slow and reversible.
Q4 (a)
Chlorine is produced by the electrolysis of brine. It has found use in water purification.
(i) Write an equation for the reaction between chlorine and
1. Cold NaOH(aq)
2. Hot NaOH(aq)
(ii) Explain the use of chlorine in water purification
Chlorine can also be used as bleaching agent.
Use chemical equations to explain the role of water in the bleaching action of chlorine.
9189/02/M/J/ 2011
Page
426
(b)
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 426
Solution
(a)(i) Please refer to main text
(ii) Chlorine reacts with water according to the equation
Cl2 + H2O Ý 2H+ + OCl- + ClOCl- is a good oxidizing agent. It oxidizes organic matter in water, including bacteria. Organic food
molecules are also oxidized and this depletes the food reserves on which bacteria feed. The
reaction also produces H+ ions which make the water acidic, leading to death of bacteria.
(b)
Chlorine reacts with water to form the oxidizing agent ClO-. Bleaching involves the oxidation of
dirt and dyes by ClO-, leading to whitening of materials, for example clothes.
Q5
Bromine has been obtained from sea salt for a number of years. For simplification, sea salt can be
considered to be sodium chloride containing some sodium bromide.
One early method of obtaining bromine is outlined in the following scheme.
sea salt
chlorine
intense
yellow
colour
A
separate with ether
B
yellow
solution in
ether
cold aqueous
potassium
C
hydroxide
ether
recycled
red fumes which
condense to a red
liquid ( bromine)
warm with acid
and distil
D
two
colourless
layers
white
paste
aqueous layer is
concentrated
(a) (i) What causes the yellow colour produced in A?
(ii) Write an equation for the reaction in A.
(b) Ether is an organic solvent immiscible with water. Explain why the yellow substance dissolves in the
ether in B but the salt does not dissolve in the ether.
(d) Suggest an equation for reaction D, the reaction of acid on the products of reaction.
9701/02/O/N/1998
Page
427
(c) Suggest an equation for reaction C.
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 427
Solution
(a)(i) Bromine (the solution is yellow because it is very dilute).
(ii) 2Br-(aq) + Cl2(aq)  Br2(aq) + 2Cl-(aq) [Recall that this is a redox displacement reaction]
(b) Bromine, being non-polar, easily dissolves in ether which is also non-polar. This is because the
dissolving process only requires the breaking of weak Van der Waals forces in the solvent (ether) and
solute (bromine). On the other hand, the salt is ionic and the solvent is not able to break the strong
forces of attraction between the ions of the salt .
(c) Br2 + 2OH-  Br- + BrO- + H2O [This is the disproportionation of a halogen in an alkali]
(d) Br- (aq) + BrO- (aq) + 2H+(aq)  Br2(aq) + H2O(l)
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise
9.3
text box.]
1(a) (i) State three different substances in everyday or industrial use which contain chlorine.
(ii) For each substance, state one use and explain how the use depends on a relevant property of
the substance.
(b)
Comment on any problems which arise from the disposal of materials which contain chlorine
9701/02/O/N/2000
2(a) Potassium chlorate (V), KClO3, is widely used in fireworks and match heads. It decomposes when
heated with a catalyst to form potassium chloride and oxygen.
Construct an equation for the decomposition of potassium chlorate (V).
(b)(i) Suggest why potassium chlorate (V) is used in fireworks and match heads.
Page
428
(ii) Suggest one substance which could be used as the combustible material in fireworks or match
heads.
(c)
How can potassium chlorate (V) be obtained from chlorine? Illustrate your answer with an
equation.
(d)
The standard redox potential of chlorate (V) ions is given below.
ClO3- + 6H+ + 6e- Ý Cl- + 3H2O
Eθ = + 1.45 V
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 428
Use the Data Booklet to predict what you would expect to observe when acidified potassium
chlorate (V) is added separately to each of the following reagents. Write a balanced equation for any
reaction that occurs.
(i) aqueous iron (II) sulphate.
(ii) acidified potassium manganate
9701/02/O/N/1997
Page
429
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript © L. MWANAWENYU 2011. Please do not photocopy or reproduce
Page 429
430
9.5 NITROGEN AND SULPHUR
9.5.1
Nitrogen
Nitrogen, a Group (V) element, has the electronic configuration 1s²2s²2p³. The valence shell is made up of
two sub shells, the 2s and the 2p (used in bonding). The total number of valence electrons is therefore five.
The electronic configuration of nitrogen has important consequences.
 Nitrogen uses three 2p electrons to bond with another nitrogen atom. This is because these three
electrons are unpaired. An atom usually makes use of its unpaired electrons in bonding. The N to
N bond is therefore a triple bond, since it is formed by the overlap of three singly occupied atomic
orbitals from each atom. Remember that in a multiple bond (double or triple), only one can be
sigma. The rest are pi. The 2s electrons, being paired, are not used in the formation of covalent
bonds in the nitrogen molecule. They remain as a lone pair on each atom.

overlap

of
 orbitals
1s


2
2s2
px p y pz

p x py
2p
2p
N
N
pz
N  N
The three unpaired electrons in a N atom can also be used to make three bonds with other atoms,
for example hydrogen (as in NH3) and chlorine (as in NCl3).
The paired 2s electrons can also be used in dative bond formation, for example in the ammonium
ion and in BF3.NH3 . Fig 9.5.1 shows formation of a dative bond between ammonia and a proton,
resulting in the formation of the ammonium ion.

H+
H
N
H
N
H
H
NH3
trigonal pyramidal
bond angle 1070
H
H H
NH4+
ammonium ion
bond angle 1090
The maximum oxidation number nitrogen
can achieve in its compounds is +5,
corresponding to the participation of all
five valence electrons. Note that N can only
form a maximum of four bonds, for
example as in NH4+. It cannot form more
than four bonds because it has no dorbitals to accept extra electron density,
that is, it cannot expand its octet.
Fig 9.5.1
Other Group (V) elements are able to expand the octet, for example phosphorous forms PCl 5.
Phosphorous and the rest of the Group (V) elements have accessible d – orbitals which allow
formation of more than four bonds.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
431

With five electrons in the outer most shell, nitrogen can form a negative ion by accepting three
electrons. This ion, known as the nitride ion, N3-, exists in ionic compounds such as magnesium
nitride, Mg3N2.
1.1 Occurrence of nitrogen
Nitrogen is a diatomic element. It is a colourless and odourless gas which occupies about 79% of the
atmosphere by volume. It is the most abundant gas in the atmosphere. Nitrogen is also present in a large
number of compounds. It is one of the major elements present in biological molecules, for example proteins.
Nitrogen is obtained industrially from the distillation of liquid air. Its direct uses include
 to provide an inert atmosphere in reactions or processes that are sensitive to oxygen, for example
during the production of transistors
 Refrigerant
 Annealing stainless steel
 in oil extraction to force up oil under pressure
1.2 Ideality
Ideality is the extent to which a particular gas satisfies the properties expected of a gas in general.
At ordinary temperatures and pressures, N₂ closely resembles an ideal gas. This is because the forces of
attraction between the molecules are very weak Van der Waals forces. Because of this, forces of attraction
between the molecules are very weak and collisions are almost perfectly elastic, because when molecules
collide, they do not stick to each other. Also, since the molecules have very little tendency to stick to each
other, they are spread out so that the volume of a single molecule is negligible compared to the volume of
the container.
Nitrogen, like all gases, deviates from ideality at low temperatures and high pressures. When a sample of
Nitrogen is cooled, its particles come closer together so that the forces of attraction between the molecules
are stronger. Also, kinetic energy of the molecules decreases so that most of the times the molecules do not
move away from each other but remain associated by intermolecular attractions. At a sufficiently low
temperature, the molecules come so close that the sample becomes liquid. This happens at -2830C. Thus
nitrogen can be obtained from air by fractional distillation of the liquid.
1.3
Preparation of nitrogen in the lab
Nitrogen can be prepared at a small scale by the thermal decomposition of sodium azide, NaN3.
2 NaN3(s) → 2 Na(s) + 3 N2(s)
(Care is needed here since sodium azide is extremely poisonous. The sodium formed can also pose a fire
risk).
This reaction is the basis of the safety air bags found in some cars. When an impact occurs, a sensor causes
a certain chemical to burn. The heat released results in the rapid decomposition of sodium azide. The bag
is rapidly filled with nitrogen gas and provides a cushion so that the driver or passenger does not hit against
hard surfaces such as the steering wheel. The processes that occur are so fast that the bag is completely
filled with nitrogen approximately 5o milliseconds after a crash.
Q
What is the oxidation state of nitrogen in sodium azide?
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
432
1.4
The Haber process for the manufacture of ammonia
Ammonia is commercially manufactured through the Haber process, which involves the equilibrium
N₂ (g) +3H₂ (g) ⇌ 2 NH₃ (g)
... ∆Hr = -94Kj
The operating conditions are
Pressure: 200 atmospheres
Temperature: between 450⁰C and 500⁰C
Catalyst: finely divided iron or powdered iron (III) oxide, Fe2O3
Justification for pressure used in the Haber process
Formation of ammonia is accompanied by a decrease in number of moles of gas. By Le Chatelier’s principle,
a high pressure should be used because this favours the side with the smaller number of moles, that is, more
ammonia would be formed. This explains why the Haber process employs a high pressure of 200 atm.
Justification for temperature used in the Haber process
The forward reaction, that is, formation of ammonia, is exothermic. To obtain a high yield of ammonia, low
temperatures would have to be used. By Le Chatelier’s principle, lowering temperature favours the
exothermic side of the equilibrium, that is, formation of ammonia. In this way, the constraint of cooling is
removed by the release of heat energy.
However, in practice, the temperature used is not low, but is moderate. This is because low temperatures
would increase the time required to attain the equilibrium yield of ammonia, that is, it reduces the rate of
reaction. To make the reaction fast, high temperatures would be required, but by Le Chatelier’s principle,
this would decrease the equilibrium yield of ammonia. The actual temperature used (about 450⁰C), is
therefore a compromise between high and low temperature.
Justification for the use of a catalyst
One way to make a reaction fast without using a catalyst is to carry it out at very high temperatures. In the
Haber process, it has already been mentioned that a very high temperature would favour the reverse
reaction which is endothermic, thus reducing the yield of ammonia. Another problem associated with the
use of very high temperatures is the high cost of fuel that would be incurred. The process also becomes
expensive in terms of constant maintenance caused by the weakening of pipes by heat.
The temperature used in the Haber process is therefore relatively low, and the reaction would be very slow
without a catalyst. A catalyst makes the reaction faster under the relatively low operating temperature.
The catalyst used in the Haber process is finely divided iron, containing very fine particles to increase
surface area.
The conditions which are employed give a yield of only 15%. However, any unreacted gases are recycled to
give an overall yield of about 98%.
Fig 9.5.2 is a flow chart which summarizes the key stages in the Haber process.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
433
Fig 9.5.2




Flow chart of the Haber process
N2 and H2 are cleaned and mixed in the ratio 1: 3. The mixture of gases is heated in a heat
exchanger.
The converter converts N2 and H2 into ammonia. The converter has several beds containing finely
divided iron or powdered Fe2O3 as catalyst. Several catalytic beds are used to increase the yield of
ammonia. Any gases which are not converted in the first beds are passed onto the next beds.
Gases emerging from the converter are ammonia and unreacted N 2 and H2. These gases contain a
lot of heat energy, so they are passed back to the heat exchanger to recover the heat and thus
minimize wastage of energy.
The gaseous mixture is passed into a condenser where it is cooled. NH3, which is the least ideal gas
in the mixture, easily liquefies since the molecules attract each other through hydrogen bonding.
The liquid ammonia is removed whilst unreacted N2 and H2 gases are recycled.
About Haber and the Haber process
In 1908, Fritz Haber, a young German research chemist discovered that heating nitrogen and hydrogen gases
at 600 atm using a suitable catalyst formed an equilibrium mixture containing ammonia. Prior to this
discovery, nitrogen from the atmosphere had no important uses in industrial chemical reactions because of its
inertness. Haber then made a prototype plant which could produce small amounts of ammonia. Haber sold his
rights to the ammonia process to BASF, a giant German manufacturing company. Carl Botsch, working for
BASF, found a method to scale up Haber’s table top ‘plant’ into a giant plant that could produce thousands of
tonnes of ammonia per year. The Haber process was also fine tuned so that it could produce satisfactory yields
at relatively low pressures and temperatures. Both Haber and Botsch won the Nobel Prizes for chemistry in
1918 and 1931 respectively (The Haber process is sometimes referred to as the Haber - Botsch process).
The production of ammonia made an interesting contribution to First World War. Now German did not have
to depend on other countries for food, since it could now greatly enhance the agricultural sector through the
manufacturing of nitrogen fertilizers. It also meant that Germany could now more than double its war effects
by producing explosives which it could not buy from other countries.
Source of hydrogen used in the Haber process
One source is the catalyzed reaction between methane and steam,
CH₄ (g) + H₂O (g)  CO (g) + H₂ (g)
The carbon monoxide produced can poison the catalyst, so it has to be removed from the hydrogen. Production
of of
hydrogen
by this method uses less energy than the electrolysis of water which was used by the BASF
Uses
ammonia
Company.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
434




Manufacture of nitrogen fertilizers such as ammonium nitrate.
As a household disinfectant. Ammonia has an antibacterial effect.
As a refrigerant
Manufacture of explosives. The ammonia is first catalytically oxidized to nitric acid. Nitric acid is
then used in the synthesis of explosives such as TNT.
The key stages in the industrial synthesis of nitric acid
1. Gaseous ammonia is oxidized in oxygen at 900⁰C and in the presence of platinum or rhodium
(or a mixture of the two metals) .
4NH₃ (g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O (g) ... ∆H = -950Kj
2. NO is then oxidized to NO₂ by adding more oxygen
2NO (g) + O₂ (g) → 2NO₂ (g) ... ∆H = -114Kj
3. NO₂ then reacts with water to form nitric acid.
3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g) ... ∆H = -117Kj.
The process yields about 60% nitric acid. The concentration of the acid is raised by fractional
distillation.
1.5 Chemical reactions of nitrogen
Nitrogen is relatively inert because of the large energy of the N to N triple bond. A large amount of energy
is needed to break this bond during chemical reactions. Nitrogen only reacts at high temperatures since the
activation energy required to start the reactions is overcome by supplying heat energy.
The formation of ammonia by the Haber process has already been discussed. The use of a catalyst in this
process allows moderate temperatures to be employed (450 to 500 θC). Other processes which involve
nitrogen as a reactant include
 The formation of nitrogen dioxide and nitrogen monoxide (NO x) in the internal combustion engine.
 The formation of NOx in the atmosphere during lightning flashes.
 Formation of magnesium nitride (Mg3N2) together with magnesium oxide during the combustion
of magnesium metal.
 Nitrogen fixing bacteria in soil have an enzyme which converts nitrogen into soluble inorganic
compounds. This process is known as nitrogen fixation.
1.6 Some important compounds of nitrogen
1.6.1
Ammonia
Ammonia is an important chemical manufactured by the Haber process. Its uses include
 Manufacture of nitric acid, whose uses include manufacture of nitrogen fertilizers (for example
ammonium nitrate), manufacture of explosives such as trinitrotoluene (TNT) and the
manufacture of organic dyes.
 As a refrigerant
 As a household disinfectant
Properties of ammonia
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
435
Ammonia is a gas with a characteristic pungent odour. It dissolves easily in water, forming a weakly alkaline
solution. When ammonia is added to water, two things happen :
 Some of the ammonia molecules associate with water molecules through hydrogen bonding. This
is a physical process.
hydrogen bonds
N
H H
H
O
H
O
H
H
H

A smaller proportion of the ammonia molecules react with water in a Brønsted Lowry
neutralization reaction, to form a solution containing NH4+ and OH- ions.
NH3 (aq) + H2O (l) Ý NH4+ (aq) + OH-(aq) ... (i)
These two processes help to explain why ammonia is soluble in water. The solution formed can be thought
of as ammonium hydroxide, NH4OH . However, this compound cannot be separated from the mixture. An
attempt to separate NH4OH by evaporation results in equilibrium (i) shifting to the left, resulting in
ammonia gas being driven out. Equilibrium (i) lies well to the left, that is, ammonia dissolves poorly in
water. The reaction involves Brønsted Lowry neutralization in which ammonia acts as base (proton
acceptor), and water as an acid (proton donor). Like all Brønsted bases, ammonia contains a lone pair which
it uses to form a dative bond with an H⁺ ion from the acid.
Precipitation of metal hydroxides by dilute aqueous ammonia
Dilute ammonia can precipitate some metal ions out of solution, for example
Fe2+ (aq)
light green
solution
NH3(aq)
Fe(OH)2 (s)
dirty green
precipitate
The hydroxide ions which precipitate the metal come from the reaction of ammonia with water.
In some cases, the concentration of the hydroxide ions in aqueous ammonia is not large enough to cause
precipitation of the metal hydroxide, for example, calcium ions may not be precipitated in this way.
Ammonia as a complexing agent
Ammonia is able to react with transition metal ions to form complexes. In these reactions, ammonia acts
as a ligand. A ligand is an electron rich species which is attracted and forms a dative bond to a metal cation.
Transition metals are able to form complex ions because of the presence of partially filled d orbitals to accept
the electron density from the ligands. When aqueous ammonia is added gradually to a blue solution
containing copper (II) ions, a light blue precipitate is formed which dissolves in excess ammonia to form a
deep blue solution. The initial reaction is the precipitation of copper (II) hydroxide by the hydroxide ions
present in dilute ammonia. When the ammonia solution is added in excess, ammonia ligands replace the
OH groups forming a soluble complex with copper (II) ions.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
436
Cu2+ (aq) + 2OH- (aq)  Cu(OH)2(s)
light blue
from NH4OH
light blue ppt
[Cu(NH3)4(H2O)2]2+ (aq)
xs NH3 (aq)
deep blue solution
solution
Reactions of ammonia with acids
Ammonia reacts with acids to form ammonium salts. In these reactions, NH 3 uses its lone pair to form a
dative bond to an electron deficient H+ ion, which contains a vacant 1s orbital.
H+
vacant 1s orbital
+
H
N
H H
N
H
ammonia



1.6.2
H
H
H
ammonium
Ammonia is trigonal pyramidal, with a bond angle of 107 0. This corresponds to three bond pairs
and one lone pair. The ammonium ion has four bond pairs and no lone pairs. Its shape is therefore
tetrahedral, with a bond angle of 1090.
The fourth bond in the ammonium ion is a dative (co-ordinate bond). That is, only one atom
provides both electrons of the bond.
In the reaction above, ammonia acts as a base because it accepts a proton. The ammonium ion is
acidic because it can release a hydrogen ion, forming ammonia (reverse reaction). NH3 and NH4+
therefore constitute a conjugate acid - base pair.
Oxides of nitrogen
Nitrogen dioxide and nitrogen monoxide (commonly referred to as NOx) have been associated with
environmental processes that lead to atmospheric pollution and depletion of the ozone layer. Sources of
NOx are both natural and attributable to human activities.
 Lightning flashes provide the activation energy required to break the very strong triple bond in
nitrogen and initiate the reaction with oxygen.
 Exhaust fumes from the internal combustion engine contain nitrogen monoxide and nitrogen
dioxide. The very high temperatures that develop in the engine provide the activation energy for
the oxidation of atmospheric nitrogen by oxygen.
N2 (g) + O2 (g)  2NO (g)
NO is produced in greater amounts than NO2. This is because there is a limited amount of
oxygen in the internal combustion engine. However, once it enters the atmosphere, NO is further
oxidized to NO2.
NO2 (g) + ½ O2 (g)  NO2 (g)
NOx has been blamed for environmental problems such as formation of acid rain and photochemical smog,
and for the depletion of the ozone layer.
NOx act as catalyst for the processes that form acid rain, for example, NO2 may participate in the following
reaction scheme.
NO2 (g) + SO2 (g)  NO (g) + SO3 (g) ... (i)
NO (g) + ½ O2 (g)  NO2 (g) ... (ii)
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
437
In reaction (i), NO2 is reduced by sulphur dioxide, which is in turn oxidized to SO 3. In step (ii), NO2 is
recovered when NO reacts with atmospheric oxygen. The NO 2 produced in (ii) is recycled back to step (i),
where it converts more SO2 to SO3.
 The formation of SO3 is a significant step in the formation of acid rain. NO2 and SO2 are only weakly
soluble in water, so their direct contribution to acid rain is slight. SO 3 has a very high
solubility in water, in which it forms a solution of sulphuric acid
SO3 (g) + H2O (l)  H2SO3 (aq)

Note that this is not a redox reaction. It is simply conversion of an acid anhydride (SO3) to an acid
hydride (H2SO4). Acid rain has adverse environmental effects, including destruction of vegetation,
buildings, metal structures and the killing of aquatic animals. Acid rain also mobilizes toxic ions in
soil, for example, Al3+ ions can be released when acid rain dissolves aluminium oxide present in the
soil. These ions can be carried into water bodies resulting in the death of aquatic organisms such as
fish.
The ability of NO and NO2 to participate in atmospheric processes is a result of their electronic
structures. Both molecules are free radicals, that is, species which contain an odd total number of
valence electrons. NO contains 11 valence electrons (5 from nitrogen + 6 from oxygen), whilst NO 2
contains 17. The result is that one electron in the outer shell is inevitably unpaired. This odd
electron causes the molecules to be very reactive (that is, to be unstable) as they seek to attain
electronic states in which all valence electrons are paired. NO is one of the chief culprits in the
destruction of the ozone layer, according to the equation
NO (g) + O3 (g)  NO2 (g) + O2 (g).
Halogen atoms, for example, Cl, which are also free radicals, also contribute to the depletion of
the ozone layer.
Preventing NOx pollution
Emission of NOx and other pollutants from the internal combustion engine can be minimized by fitting a
catalytic convertor into the exhaust pipe of the vehicle (Fig 9.6.4). The catalytic convertor converts NO,
NO2, CO, unburnt carbon (soot) and unburnt hydrocarbons to products which are harmless.
Fig 9.6.4
The catalytic convertor
CO, soot (carbon) and hydrocarbons (HC) are oxidized to carbon dioxide and water.
C(s) + O2 (g) → CO2 (g)
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
438
2CO(s) + O2 (g) → 2CO2 (g)
CxHy (g) + (x + y/4) O2 → xCO2 (g) + y/2 H2O
Meanwhile, NO and NO2 are reduced to nitrogen gas by combining with CO.
2NO (g) + 2CO (g) → N2 (g) + 2CO2
This reaction therefore removes two pollutants at once.
The convertor makes use of a mixture of expensive metals (platinum, palladium and rhodium) coated on
the surface of a honeycomb structure to maximize surface area. The catalysis involved is therefore
heterogeneous.
 The catalytic convertor has not yet gained popularity in developing countries because it is expensive
(it uses three of the most expensive metals)
 The convertor is poisoned by lead. Cars fitted with catalytic convertor must run on unleaded petrol.
(In fact, the use of leaded petrol has greatly declined because of the pollution problems associated
with lead compounds released from the internal combustion engine)
1.6.3 Nitric acid
This is a very important industrial chemical prepared by the catalytic oxidation of ammonia. The overall
reaction is
NH3 (g) + 2O2 (g)  HNO3 (aq) + H2O (l)
The process is carried out at 9000C, using a platinum or rhodium (or a mixture of both) catalyst. Some uses
of HNO3 have already been outlined. In addition, HNO 3 acid is used to acidify solutions where other acids
would result in precipitation of some ions, for example, sulphuric acid may cause precipitation of Ba 2+ and
Pb2+ as BaSO4 and PbSO4 respectively.
Pure nitric acid is colourless but it soon attains a yellowish taint due to formation of NO 2 by decomposition.
Dilute HNO3 shows the usual chemical properties expected of a strong acid. However, concentrated HNO3
is also an oxidizing agent. In fact, it is strong enough to oxidize copper metal to copper (II). HNO 3 is itself
reduced to NO2 (red brown fumes).
Cu(s) + 4HNO3 (aq)  Cu (NO3)2(aq) + 2NO2 (g) + 2H2O (l)
At the beginning of the experiment, red brown fumes with a disagreeable and choking odour are produced
(Fig 9.65). This gas is NO2, which is poisonous; the experiment should be carried out in a fume hood.
Fig 9.6.5
metal.
Prepublication manuscript
The action of concentrated HNO3 on copper
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
439
A greenish solution is formed, which contains a soluble complex of Cu2+ and nitrate ions.
The concentration of HNO3 soon decreases, and so does the production of NO 2. In fact, at low concentrations of HNO3, NO is produced instead of NO2. The greenish solution is converted to a light blue solution
of copper (II) nitrate by the addition of more water (dilution).
1.7 Organic nitrogen fertilizers
Nitrogen in the form of soluble nitrate ions is an important nutrient required for normal growth of plants.
Common nitrogen based fertilizers include ammonium sulphate, ammonium nitrate and urea
(CO(NH2)2). All three fertilizers are made from the neutralization of concentrated ammonia by an acid.
NH3 + H2SO4  (NH4)2SO4
NH3 + HNO3  NH4NO3
2 NH3 + CO2 (g)  CO(NH2)2 + H2O
NH4NO3 has the following advantages over the other two fertilizers


It dissolves well in soil water, thus immediately releasing nutrients to plants.
It has a greater content of nitrogen.
Its major disadvantages are




It is explosive
It easily absorbs moisture to form clumps
The two factors above make it difficult to store
Because of its high solubility, it may dry up crops by withdrawing water from tissues (through
osmosis).
It does not stay for a long time in the soil.

The use of nitrogen fertilizers has been associated with the pollution of water bodies. Eutrophication is a
form of water pollution in which there is a sudden bloom of water plants and algae due to increased levels
of nitrate in water. This proliferation of plants and algae cause depletion of oxygen in water in two ways.


A thick layer of vegetation on the surface of water prevents sunlight from reaching the bottom.
As a result, plants that grow at the bottom of the water bodies cannot photosynthesize to produce
oxygen.
Some algae and water plants soon die, and their decay uses up oxygen in water.
The human body has a tolerance to moderate concentrations of nitrate ions in water. However, babies are
poisoned by concentrations of NO3- that are greater than 100 ppm. The ability of their blood to carry oxygen
is greatly reduced, causing the ‘blue baby syndrome’.
9.5.2
Sulphur
Sulphur, in Group (VI), has the electronic configuration 1s 2 2s2 2p6 3s2 3p4. This configuration has
important consequences on the chemistry of sulphur.
 Having six electrons in the outer most shell, it can easily form a negative ion by accepting two
electrons from a metal. A large number of metals form sulphides, some of which are important ores
found in the earth’s crust.
 Sulphur shows a range of oxidation states, up to +6, corresponding to participation of all six valence
electrons.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
440
Sulphur is also able to expand its octet because of the availability of energetically accessible d orbitals to
accommodate some of the electron density. Examples of compounds in which sulphur has more than an
octet of electrons in its outer shell include SF 6 and SO3, whose structures are shown below.
F
F
F
O
S
S
F
F
F
SF6 : octahedral
bond angle = 900
O
O
SO3 : trigonal planar
bond angle = 1200
2.1 Occurrence of sulphur
Elemental sulphur, which is a yellow solid, exists as S 8 molecules. Sulphur is found in many compounds;
including biological molecules such as proteins (not all proteins contain sulphur). It is also found in volcanic
eruptions, meteorites and medicines such as Epsom salts (MgSO4). H2SO4 is one of the most important and
well known compounds of sulphur.
2.2 Reaction with O2
Sulphur burns readily with a blue flame, forming sulphur dioxide, which is a colourless gas with a choking
smell.
1/8 S8 (s) + O2 (g) → SO2 (g)
This reaction occurs in volcanoes, and partly accounts for the presence of SO 2 in the atmosphere.
SO2 is an important raw material in the manufacture of sulphuric acid. It is also used in the food industry
as a preservative. It is added in very small amounts to some sauces and drinks to kill bacteria. Its
antibacterial action can be explained in two ways
 It reacts with water to form an acidic solution of sulphrous acid, H 2SO3. This lowers pH and causes
bacteria to die.
 Some of the sulphrous acid is oxidized by oxygen to form sulphuric acid. The removal of oxygen
slows down decay processes that are caused by bacteria.
The concentration of sulphrous and sulphuric acids in the food is too little to have any harmful effects on
humans.
2.3 Sulphuric acid
Sulphur is an important raw material in the manufacture of sulphuric acid by the contact process.
The raw materials for this process are
 Sulphur dioxide from the burning of sulphur or roasting of metal sulphides (such as iron pyrite) in
air.
 Oxygen from the fractional distillation of liquid air.
The two gases are cleaned, mixed in the ratio of 2 volumes of SO₂ to 1 volume of oxygen and passed over a
catalyst. One catalyst commonly used is vanadium pentoxide (vanadium (V) oxide - V₂O₅) because it is
cheaper than metals such as platinum. Vanadium pentoxide is also not easily poisoned.
The following reaction takes place
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
441
2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g) ... -197Kj
The operating conditions used are
Temperature: 450⁰C
Pressure
: 1.0 atm (atmospheric pressure)
Catalyst: vanadium pentoxide.
By Le Chatelier’s principle, high yields of SO₃ would be achieved by using low temperature since the forward
reaction is exothermic. However, this would slow the reaction, so in practice a compromise temperature of
about 450⁰C is used.
The formation of SO₃ is accompanied by a decrease in the number of moles of gas . By Le Chatelier’s principle, high yields of SO₃ would be obtained by carrying out the process at high pressures. In practice, the
reaction is carried out at atmospheric pressure. This is because the temperature used, 450⁰C, gives a very
high yield (about 97%) at atmospheric pressure, so there is no need to use high pressures.
The formation of SO₃ is an exothermic process. The use of heat exchangers between catalyst beds keeps the
temperature constant at 450⁰C. If this was not done, there would be a build up of heat, with two consequences
 the yield of SO3 decreases since high temperatures favour the reverse reaction, which is endothermic.
 pipes and equipment are gradually weakened by heat, so this would increase maintenance costs.
The SO₃ gas produced is then absorbed in absorption towers containing concentrated sulphuric acid. This
results in the formation of oleum.
SO₃(g) + H₂SO₄(l) → H₂S₂O₇(l)
oleum
Sulphuric acid, when needed, is then obtained by hydrolysis of oleum in water.
H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(aq)
One alternative is to absorb the sulphur trioxide in water, which directly produces sulphuric acid.
SO3(g) + H2O(l)  H2SO4(aq)
This process is not satisfactory since the reaction is so exothermic that the heat produced causes the
sulphuric acid to form as a vapour. Condensing the vapour back to liquid is difficult. The flow diagram is a
flow chart summarizing the contact process.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
442
Q
A
What is the justification for the operating temperature used in the contact
process.
2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g) ... -197Kj
Since the forward reaction is exothermic, the yield of SO 3 would be increased by cooling.
However, a very low temperature would slow the reaction. The actual temperature used
is therefore a compromise between the low temperature required to increase yield of
SO3 and the high temperature required to increase the rate of reaction.
Properties of sulphuric acid

Concentrated sulphuric acid is a viscous and oily liquid. Its viscosity is a result of hydrogen bonds
which are present between its molecules. Very concentrated sulphuric acid also contains some
molecules of oleum. The following diagrams show hydrogen bonding in sulphuric acid, and the
structure of oleum.
O
O
S
H
O
O
H
O
O
hydrogen bond
O
O
S
H

oleum
O
O
H
S
HO
O
O
x
O
S
OH
Suggest the size of bond angle x.
Sulphuric acid is denser than water, so when it is added to water, it tends to sink, meanwhile dissociating to release hydrogen ions. The reaction of sulphuric acid with water is very exothermic and
can be explosive. Risk is kept to a minimum by adding the acid to water, because as it sinks, the
reaction occurs gradually. Adding water to concentrated sulphuric acid is dangerous. Water, being
less dense, will briefly settle on the surface of the acid. A violent reaction occurs at the junction of
the two solutions, causing the liquid at the surface to boil, throwing out a hot and
acidic spray.
Always add concentrated sulphuric acid to water, not water to concentrated sulphuric acid.
Uses of sulphuric acid





As electrolyte in lead acid accumulators
Washing metal electrodes before they are used in electrolytic processes.
Manufacture of ammonium sulphate fertilizer.
Manufacture of explosives
Manufacture of paints and dyestuffs.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
443
2.4 Pollution by SO2
SO2 is a well known air pollutant. It irritates the respiratory system and may trigger attacks in asthmatic
people. However, the major environmental hazard associated with SO 2 is its contribution to the formation
of acid rain. This can happen in two ways:
 SO2 dissolves directly in rain water, to form a dilute solution of sulphrous acid, H 2SO3.
SO2 (g) + H2O (l) → H2SO3 (aq)
Note that this is not a redox reaction. The oxidation state of sulphur is the same in SO 2 and H2SO3.
As the rain passes down through the atmosphere, some H 2SO3 is oxidized to sulphuric
acid
H2SO3 (aq) + ½ O2  H2SO4 (aq)

Some sulphur dioxide molecules are oxidized in the atmosphere, to form sulphur trioxide. The
reaction is catalyzed by oxides of nitrogen, as already discussed.
SO2 (g) + ½ O2 → SO3 (g)
This is reaction is more important in the formation of acid rain because sulphuric acid is more
soluble in water than sulphrous acid (sulphrous acid tends to dissociate to form SO 2 and water).
SO3 then dissolves in rain water to form sulphuric acid.
Some effects of acid rain have already been discussed.
Sources of atmospheric sulphur dioxide
Air naturally contains sulphur dioxide due to natural phenomena such as volcanic eruptions. However, in
recent years, human activities have greatly increased the concentration of this gas in the atmosphere. These
activities include
 the combustion of sulphur containing fuels, for example coal and oil products. Coal and oil were
made millions of years ago from the remains of plants and animals. The presence of sulphur in
these fuels can be traced back to the proteins that were present in the plant and animal matter.
 Industrial activities, such as the extraction of metals by roasting of sulphur containing ores such
as copper iron sulphide, CuFeS.
Methods of minimizing SO2 pollution


Cleaner sources of fuels or energy should be used, for example, ethanol.
Exhaust gases from industries and power stations can be treated with chemicals that can absorb
SO2 before it is emitted into the atmosphere. A Flue Gas Desulphurization (FGD) plant, which
contains heated powdered limestone (powdered to increase surface area), can be fitted to the
chimney stacks. Flue gases containing sulphur dioxide, for example from the combustion of coal,
are absorbed into the hot limestone powder and react with CaCO3, to form calcium sulphite.
SO2 (g) + CaCO3(s) → CaSO3(s) + CO2 (g)
Further oxidation of calcium sulphite occurs, to produce CaSO 4, which can be used in the manufacture of plaster board or cement.
CaSO3(s) + ½ O2 (g) → CaSO4(s)

Sulphur in crude oil can be removed by pumping in hydrogen and using a suitable metallic salt
catalyst. Under these conditions, the organic sulphur containing impurities are reduced by hydro-
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
444
gen to hydrogen sulphide.
2RSH (g) + H2 (g) → R2 (g) + 2H2S ... (i)
The hydrogen sulphide is then burnt in a limited amount of oxygen, which converts it to a
mixture of sulphur and sulphur dioxide
4H2S (g) + 4O2 → 4H2O + 2S(s) + 2SO2 ... (ii)
The sulphur dioxide produced in this reaction is reacted with H 2S that has not reacted in step (i),
to produce more sulphur. This reaction is catalyzed by Al 2O3.
SO2 (g) + 2H2S (g) → 2H2O (l) + 3S(s)
Sulphur which is obtained in this process has important uses, for example, the manufacture of
sulphuric acid.
Questions, solutions and discussions
Q1
Sulphuric acid is used in many industrial processes of major importance.
The first stage in the manufacture of sulphuric acid is to pass air over burning sulphur.
The emerging gas has the following composition by volume.
sulphur dioxide 10%
sulphur trioxide 0.2%
oxygen 10%
nitrogen etc. 79–80%
(a) (i) Write an equation for sulphur burning in air.
(ii) Suggest why the air is passed so fast that only half the oxygen is used.
The emerging gas is passed over a catalyst maintained at 450–550 °C in the reaction
chamber.
(b) Name the catalyst used in the Contact process.
(c) The high yield is only achieved under certain conditions.
After each condition explain why this leads to an increased yield of sulphur trioxide.
(i) There needs to be an excess of air in the reacting gas mixture.
(ii) The catalyst needs to be cooled.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
445
(iii) The air used to burn the sulphur must be as clean as possible.
Sulphur trioxide is formed in 98% yield; 2% of sulphur dioxide remains unconverted.
2SO2(g) + O2(g) ∏ 2SO3(g)
ΔH = –197 kJ mol–1
(d) In modern plants, nearly all the SO2/SO3 mixture is absorbed but up to 0.05% by
volume of SO2 may be allowed to pass into the atmosphere through a chimney stack.
Give two reasons why SO2 should not be discharged into the atmosphere.
(e) (i) When concentrated sulphuric acid is warmed with solid sodium chloride, misty fumes are
produced.
Identify the fumes
Write an equation for the reaction
(ii) When concentrated sulphuric acid is warmed with solid sodium iodide, purple
fumes are produced.
Identify the fumes.
9701/02/M/J/02
Solutions
(a) (i) S + O2  SO2
(ii) The unused oxygen is required for the next stage, which is the conversion of SO 2 to SO3.
(b)
(c)
Vanadium (V) oxide, V2O5
2SO2(g) + O2(g) ∏ 2SO3(g)
ΔH = –197 kJ mol–1 … (I)
(i) A large concentration of oxygen drives equilibrium (I) to the right, increasing the yield of
SO3.
(ii) Since the reaction is exothermic, by Le Chatelier’s principle, high temperature would
decrease the equilibrium yield of SO3 by favouring the reverse reaction, which is
endothermic. The catalyst therefore needs to be cooled so that the forward reaction is
favoured.
(iii) The catalyst is easily poisoned by impurities in air.
(d)
Sulphur dioxide is a respiratory irritant and may trigger attacks in asthmatic people.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
446
Sulphur dioxide leads to the formation of acid rain which destroys buildings, metal structures,
vegetation and aquatic life.
(e)(i)
HCl
NaCl + H2SO4  HCl + NaHSO4
OR 2NaCl + H2SO4  2HCl + Na2SO4
(ii)
Iodine
Q2
Ammonia is manufactured from nitrogen and hydrogen by the Haber process.
Hydrogen is usually obtained by reacting methane and steam; the by-product is carbon
monoxide.
(a) Construct a balanced equation for this production of hydrogen.
The reaction between nitrogen and hydrogen is exothermic and incomplete.
(b) (i) Write an equation for the Haber process.
(ii) State the three conditions necessary for the efficient working of a Haber process
plant.
(iii) Draw a flow diagram to show how the gases pass through the plant. The part
where the ammonia is formed should be called the converter. Label the flow diagram to
explain the process.
(c)
Explain why the pressure you have quoted in (b)(ii) is used.
(d)
Most of the ammonia produced which is not used as fertilizer, is oxidized to nitric acid,
HNO3.
Construct an equation for the oxidation of ammonia by atmospheric oxygen to form
nitric acid.
(e)
Urea, CO(NH2)2, is a naturally occurring substance which can be hydrolyzed with water
to form ammonia according to the following equation.
H2O(l) + CO(NH2)2(aq) → CO2(aq) + 2NH3(aq)
The standard enthalpy changes of formation of water, urea, carbon dioxide and ammonia
(in aqueous solution) are given in the following table.
Use these data to calculate the standard enthalpy change for the hydrolysis of urea.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
447
Compound
ΔHθf / Kj mol-1
H2O (l)
-287.o
Co (NH2)2(aq)
-320.5
CO2 (aq)
-414.5
NH3(aq)
-81.0
Solutions
(a)
CH4 + H2O  3H2 + CO
(b)(i)
N2 + 3H2 ∏ 2NH3
(ii)
- pressure : > 200 atm, temperature : 4500C to 5000C, Catalyst : Fe or Fe2O3
(iii)
(c) N2 + 3H2 ∏ 2NH3
The equation shows that the product side has less moles of gas than the reactant side. The reaction is
therefore accompanied by a decrease in pressure. In the Haber process, high pressure is employed as
this favours the side with the smaller number of moles, that is, the formation of ammonia. According
to Le Chatelier’s principle, increasing pressure favours the side which results in a decrease in
pressure, that is, the side with less moles of gas.
(d)
(e)
NH3 + 2O2  HNO3 + H2O
31 Kj mol-1. The working is left as an exercise to the reader.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
448
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
9.5in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
1.
When hydrocarbons such as petrol or paraffin wax are burned in an excess of air in a laboratory,
carbon dioxide and water are the only products.
When petrol is burned in a car engine, nitrogen monoxide, NO, is also formed.
(a) Explain how NO is formed in an internal combustion engine but not formed when a small sample
of petrol is burnt in an evaporating basin.
The engines of modern motor cars have exhaust systems which are fitted with catalytic converters
in order to reduce atmospheric pollution from substances such as NO.
(b)(i) State three more pollutants, other than CO2 and NO, that are present in the
exhaust gases of a car engine.
(ii)
What is the active material present in the catalytic converter?
(iii) Write one balanced equation to show how NO is removed from the exhaust gases of a car
engine by a catalytic converter.
NO is also formed when nitrosyl chloride, NOCl, dissociates according to the following equation.
2NOCl (g) ∏ 2NO(g) + Cl2(g)
Different amounts of the three gases were placed in a closed container and allowed to come to
equilibrium at 230 °C. The experiment was repeated at 465 °C.
The equilibrium concentrations of the three gases at each temperature are given in the following
table.
Concentration / mol dm-3
Temperature / 0C
NOCl
NO
Cl2
230
2.33 x 10-3
1.46 x 1o-3
1.15 x 10-2
465
3.68 x 10-4
7.63 x 10-3
2.14 x 10-4
(c) (i) Write the expression for the equilibrium constant, Kc , for this reaction. Give the units.
(ii) Calculate the value of Kc at each of the temperatures given.
(iii) Is the forward reaction endothermic or exothermic? Explain your answer.
(d)
The temperature of the equilibrium was then altered so that the equilibrium concentrations of
NOCl and NO were the same as each other.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
449
What will be the effect on the equilibrium concentration of NOCl when the following changes
are carried out on this new equilibrium? In each case, explain your answer.
(i) The pressure of the system is halved at constant temperature.
(ii) A mixture of NOCl (g) and NO (g) containing equal numbers of moles of each gas introduced
into the container at constant temperature.
9701/02/M/J/08
2
Nitrogen makes up about 79% of the Earth’s atmosphere. As a constituent element of proteins, it is
present in living organisms.
Atmospheric nitrogen is used in the Haber process for the manufacture of ammonia.
(a) Write an equation for the formation of ammonia in the Haber process.
(b)
The Haber process is usually carried out at a high pressure of between 60 and 200 atmospheres .
State two further important operating conditions that are used in the Haber process.
For each of your conditions, explain why it is used.
(c)
State one large-scale use for ammonia, other than in the production of nitrogenous fertilizers.
(d)
The uncontrolled use of nitrogenous fertilizers can cause environmental damage to lakes and
streams. This is known as ‘eutrophication’.
What are the processes that occur when excessive amounts of nitrogenous fertilizers get into
lakes and streams?
In many countries, new cars have to comply with regulations which are intended to reduce the
pollutants coming from their internal combustion engines.
Two pollutants that may be formed in an internal combustion engine are carbon monoxide, CO, and
nitrogen monoxide, NO.
(e)(i) Outline how each of these pollutants may be formed in an internal combustion engine.
(ii) State the main hazard associated with each of these pollutants.
Pollutants such as CO and NO are removed from the exhaust gases of internal combustion engines by
catalytic converters which are placed in the exhaust system of a car.
(f) (i) What metal is most commonly used as the catalyst in a catalytic converter?
(ii) Construct one balanced equation for the reaction in which both CO and NO are removed
from the exhaust gases by a catalytic converter.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
450
9.6 AN INTRODUCTION TO THE
CHEMISTRY OF THE TRANSITION
ELEMENTS
The Transition elements are found in the d-block of the Periodic Table. They have a number of chemical
and physical properties which distinguish them from main group metals such as calcium.
9.6.1
What is a transition element?
A transition element has been defined by some as any element which occupies the d -block of the Periodic
Table. However, this is hardly a definition but a general description of location.
A more informative definition of a transition element is:
... a d-block metal that forms at least one ion or compound in which the d orbitals are
partially filled.
This definition excludes some metals which are found in the d- block of the periodic table, such as Sc and
Zn. Scandium forms +3 compounds in which all the d- orbitals are empty (d0). Zinc forms +2 compounds
in which all the d- orbitals are fully occupied (d10). Compounds of these elements tend to be white when
solid or colourless in solution. In this regard they differ from the compounds of transition elements which
are coloured.
9.6.2
Electronic structure of the transition elements
The transition elements are characterized by occupancy of the d orbitals. From Sc to Zn (first row of the
transition elements), the number of d electrons increases by one from one element to the next.
There are five degenerate orbitals in a d sub shell, designated dxy, dxz, dzy, dz2 and dx2-y2. Each orbital
carries a maximum of two electrons, so the d sub shell has a maximum capacity of 10 electrons. This maximum occupancy is found at copper and zinc.
The first three orbitals (dxy, dxz, dzy) each have four lobes that are aligned between the axis, for example,
the dxy has lobes oriented between the x and y axis.
The remaining two d orbitals form their own set (in
terms of geometry) in that they have lobes that are
directed along the axis, as shown below.
Fig 9.6.1 The dxy, dxz and dzy orbitals
have lobes that are directed between the
axis.
Prepublication manuscript
Fig 9.6.2
The dz² and dx2-y2 orbitals have lobes that are
©L.Mwanawenyu. Please
do not photocopy or reproduce in any way
directed along the axes.
451
A transition element has the general configuration [Ar] 4sx3dn, where x is 2, (except in Cr and Cu where
it is 1) and n = 1, 2, 3 ... 10 (but cannot be 4 or 9).
Element
Configuration
Illustration
Sc
[Ar] 3d14s2
Ti
[Ar] 3d24s2
[Ar]
4s2
V
[Ar] 3d34s2
[Ar]
4s2
Cr
[Ar] 3d54s1
[Ar]
4s1
Mn
[Ar] 3d54s2
[Ar]
4s2
Fe
[Ar] 3d64s2
[Ar]
4s2
Co
[Ar] 3d74s2
[Ar]
4s2
Ni
[Ar] 3d84s2
[Ar]
4s2
Cu
[Ar] 3d104s1
[Ar]
4s1
Zn
[Ar] 3d104s2
[Ar]
4s2
[Ar]
4s2
Table 9.6.1 Electronic
configurations of the
transition elements from
Sc to Zn
The d4 and d9 configurations do not exist in a neutral transition metal atom. These two configurations would
have been expected in Cr and Cu respectively. Instead of having d⁴, Cr has d⁵. Instead of having d⁹, Cu has
d10.
Table 9.6.1 shows the electronic configurations of the first row transition elements, from scandium to zinc.
Which one is the valence shell?
The valence shell is the 4s, not the 3d. There are two principal quantum numbers involved here, 4 and
3. The shell with the higher principal quantum number is the valence shell, which is the 4s in this case.
Some readers may recall that when arranging the orbitals in terms of energy, the 4s comes before the 3d
... 4s 3d...
As explained in section 2, readers should note that this order is what we would expect if we arranged the
empty orbitals in terms of energy. However, when electrons start to enter the d orbitals, they repel the 4s
electrons. This causes the 4s shell to move away from the nucleus, that is, it becomes higher in energy than
the 3d sub shell. The fact that the 4s is the valence shell explains why the +2 oxidation state is so common
among the transition elements. It corresponds to the loss of the two 4s electrons, which are furthest from
the nucleus and experience the least attraction from the nucleus.
An important note
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
452
Even though the 4s shell is higher in energy, it is so energetically close to the 3d that in chemical reactions, the two
shells behave as one shell. This has important consequences to the chemistry of the transition elements, as explained
later.
The configurations of Cr and Cu
We would have expected the following configurations for Cr and Cu, respectively
Cr: [Ar] 4s23d4
Cu: [Ar] 4s23d9
The actual configurations are formed by transferring an electron from the 4s orbital to a 3d orbital. By so
doing, Cr attains a configuration in which the d orbitals are exactly half filled (with 5 electrons). Copper
attains a configuration in which the d orbitals are exactly filled with 10 electrons.
There is an extra stability associated with an exactly half filled or exactly filled sub shell. This
stability should be explained in terms of the resulting symmetric occupancy of the sub shells by
electrons, that is, there are no regions which carry more electrons than others.
9.6.3 Physical properties of the transition elements
Table 9.6.2 summarizes some of the physical properties of the transition elements.
Ionization energies
Element
Density/
gcm-3
m.p./0C
b.p.0C
Metallic
radius/nm
1st
2nd
3rd
Sc
2.99
1541
2831
0.164
631
1235
2389
Ti
4.50
1660
3287
0.147
658
1310
2653
V
5.96
1890
3380
0.135
650
1414
2828
Cr
7.20
1857
2670
0.129
653
1592
2987
Mn
7.20
1244
1962
0.137
717
1509
3249
Fe
7.86
1535
2750
0.126
759
1561
2958
Co
8.90
1495
2870
0.125
758
1646
3232
Ni
8.90
1455
2730
0.125
737
1753
3394
Cu
8.92
1083
2567
0.128
746
1958
3554
Zn
7.14
420
907
0.137
906
1733
3833
Table 9.6.2
Physical properties of the transition elements
In general, transition elements have the following physical properties
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
453
1.
2.
3.
4.
5.
6.
Generally small atomic radii.
High densities and melting points.
Hard and rigid
High ionization energies.
Excellent electrical conductivities
Paramagnetic
3.1 Transition metals generally have small atomic radii and
high densities
Explanation


Large nuclear charge. The large number of protons in the nucleus results in shells being
attracted strongly towards the nucleus. This causes the atoms to shrink.
Poor shielding effect of the d orbitals. Because of their shape, the d orbitals poorly shield the
outer 4s shell from the nucleus. As a result, the outer shell feels a strong attraction by the nucleus,
to which it is attracted, resulting in shrinking of size.
The high densities of the transition metals are a direct consequence of the small atomic radii and heavy
nucleus (large number of protons and neutrons).
 the atoms have large masses
 and yet their size is small.
Now, density is given by m/V. This formula shows that
1. density varies directly as mass. The larger the mass, the higher the density.
2. density varies inversely as volume. The smaller the volume, the higher the density.
The atoms of transition elements are quite small compared to their mass. The effect of small atomic radii
(volume) and large mass is to increase density, that is, each atom has a large mass packed in a small
volume.
Trend in atomic radii across the first row transition elements
If you study Table 9.6.2 you will notice that the radii of the transition elements do not change much across
the period, from Sc to Zn. We expect atomic radii to decrease across the row in response to increasing
nuclear charge. As the number of protons in the nucleus increases, so does the attractive influence of the
nucleus for the outer most shell, so the atoms shrink in size. The decrease in atomic radii does occur but
not as rapidly as happens in the non transition periods such as Period 3.
This slow and gradual decrease in atomic size is due to the fact that across the row, electrons enter a
penultimate (underlying) 3d shell, thus shielding the outer 4s sub shell from feeling the full attractive
influence of the nucleus. In other words, the increase in nuclear charge has a small effect on the outer 4s
shell, because this shell is prevented by the 3d electrons from experiencing the full effect of the nuclear
charge.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
454
The outer 4s shell is fully occupied
so electrons enter the penultimate
3d shell
Fig 9.6.3 In the transition elements, electrons enter the penultimate 3d shell. The 4s
shell is therefore shielded by the 3d electrons.
3.2
Transition elements have relatively high ionization
energies which change only gradually across the period
Ionization energy measures how strongly valence electrons are held to the atom. In transition elements, the
outermost electrons are firmly held for two reasons
 The atoms are small, so the valence electrons are close to the nucleus and they feel a powerful
attraction by protons.
 The charge density in transition metal atoms is high. The large number of protons in the
nucleus attracts the valence electrons strongly.
Because of these reasons, the valence electrons are difficult to remove; a very large amount of energy would
be required to do so. Notice that the variation in first and second ionization energy across the first row of
transition elements is only gradual, that is, the increase in ionization energy from one element to the next
is very small, compared to non transition elements.
Generally, ionization energies increase across the first row transition elements, in response to increasing
nuclear charge. Each additional proton increases strength of attraction between the nucleus and the
valence electrons. This increase in ionization energy is not so sharp for a reason already explained:
Across the period, electrons are being added into a penultimate 3d sub shell which shields the outermost
4s electrons from feeling the full attractive influence of the nucleus. This screening (shielding) of the
outer shell by the 3d electrons implies that adding a single proton to the nucleus does not have a large
effect on the size of the attractive force felt by the valence shell.
However, the change in third and fourth ionization energies is rapid across the period is rapid. This is
because this time electrons are being removed from the d-sub shell, which experiences very little shielding
from the remaining electrons (Fig 9.6.4).
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
455
ionization energy /
Kjmol-1
IE (3)
3500
2000
IE (2)
IE (1)
500
21
25
Sc
Mn
proton number
30
Zn
Fig 9.6.4 Trend in first, second
and third ionization energies
across the first row of transition
elements
Note that the third ionization energy of Fe is lower than expected because the electron being removed is
from a paired orbital where it experiences repulsion from the other electron.
3.3 Transition elements are excellent conductors of electricity
Conductivity in a metal is a result of delocalized electrons. These mobile electrons move in response to a
potential difference between two points of a conductor. The greater the number of electrons in the delocalized sea, the greater the conductivity. In transition elements, the 4s and the 3d shells, being very close in
energy, act as if they where one shell.
Electrons are thus lost from both the 4s and the 3d sub shells and contributed to a common pool. There
are therefore a large number of electrons in the delocalized sea to conduct electric current.
However, some non transition metals like Mg may have electrical conductivities that are comparable to or
even greater than that of some transition metals. This is because of the small nuclear charge in these non
transition metals. The delocalized electrons in the metals easily move because they experience a relatively
small attraction from the positive charge on the metal atoms and the protons in the nuclei of these atoms.
On the other hand, even though a transition metal has a large number of delocalized electrons, mobility of
these electrons is restricted by attraction to the large positive charge on the metal atoms and the large
number of protons in the nucleus of the atom.
3.4 Transition metals have high melting and boiling points
Melting or boiling a metal requires the breaking of metallic bonds. The melting point or boiling point of a
metal is therefore a measure of the strength of these bonds. Transition metals have high melting points and
boiling points due to the presence of strong metallic bonds. The strength of these bonds can be explained
in terms of
 the large number of delocalized electrons in the metal lattice. Remember that these electrons come
from both the 4s and the 3d sub shells of the atoms.
 the large charge that remains on the metal atoms after they have contributed their 4s and 3d
electrons into the pool of delocalization.
These two factors result in very strong electrostatic attractions between the positive metal atoms
and the sea of delocalized electrons.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
456

the large nuclear charge of the atoms. Protons in the nucleus also have an attraction for the
delocalized electrons. This attraction is strong in the transition metals due to the large number of
protons in their nuclei.
We expect melting and boiling points of the transition elements to increase across the period in response
to increasing nuclear charge and decreasing atomic radii. A clear trend in melting points is not observed
because the metals tend to have different crystal structures.
3.5 Transition elements and their compounds are paramagnetic
Paramagnetism is the observation that some metals are attracted to a magnet.
The opposite of paramagnetism is diamagnetism, in which a substance is slightly repelled by the magnetic
field of a magnet. Non transition metals such as magnesium and aluminium tend to be diamagnetic.
Paramagnetism in transition metals is a result of unpaired electrons in the d orbitals. This explains why
copper and zinc are diamagnetic. Their d-electrons are all paired.
9.6.4
Chemical properties of the transition elements
Transition elements show the following chemical properties which are not usually shown by non-transition
elements.
4.1
1.
They exhibit variable oxidation states in their compounds
2.
They tend to form complex compounds
3.
They tend to form coloured compounds
4.
They and their compounds show catalytic activity
Variable oxidation states
Transition elements tend to exhibit different oxidation states in their compounds, for example, manganese
forms stable compounds whose oxidation states range from +2 to +7. Compare with a non-transition
element such as calcium which has only one oxidation state, in this case, +2, in its compounds.
This tendency of transition metals to exhibit variable oxidation states is caused by the close proximity in
energy of the 3d and the 4s sub-shells. These shells are so close in energy that they behave as one shell.
Consequently, electrons may be lost from both shells. Different oxidation states would be formed
corresponding to the total number of electrons lost from the 4s and 3d sub-shells.
Compare vanadium with calcium. In Ca, only the two valence electrons in the 4s shell may be lost, forming the +2 oxidation state. In Vanadium, the total number of electrons in the 4s and the 3d sub shells is
five. A range of oxidation states, from +2 to +5, is therefore possible.
The following points should be noted:

In all transition compounds, except copper, the +1 state is very unstable.
To form the +1 state, one electron would have to be lost from the 4s sub shell. This would leave a
single electron in the valence shell. Such a configuration is very unstable. It explains why metals
which have a single valence electron, such as Na, are very reactive.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
457

Electrons are always lost first from the 4s sub shell, because it is the valence shell. As a result, the
+2 state, corresponding to loss of both valence electrons, is common to all transition elements.
However, note that Cr and Cu have a 4s1, not 4s2 configuration.

Only unpaired d- electrons may be lost during reactions
A key reason why substances react is to try and pair up electrons. By being paired, electrons attain
stability. In the transition elements, if the d-electrons are already paired, they do not show any
further tendency to react. The range of possible oxidation states therefore decreases when one or
more d orbitals contain paired electrons.
Table 9.6.5 shows that the number of possible oxidation states increases from scandium and reaches a
maximum at manganese. This is a result of increasing number of unpaired electrons in the d sub shell. After
Mn, the number of possible oxidation states start to decrease due to pairing of electrons in the d orbitals.
Element
Configuration
Sc
[Ar]3d1
4s2
Possible oxidation
states
+3
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
[Ar]3d2
4s2
[Ar]3d3
4s2
[Ar]3d5
4s1
[Ar]3d5
4s2
[Ar]3d6
4s2
[Ar]3d7
4s2
[Ar]3d8
4s2
[Ar]3d10
4s1
Ar]3d10
4s2
+1
+1
+1
+1
+1
+1
+1
+1
+1
+2
+2
+2
+2
+2
+2
+2
+2
+2
+3
+3
+3
+3
+3
+3
+3
+3
+4
+4
+4
+4
+4
+4
+4
+5
+5
+5
+5
+5
+6
+6
+6
+7
Table 9.6.5 Range of possible oxidation states for the transition elements. The stable oxidation states
for each element are highlighted.
The electronic configurations of transition metal ions
To work out the electronic configuration of the cation of a transition metal, we need to know the configuration of the neutral atom. The size of positive charge on the ion is then equal to the number of electrons
lost from the neutral atom. You should remember that the first two electrons are lost from the 4s sub shell,
for example, the electronic configuration of Sc³⁺ is 1s²2s²2p63s2 3p64s03d0, corresponding to the loss of two
electrons from the 4s and one electron from the 3d shell. Usually empty orbitals are not shown, so the
electronic configuration can be written as 1s²2s²2p63s2 3p6. This ion is stable since it has the electronic
configuration of a noble element, Ar.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
458


The lower oxidation states form aqua complexes with water, for example, the species usually abbreviated as Cu²⁺(aq) is in fact the hexaaqua complex, [Cu (H2O)6] 2+.
The higher oxidation states, from the +4 and upwards, cannot exist as hexaaqua complexes. These
ions would have a charge density so high that they would immediately polarize water molecules,
releasing hydrogen ions and bonding with oxygen to form anions. In fact, in aqueous solution, the
higher oxidation states may exist as oxo-anions (Table 9.6.4), for example, vanadium forms the
oxo-anions VO₂⁺ and VO²⁺, in which vanadium has oxidation numbers +5 and +4 respectively. In
solution, these oxo-anions exist with counter ions such as the chloride ions, to form, for
instance, VO₂Cl (aq), which is a soluble salt.
Compounds of transition metals in which the metal has a high oxidation state also exist as insoluble
compounds such as MnO2.
Oxo - anion

Q

4.2
Oxidation state of
the transition
element
Colour of oxo-anion
in aqueous solution
VO2+
+4
Blue
VO2+
+5
Yellow
MnO4-
+7
Cr2O72-
+6
Orange
CrO42-
+6
Yellow
Violet/purple
Table 9.6.4 Oxo-anions of the
transition elements
The lower oxidation states have a tendency to be oxidized to higher oxidation states, that is, they
often react as reducing agents. A well known example is Fe (II). This can reduce a number of
substances, for example, Cl₂. In these reactions, the Fe (II) is oxidized to Fe (III).
Use the Data Booklet to show that the reaction between chlorine and aqueous Fe 2+ (aq) is
feasible under standard conditions. Also check if a similar reaction can take place with
iodine. Explain any differences that would be observed in these two situations.
The higher oxidation states have a tendency to be reduced to lower oxidation states. In this
process, they act as oxidizing agents. Commonly used oxidizing agents include KMnO₄ and
K₂Cr₂O₇. In these compounds, the transition element has its highest possible oxidation state.
Transition elements tend to form complex compounds
A complex compound is one which is made up of electron rich species known as ligands that are
bonded to a central metal cation through dative bonds.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
459
An example of a complex ion is [Co(NH 3)6]3+, in which ammonia ligands form dative covalent bonds to a
central Co3+ ion.
The square bracket shows the co-ordination sphere, that is,
the central metal ion and its ligands. A charged complex
requires counter ions to balance its charge, for example, the
complex shown may be associated with three chloride ions.
The counter ions are written outside the square bracket, as
shown in Fig 9.6.7. The counter ions are not part of the coordination sphere.
3+
NH
..
3
..NH
.
.NH
3
3
Co
H3N
..
.
.NH
..NH
3
3+
NH
..
3
3
Fig 9.6.6 Structure of a typical complex,
showing the use of lone pairs by
ligands in
[Co(NH3)6]3+. 3Clforming dative bonds to the central metal
ion.
..NH
.
.NH
3
=
3
Co
H3N
[Co(NH3)6]3+.3Cl-
Fig 9.6.7 The salt
showing the co-ordination sphere
and its counter ions
3Cl-
..
.
.NH
..NH
3
3
In the Fig 9.6.7 NH₃ is involved in the co-ordination sphere because it is a stronger ligand than chloride
ions. A salt in which one of the ions is actually a co-ordination complex, for example, [Ni (NH3)₆]3⁺, can be
analyzed by adding to the aqueous solution a reagent which will react with the counter ions. The reagent
will not be able to react with the ligands of the co-ordination sphere because they are firmly held to the
central metal ion by strong covalent bonds. For example, adding aqueous silver nitrate to the salt shown in
Fig 9.6.7 will result in the chloride ions being precipitated out of solution as AgCl. The number of moles of
chloride present in solution can then be determined from the mass of AgCl precipitated. Once the
number of counter ions present per each co-ordination sphere is known, it becomes possible to determine
the overall charge on the co-ordination sphere. For example, if it is shown that there are three chloride
counter ions, then the charge on the co-ordination sphere must be +3.
A ligand can be neutral or negatively charged. All ligands, neutral or negatively charged, contain at least
one atom which bears a lone pair of electrons. It is this lone pair that is used to make a dative bond to the
central metal cation. The central cation itself must have the ability to accept and stabilize extra electron
density.
Why transition metals can form complex ions
Transition elements have partially filled d orbitals which can be used to accept electron density from the
ligands.
It must be mentioned here that a few non-transition metals can also form complex ions. The prerequisite
for complex ion formation is availability of vacancies to accommodate extra electron density, or simply the
ability to attract and form stable compounds with ligands. Non transition elements which can form complex
ions include Al and Pb.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
460
Classification of ligands
A ligand is an electron rich species, either neutral or negatively charged, and carrying one or more lone
pairs of electrons which can be used to form one or more dative bonds to a central metal cation.
Ligands can be classified in terms of the number of dative bonds they can form, per molecule of the ligand,
to the central metal cation (Table 9.6.5). For instance, a bidentate ligand has two sites that it can use to
form two dative bonds during complex formation. Bidentate literally means ‘two toothed’. Note all lone
pairs on a ligand may be used in complex formation. Consider water which carries two lone pairs on an
oxygen atom. Each water molecule can only form one dative bond, not two, to the central metal ion. The
lone pairs in water are on the same atom and are therefore too close to allow formation of two dative bonds
to the metal ion.
In general, if an atom in the ligand contains two or more lone pairs of electrons, only one may be used in
dative bond formation. For lone pairs to form dative bonds to the central metal ion, they must be sufficiently
separated from each other, that is, they must be on different atoms, and they must be correctly positioned.
A polydentate ligand can form more than two dative bonds with the central metal ion. Sometimes the shape
of the ligand changes to bring the bonding atoms into correct orientation relative to the metal ion.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
461
Type of ligand
Examples
Monodentate
Examples of complexes that contain such
ligands
H2O, Cl-, NH3, OH-, F-,
[CuCl4]2- yellow
CO, CN-
[Co(NH3)6]3+ golden brown
[ Fe(H2O)6]2+ light green
C2O42- (oxalate = ethanedioate)
Bidentate
O
O
O
C
binding sites
O
C
O
O
2+
Cu
C
C of ligands in terms of maximum number of dative bonds that can be
Table 9.6.5 Classification
C
C
O
O ion
made to the central metal
per molecule of the ligand.
O
O
O
O
H2NCH2CH2NH2 (ethanediamine )
H
H
C
H
C
NH2
binding sites
NH2
H
Hexadentate
(polydentate)
[Cu(C2O4)2]
EDTA (ethanediamine -tetraacetate)
CH2COOH
N
H
C
H
C
H
CH2COOCH2COO-
N
[Cu(EDTA)]2Deep blue
CH2COO-
1. Complexes can be positively charged, negatively charged, or they can be
neutral
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
462
The net charge on the complex depends on the charge on the ligands and the metal cation. Consider the
hexa-aqua complex [Cu (H₂O)6] ²⁺, which contains a +2 metal cation and 6 neutral water ligands. The net
charge on the complex is equal to charge on the metal ion, since the ligands do not neutralize the positive
charge carried by the cation. Now suppose that the water ligands are replaced by four chloride ligands to
form the complex CuCl4x . The net charge x on the complex is determined by adding the total negative
charge from the ligands to the positive charge on the metal cation.
x = charge on cation + total charge from ligands = 2 + (-4) = -2
The complex therefore has the formula CuCl42-. All it means is that negative charge in the complex is in
excess by 2. Neutral complexes are usually insoluble in water, and from the way their formulae are
written, it may not be apparent that they are complexes. Consider what happens when NaOH (aq) is added
to an aqueous solution containing Cu 2+ ions. The reaction taking place is usually conveniently
written as
Cu2+ (aq) + 2OH-(aq) → Cu(OH) 2(s) (light blue precipitate)
Written like this, the equation is misleading in two ways:
 It does not show that this is in fact a ligand exchange reaction.
 It does not easily reveal that Cu(OH)2(s) is a neutral complex (insoluble, so it appears as a
precipitate).
The reader now knows that a formula such as Cu2+ (aq) is a convenient way of representing an aqua
complex, in this case, [Cu(H2O)6]2+. The reaction of Cu(II) with OH- ions therefore involves ligand
replacement; two water molecules in the aqua complex are replaced by OH - ligands.
[Cu(H2O)6]2+(aq) + 2OH-(aq) → Cu [(OH) 2(H2O)4](s) (light blue precipitate)
The product formed, Cu [(OH) 2(H2O) 4] (s), is a neutral complex. The net charge on the complex is given
by
charge on metal ion + total charge from two OH- ligands + total charge from four water ligands
= 2 + (-2) + 4(0) = 0
In a neutral complex such as Cu [(OH) 2(H2O)4], it is normal practice to ignore the water ligands and just
show the formula as Cu(OH) 2.
Q
A
Determine the charge on the metal ion in the following complex
[Mn (C2O4)3]3-
Recall that the ligand (oxalate) in this complex is has a charge of -2 (see Table 9.6.5).
Since there are three ligands, they contribute a total charge of -2 x 3 = -6. Let the charge on the
metal be x, then
Net charge on complex = total charge from ligands + charge on metal
But net charge on the complex is -3, so
-3 = -6 + x, x = +3
The charge on the metal is therefore +3. In fact, this is the oxidation state of manganese in this
complex.
2. The co-ordination number of a complex determines its shape
The co-ordination number of a complex refers to the total number of dative bonds around the central
metal ion.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
463
The co-ordination number need not be equal to number of ligands, for example, only one molecule of EDTA,
which is a hexadentate ligand, can make six bonds to the cation, giving a co-ordination number of six. The
smallest possible co-ordination number in a complex is 2, and this gives rise to linear complexes, for
example the diamine silver (I) ion, [Ag (NH3)2]2+. The number of dative bonds (bond pairs) around the
central metal ion determines the shape of the complex. This is a direct application of the VSEPR method
for the determination of molecular geometries. The reader needs to remember that dative bonds differ only
from other covalent bonds in how they are formed. Once formed, a dative bond cannot be distinguished
from any other covalent bond, in other words, it becomes just another bond pair of electrons. In the diamine
silver (I) complex shown below, there are two bond pairs around the central Ag+ ion. These bond pairs must
be spaced out at 1800 relative to each other. This is the optimum geometry that minimizes inter-electronic
repulsion between the two bond pairs. The complex is therefore linear.
H3N
1800
Ag
NH3
Co-ordination number 4
A co-ordination number of four gives rise to tetrahedral or square planar complexes. Examples are given
below. Note that [Pt(NH3)2Cl2] is a mixed ligand complex. There are two NH3 ligands and two Cl- ligands.
[Pt(NH3)2Cl2]
[CuCl4]2-
NH3
Cu
Pt
Cl
2-
Cl
Cl
Cl
NH3
square planar complex - bond angle 900
Cl
Cl
tetrahedral complex - bond angle 1090
Fig 9.6.8 Complexes with a co-ordination number of may show a square planar
or tetrahedral geometry.
Co-ordination number 6
This gives rise to an octahedral geometry (bond angle, 900). Examples of such complexes include
[Cu (EDTA)] 2- , [Mn (C2O4)3]3- and [Cu(en)3]2+ where en = ethanediamine. Fig 9.6.9 shows the geometry
of [Mn(C2O4)3]3-. Note that each ligand in the complex is bidentate so there is a total of 3 ligands x 2 = 6
bonds around the central metal ion. [Cu (EDTA)] 2- contains one ligand, EDTA, which makes six bonds to
the central metal ion.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
464
O
O
O
C
C
O
O
O
Pt
C
O
C
O
Fig 9.6.9 An octahedral complex of Mn
containing three C2O42- ligands
O
O
C
O
C
O
Paramagnetism in complexes of transition elements
Many complexes of transition elements are paramagnetic. This occurs if there are unpaired d electrons in
the transition metal ion. The strength of paramagnetism depends on the number of unpaired electrons. The
greater the number, the greater the paramagnetism.
Q
A
Work out the number of unpaired electrons in each of the following complexes. Hence
arrange the complexes in order of increasing paramagnetism.
(a)
[Mn(C2O4)3]3-
(d)
VO(H2O)5(OH)]+
(b)
CuCl₄²⁻
(e)
(c) [V(H₂O)₆]²⁺
[ Ni(NH₃)₆]²⁺
(a) First determine the oxidation number (charge) on the central metal ion, as this tells us
how many electrons were lost by the metal in forming the complex.
Let the charge on the metal be x.
x + 3(-2) = -3
x = +3
Therefore Mn lost 3 electrons in forming the complex. We deduct these three electrons
from the ground state configuration.
[Ar] 3d5 4s2
-3e-
[Ar] 3d
4
ground state
When a transition metal forms a compound, the 3d orbitals split in energy. In octahedral
complexes, the splitting results in the dxy , dxz and dyz occupying the lower energy level,
whilst the dz2 and the dx2- y2 occupies the higher energy level. Mn in [Mn(C2O4)3]3- has
four d electrons, which are distributed in the orbitals as shown below (the lower orbitals
are preferably occupied since they are more stable).
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
465
d x 2 - y2
dxy
dxz
d z2
dyz
In this case, electrons enter only the lower set of d orbitals because C2O42-, like most bidentate
and polydentate ligands, is a strong ligand which results in a large separation between the
lower and the higher d orbital sets. A large amount of energy would be required to place an
electron in the higher orbital set. In the rest of the complexes in the question, the ligands are
weak and so the energy difference between the two energy levels is small. Electrons therefore
first enter singly in all the orbitals (both lower and higher) , thereafter pairing may take place,
starting with the lower set of orbitals.
The complex [Mn (C2O4)3]3- therefore contains two unpaired electrons. It is paramagnetic.
The reader should determine the number of unpaired electrons in the remaining
complexes and determine which one is the most paramagnetic.
Stability of complexes
Stability of a complex depends on the strength of the dative bonds which the ligands make to the central
ion. In turn, the strength of these bonds depend on
 the nature of the ligand
 charge on the ligand and
 the charge on the central transition metal ion
In comparing the stabilities of complexes of the same transition element in the same oxidation state, the
important factor determining stability is the nature of and charge on the ligand.
Stability constants, Kstab
Different ligands have different affinities for the central metal cation. When a salt of a transition metal
dissolves in a large amount of water , water molecules act as ligands and form a complex with the transition
metal ion, for example [Cu(H2O)6]2+. Now suppose that a different ligand, for example, NH3, is added to the
solution containing the complex [Cu (H2O)6]2+. The introduced ligand may or may not be able to replace
the water ligands in the complex. If the new ligand can form a complex which is more stable than the aqua
complex, then ligand exchange takes place. Water ligands are replaced and a more stable complex is formed.
However, if the introduced ligand forms a complex which is less stable than the aqua complex, no ligand
exchange takes place.
Stability constants are equilibrium constants that give a measure of the ease with which a particular
ligand can replace water from a specified aqua complex.

A very large Kstab value indicates that the ligand easily replaces water ligands in the specified
complex, to form a more stable complex.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
466

Stability constants are true equilibrium constants, so they are affected by temperature.
The values are therefore quoted at 298 K.
The Kstab for the complex [Co(NH3)6]3+ is 5 x 1033. This value corresponds with the following ligand
exchange equilibrium
[Co(H2O)6]3+(aq) + 6NH3(aq) Ý [Co(NH3)6]3+ + 6H2O(l)
The equilibrium constant, Kstab for this reaction is given by
Kstab
=
[{Co(NH3)6}]3+[H2O]6
... (i)
[{Co(H2O)6}]3+[NH3]6
Since the Kstab value is much greater than zero, we infer that the numerator in equation (i) is much larger
than the denominator. In other words, the concentration of the amino complex at equilibrium is much
greater than that of the aqua complex. This shows that the reaction is favoured in the forward direction,
that is, NH3 ligands can easily replace water ligands in [Co(H2O)6]3+(aq) , to form a more stable complex,
[Co(NH3)6]3+. Stability constants are therefore useful in predicting and explaining ligand exchange
reactions. For example, we can use the stability constants given in Table 9.6.6 to predict some reactions of
copper (II) ions in aqueous solution.
Consider what happens when aqueous NH3 is added gradually to a blue solution of copper (II) sulphate,
until in excess, followed by a solution of EDTA.
Upon addition of a little aqueous ammonia, the test solution forms a sky blue precipitate. This precipitate
dissolves in excess ammonia to form a deep blue solution. Upon addition of EDTA, the solution turns even
deeper blue.
Explanation
Aqueous ammonia is actually an equilibrium mixture of NH₃ molecules, water molecules, hydroxide ions
and ammonium ions.
NH₃ (aq) + H₂O (l) ⇋ NH₄⁺ (aq) + OH⁻ (aq) ... (i)
Three of the species in this equilibrium are all ligands and they will have competition for the central metal
ion. The sky blue colour of CuSO₄ solution is due to the aqua complex, [Cu(H₂O)₆]²⁺. When a little aqueous
ammonia is added, the hydroxide ions in equilibrium (i) displaces water ligands from the complex in a
ligand exchange reaction, resulting in formation of a sky blue precipitate of copper (II)
hydroxide.
[Cu(H₂O)₆]² ⁺ + 2OH⁻(aq) ⇋ [Cu(H₂O)₄(OH)₂] + 2H₂O(l) ... (ii)
This reaction occurs because OH- ions are ligands than H2O. However, note that not all of the water ligands
are replaced. The precipitate formed, [Cu(H₂O)₄(OH)₂], is usually wrongly written as Cu(OH)₂ . When NH3
is added in excess, the blue precipitate redissolves, giving a deep blue solution. This time the increase in the
concentration of free NH3 molecules results in a displacement of the OH- ligands , forming the deep blue
complex [Cu(NH3)₄(H2O)₂]2+.
Addition of EDTA results in a further ligand exchange reaction. If you check in Table 9.6.6, you will find
that [Cu(EDTA)]2-, has a very large Kstab . EDTA is a stronger field ligand than H2O , OH- and NH3. EDTA
is therefore able to replace NH3 in the complex [Cu(NH3)₄(H2O)₂]2+ to form the very stable octahedral
complex [Cu(EDTA)]2-. The scheme below shows ligand exchange reactions involving complexes of copper.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
f
he formation of
sponding
e only ligand .
s that are easy
467
[Cu(H2O)6]2+
sky blue solution
NH3(aq)
[Cu(OH)2(H2O)4](s)
xsNH3(aq)
light blue precipitate
[Cu(NH 3)4(H2O)2]2+
deep blue solution
EDTA(aq)
[Cu(EDTA)]2-(aq)
deeper blue solution
It might be a useful exercise to try and rank ligands acComplex
Kstab
Log Kstab
cording to the ease with which they can replace water in a
ligand.
However,
this
not
215 easy because
[Fe(H
2.2 xis10
15.3 the ease with
2O)F5]
which a ligand can replace water depends on the
-2
complex.
Compare8the
constants
[FeCl
x10stability
-1.1 for the
4]
complexes [FeCl4]- and CuCl42-. Note that the stability
424
[Fe(CN)
1 x 10In
24
6] are different.
constants
the first complex,
Cl- ligands
are not likely to replace water because of the very small
3[Fe(CN)
1 xconstant.
1031
31 the K for
6]
value of
the stability
However,
stab
- ligands can
CuCl42- is3+well above zero,
indicating
that
Cl
[Co(NH3)6]
5 x 1033
33.7
easily replace water in this specific complex.
3Despite
to make some
[Co(CN)
1 x 1064 it is still possible
64
6] this complication,
useful generalizations:
CuCl424.2 x 105
5.6
 Complexes containing CN- ligands usually have
very large stability constants. In most cases, CNligands act as strong field ligands, that is, they
2+
to the central metal
[Ni(en)3] form very6strong
x 1018dative bonds
18.8
ion. The resulting complexes are very stable. We
2+
19
- can
[Cu(H2O)2(en)
x 10
19.6 water in many
2] that 4
infer
CN
easily replace
complexes
 Bidentate and hexadentate ligands form complexes
that have18large stability constants. Such li2[Cu(EDTA)]gands,
x 10
18.8
for6 example
en and
EDTA can easily replace water ligands in a large number of complexes.
Chelating agents
EDTA is an example of a chelating agent. A chelating agent is a ligand which forms a complex which is
so stable that further substitution of the ligand by monodentate ligands is unlikely. Chelating agents are
usually bidentate or polydentate ligands such as EDTA. A complex formed in this way is known as a
chelate, for example [Cu(EDTA)]². Chelating agents such as EDTA2- have found use in water purification
where they remove certain poisonous metal ions by forming stable complexes with the ions.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
468
4.3 Transition metals tend to form coloured ions and compounds
Formation of coloured compounds is common among, but not unique to transition metals. The colour of a
transition compound largely depends on



the strength of the ligand
which metal ion is present in the complex
the charge on the central metal ion.
The origin of colour in transition metal compounds
The neutral transition atom has degenerate (same energy) d orbitals. Upon compound formation, the d
orbitals split in energy, forming a higher set and a lower set of orbitals. This separation in energy is a result
of the electrostatic repulsion which the d orbitals experience from the approaching ligands. In octahedral
complexes, which are by far the most common, the dxy, dxz and dzy makes the lower set (more stable). This
is because their lobes are oriented between the axes, so they feel a smaller repulsion from the ligands, which
approach along the axes. This set is referred to as the t₂g set. The dz² and the dx²-y² (eg set) are higher in
energy. This is because their lobes are along the axes. They experience a strong repulsion from the ligands,
which also approach along the axes.
eg set
In the neutral transition
atom, all five d orbitals are
degenerate.
t2g set
All five d orbitals are raised in energy due to
repulsion by ligands. The eg set experiences
the greater repulsion since ligands (L)
approach along the axes, in line with the
lobes of the orbitals. These orbitals are
therefore higher in energy (less stable).
The t2g set is lower and more stable. This is
because its lobes are oriented between axes,
and so experiences less repulsion from
ligands, which approach along axes.
Fig 9.6.10 Splitting of d orbitals in an octahedral complex. ∆0 is the energy
difference between the higher and the lower set of orbitals.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
469
The separation in energy between the two sets of d orbitals is ∆0. The size of ∆0 for a particular transition
metal ion depends on the strength of the ligand. Strong field ligands result in large values of ∆0.
The splitting in a tetrahedral complex is an inversion of the case for octahedral complexes. The t 2g set is
higher in energy than the e g, and again this has to do with the direction in which ligands approach the d
orbitals.
dxy dxz dzy

d z 2 d x 2 - y2
dxy d
xz
dzy
Fig 9.6.11 Splitting in a
tetrahedral complex
d z 2 d x 2 - y2
When light strikes the transition compound, some wavelengths from the visible spectrum are absorbed. The
absorbed energy is used to promote electrons from the lower set of orbitals to the higher set. Light which is
reflected back to the eye therefore has fewer wavelengths, which mix to give a colour which is observed. The
colour observed is complimentary to the wavelengths absorbed, for example, red is complimentary to blue.
Cu2+ (aq) (= [Cu(H2O)6]2+ appears blue because it absorbs from the red region of visible light (actually ,
orange and yellow are also absorbed since the absorption is not sharp; a range of wavelengths is absorbed).
A necessary condition for a complex to be coloured is the presence of vacancies (unpaired electrons or
vacant orbitals) in the higher set of orbitals. These vacancies can accommodate electrons that are promoted
from the lower set of orbitals by absorption of light energy. In [Cu(H 2O)6]2+ , Cu2+ has a d9 configuration.
This hexaaqua complex has the following arrangement of electrons in the d orbitals of Cu 2+.
2
2
d z2 d x - y
vacancy
one electron can be promoted form
any lower orbital
dxy
dxz
Fig 9.6.12 Splitting of orbitals
in Cu2+, which has a d9
configuration.
dzy
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
470
V
I
B
G
Y
O
R
1. White (visible) light has seven components (wavelengths). If the
components were viewed separately they would have the following
colours: Violet, Indigo, Blue, Green, Yellow, Orange and Red.
Mixing these seven colours gives white light.
2
1
V
IB
B
3
G
Y
O
4
3. The eye sees an object as a result of light reflected to it from the
object. In the case of [Cu (H2O)6]2+, the light reflected to the eye has a
missing component, the red wavelength.
4. The colour seen by the eye is a compliment of the absorbed
colour(s). In the case of [Cu (H2O)6]2+, the eye sees blue, which is
complimentary to red.
eye
Fig 9.6.13
2. If a compound has vacancies in the higher set of d orbitals, one or
more components of white light (energy) are absorbed and used to
promote one or more electrons from the lower energy level to the
higher. [Cu (H2O)6]2+(aq) , which is an octahedral complex, absorbs
the red component from white light.
The origin of the blue colour in Cu2+ (aq)
From this account, it should be easy to see why some compounds of the d block of the Periodic Table are
white when solid and colourless in solution. Complexes of Zn (II) such as [Zn(H₂O)₆]²⁺(aq) are colourless.
The reason is that all the d orbitals (the lower and upper sets) are fully occupied (Zn²⁺ = d₁0). There are no
vacancies in the upper energy level to accommodate electrons from the lower energy orbital (Fig 9.6.14(a)).
Compounds of Sc (III) lack colour for a different reason. The d orbitals are completely empty (d0). Since
there are no possible electron transitions, light is not absorbed (Fig 9.6.14(b)). The light which is reflected
back to the eye has a full complement of wavelengths, so the eye will see it as white light, that is, the
compound would appear colourless or white.
2
2
d z2 d x - y
dxy
dxz dzy
(a)
Prepublication manuscript
2
2
d z2 d x - y
dxy
dxz dzy
Fig 9.6.14 (a) Zn (II) compounds are colourless or
white because they have a d10 configuration .
(b) Compounds of Sc (III) are colourless for a
different reason; the d orbitals are completely
empty. In both cases d-d transitions are not
possible
(b)
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
471
Colours of common transition metal compounds
Remember to treat the ions Mn+ (aq) as compounds of the formula [M(H2O)6]n+. All ions in the following
table are in aqueous solution.
Element
Compound/ion
colour
Oxidation
state
of element
Comments
Vanadium
VO2+
VO2+
V3+
V2+
Yellow
Blue
Green
Violet
+5
+4
+3
+2
Vanadium is named after Vanadis,
the Scandinavian goddess of beauty.
This is because all of its stable
oxidation states exhibit attractive
colours in aqueous solution.
Chromium
Cr3+
Green
+3
Acidic in solution due to salt
hydrolysis
CrO42-
Yellow
+6
Cr2O72-
Orange
+6
Dominant in alkaline solution
Dominant in acidic solution
Manganese
Cu
Mn2+
Pale
pink
+2
Easy to miss the colour if solution is
too dilute
MnO4-
Purple
+7
Intense colour is due to charge
transfer, not d-d transitions
MnO2(s)
Black
+4
A catalyst for the decomposition of
hydrogen peroxide.
Cu2+
Light
blue
+2
CuO(s)
Black
+2
CuCl42-
Yellow
+2
Deep
blue
+2
red
+1
[Cu(NH3)4(H2O)2+]
Cu2O(s)
Prepublication manuscript
CuCl42- is formed when concentrated
HCl or concentrated NH4Cl is added
to a solution containing Cu2+.
A precipitate of red copper (I) oxide,
Cu2O, is formed during the test for
aldehydes, for example glucose.
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
472
Fe
4.4
Fe2+
Light
green
+2
Fe3+
Yellow
brown
+3
Acidic in aqueous solution due to salt
hydrolysis
[Fe(CN6)]4-
Yellow
+2
Gives dark blue colour with Fe3+(aq)
[Fe(CN6)]3-
yellow
+3
Gives dark blue precipitate with Fe2+
Catalysis by transition elements and their compounds
Transition metals and their compounds are associated with catalytic activity. The explanation lies in the
presence of partially filled d orbitals and the ability of the transition element to change oxidation state
during the catalysis.
Heterogeneous catalysis is when the catalyst and the reactant are in different phases. Well known cases
are when the catalyst is solid and the reactants are gaseous. An example is the use of solid iron in the Haber
process for the manufacture of ammonia from nitrogen and hydrogen gases. The partially filled d orbitals
in the transition metal are used to temporarily accommodate electron density from the reactants during the
reaction. The adsorption of reactant molecules onto the surface of the catalyst weakens their bonds, thus
lowering the activation energy of the reaction. By adsorbing the reactants, the catalyst also brings reactant
molecules together so that bonds can form.
The contact process for the manufacture of sulphuric acid also involves heterogeneous catalysis. The
catalyst, solid vanadium (V) oxide, V2O5, participates in the conversion of SO2 to SO3.
SO2 (g) + ½ O2 (g)
Ý
SO3 (g)
450-5000C
1 atm
The mode of action of the catalyst here is thought to involve dissociation to produce O2, which then
facilitates the oxidation of SO2 to SO3.
V2O5(s)
Ý 2V2O4(s) + O2 (g)
Note that the oxidation of V decreases from +5 to +4, that is, vanadium undergoes reduction. The catalyst
is recovered when V2O4 combines with oxygen from the atmosphere, resulting in the oxidation of
vanadium back to the +5 state.
V2O4(s) + ½ O2 (g) Ý V2O5(s)
The ability of vanadium to change its oxidation state is important in this catalysis.
A homogenous catalyst is one which is in the same phase as the reactants. A commonly quoted example
is the catalysis of the reaction between peroxodisulphate and iodide ions by Fe (III) or Fe (II) ions in
aqueous solution (see section 8 for details of this reaction).
S₂O₈²⁻ (aq) + 2I⁻(aq) → SO₄²⁻(aq) + I₂(aq)
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
473
This redox reaction is judged to be energetically feasible by consideration of its E cell value, which is large
and positive. However, the reaction is very slow because the ions involved are likely charged and so instead
of colliding to form products, they tend to repel. The reaction therefore has very high activation energy.
That is, a very large amount of energy must be supplied to force the reactants to collide. The
catalyst provides a new reaction pathway involving oppositely charged ions which therefore collide easily
by mutual attraction. The catalyst is able to carry out the catalysis because it has the ability to vary its oxidation state. During the course of the reaction, the catalyst reacts just as if it were one of the reactants, and by
so doing it changes its oxidation state. At the end of the reaction, the catalyst reverts back to its original
oxidation number.
The ability of a transition element or its compounds to participate in catalysis can be explained in
two ways.

The transition metal has the ability to switch between oxidation states. This allows the
catalyst to participate as a reactant in the catalysis, and to be converted back to the
original form at the end of the reaction.

The presence of partially filled d orbitals may allow the reactant molecules to temporarily
adsorb to the surface of the catalyst.
9.6.5 UV - Vis spectroscopy of transition complexes
UV –Vis spectroscopy (ultraviolet-visible spectroscopy) is a useful procedure for the
determination of chemical species that can absorb the ultraviolet or visible portion of the electromagnetic
spectrum. Absorption of UV or visible radiation by complexes is a result of the presence of unpaired
electrons in the d orbitals.
As already explained, energy is absorbed in promoting one or more electrons from the lower set of d orbitals
to the higher set in a complex. Every complex absorbs only within a certain range of energies, so its
absorption spectrum acts like a fingerprint that can help to identify it. The absorption of energy is not sharp,
that is, a complex will not absorb a single frequency of energy, but will absorb within a well defined range.
However, each complex will has a particular wavelength of energy at which it will absorb the most. This
point of maximum absorption can be used to identify the complex present in the solution. Fig 9.6.15 shows
the key features of a UV - Vis spectrometer. This instrument measures the amount of light
absorbed by a compound as a function of wavelength. Remember that wavelength is related to energy.
The longer the wavelength, the smaller the energy of the radiation, and vice versa.

Suppose that a reaction occurs in which a complex A undergoes ligand exchange to form complex
B, which has a greater absorption. First the absorption spectrum of pure A is obtained. The reaction
is then allowed to proceed. In this case, the recorder will trace out increasing absorbances due to
the formation of product B. At the end of the reaction, a maximum value of absorbance is obtained
due to the presence of a maximum yield of the product B. The graph then levels out as there is no
further reaction taking place. The rate of reaction is the same as the rate of change of absorbance
due to the formation of the product. To find this rate, values of absorbances at
different times are recorded. The rate of reaction at a given time is determined by taking the slope
of the tangent at that time.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
474
Fig 9.6.15 Principles of operation of a UV-Vis spectrometer (also known as a
spectrophotometer)
1. The source produces light with a range of wavelengths.
2. The monochromatic selects light of the same wavelength.
3. This light is split into two beams, a reference beam Ir and a sample beam Is. The sample beam passes
through a sample of the compound (complex) being analyzed. The reference beam passes through a
reference cell which contains a solvent only, for example water. The solvent used must be the same as the
one used in the sample cell.
4. The sample absorbs some energy, and the detector measures the intensity ratio of the transmitted
reference (Ir) beam to the sample beam, Is. The logarithm of this ratio is known as the absorbance, A.
A = log (
𝐈𝐫
𝐈𝐬
)
If the sample does not absorb at a certain wavelength, then Ir = Is, so A = log (1/1) = 0
If the sample absorbs, then Ir > Is, and absorbance is greater than zero.
5. The spectrometer continuously scans the wavelengths in the UV-Vis spectrum, and a chart recorder
draws a graph known as a UV-Vis spectrum. This graph measures absorbance as a function of
wavelength .
UV-Vis spectroscopy can be used in the following ways


identifying a complex present in solution, since different complexes have different maximum
absorbances.
determining rates for reactions which are accompanied by ligand exchange or complex formation.
The product of the reaction will have a different absorption spectrum from the reactant.
Example
Zinc forms a complex with the ligand PAR
Zn2+ (aq) + 2PAR  [Zn(PAR)2] ... (i)
The rate of reaction (i) has been studied.
Both PAR and its zinc complex absorb radiation in the UV-visible region. The following diagram shows
their absorption spectra.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
475
Fig 9.6.16 Absorption spectra of [Zn(H2O)6]2+ and the complex
[Zn(PAR)2].
During the course of the reaction, the absorbance of the solution increases due to the formation of the
complex Zn-PAR. When the reaction is complete, the UV-Vis spectrum reaches a maximum of about 1.2,
which is the maximum absorbance of the complex. An absorbance - time graph therefore shows the rate of
formation of the complex. The rate of reaction at time t is the slope of the tangent at that point (Fig 9.6.17).
The experiment can be repeated using different concentrations of Zn 2+ or the ligand PAR, and noting the
change of the slope (rate) at time t.
y
absorbance
x
Fig 9.6.17 Absorbance - time graph
for the reaction between Zn2+(aq) and
PAR.
0
t1
time
rate of reaction at time t = rate of change of absorbance
= y/x
Transition metal ions of biological importance
Some transition metal ions are important for the efficient running of biological processes. Examples include
Cu2+ in the respiratory system, Co 2+ in vitamins and Fe2+, which is responsible for the oxygen binding
properties of haemoglobin (See section 4.1 ).
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
476
Questions, solutions and discussions
Q1
(a) (i) State one alloy that contains nickel, and give its use.
(ii) State an example of an industrially important reaction that uses nickel as a catalyst.
(b)
The ‘crown ether’ A and the ‘crown thioether’ B are polydentate ligands that are used to
remove harmful metals from the environment.
Compound A complexes with ions of
metals on the left of the Periodic Table,
and can be used to extract radioactive
strontium ions from groundwater.
(Complexes formed by the ions of Group
II metals are similar to the complexes
formed by transition metal ions.)
Compound B complexes well with ions
of metals that are towards the right of
the Periodic Table, and can be used to
extract nickel, copper and lead ions from
solutions of industrial waste.
(i) Explain what is meant by a polydentate ligand.
(ii) What features of both compound A and compound B make them suitable to act as ligands?
(iii) Suggest the co-ordination number in, and the shape of, the complex formed between A
and Sr2+ ions.
(iv)
(c)
Suggest the co-ordination number in, and the shape of, the complex formed between B
and Ni2+ ions.
The complex [Ni(R3P)2Br2] can exist in two isomeric forms. Draw structural formulae to
show the shapes of these isomers, and describe the type of isomerism shown. [The R group is
an organic group such as methyl, –CH3.]
9701/06/O/N/2006
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
477
Solutions
(a) (i) AlNiCo – used in magnets [ alloy of Al, Ni and cobalt- used to make super magnets]. This alloy is
corrosion resistant.
Nichrome –used for high resistance wire
Cupro-nickel – used to make coins
(ii) Hydrogenation of vegetable oils/ hydrogenation of alkenes to alkanes.
(b) (i) Polydentate – forms more than one dative bonds per molecule of ligand.
(ii) They contain oxygen and sulphur atoms which bear lone pairs of electrons.
(iii) Coordination number 6, giving rise to an octahedral geometry. [ since there are six binding sites = 6
oxygen atoms].
(iv) Coordination number 4 , giving rise to a tetrahedral / square planar geometry.
(c) A form of stereoisomerism known as geometric or cis-trans isomerism is exhibited.
Br
R3P
Ni
Ni
R3P
Br
Br
R3P
trans
Q2
Br
R3P
cis
(a) Explain what is meant by the term transition element.
(b) Complete the electronic configuration of
(i) the vanadium atom, 1s22s22p6
(ii) the Cu2+ ion. 1s22s22p6
(c) List the four most likely oxidation states of vanadium.
(d) Describe what you would see, and explain what happens, when dilute aqueous ammonia is added
to a solution containing Cu 2+ ions, until the ammonia is in an excess.
(e) Copper powder dissolves in an acidified solution of sodium vanadate (V), NaVO3, to produce a
blue solution containing VO 2+ and Cu2+ ions.
By using suitable half-equations from the Data Booklet, construct a balanced equation for this
reaction.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
478
Solutions
2 (a) A d-block element which forms at least one ion in which the d orbitals are partially filled.
(b) (i) (1s2 2s2 2p6) 3s2 3p6 3d3 4s2
(ii) (1s2 2s2 2p6) 3s2 3p6 3d9
(c)
+2, +3, +4, +5
(d) Pale blue solution forms a light blue (cyan) precipitate which dissolves in excess NH3 (aq) to give a
deep blue solution.
Blue precipitate is Cu(OH)2
The deep blue solution contains the complex [Cu(NH3)4(H2O)2]2+ formed by ligand replacement .
(e) Exercise to the reader
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
9.6in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
1
(a) Explain the following observations using relevant Eθ values from the Data Booklet, and ideas
of the relative stabilities of complexes in the presence of different ligands.
(i) Solutions containing Fe2+ (aq) slowly oxidize to Fe3+ (aq) in air.
(ii) The precipitate formed when aqueous sodium hydroxide is added to a solution containing
Fe2+ (aq) rapidly turns brown in air.
(b) A 6.95 g sample of FeSO4.7H2O was dissolved in water and the volume of solution made up to
250 cm3. The solution was stored in an open container for some time and suffered partial
oxidation. On titration of a 25.0 cm3 portion of the partially oxidized solution with 0.020
mol dm–3 KMnO4, it was found that 20.5 cm3 of oxidant were required to reach the endpoint.
Calculate the percentage of Fe2+ ions that had been oxidized.
2
Iron metal and its compounds are useful catalysts in certain reactions.
(a) Apart from its catalytic activity, state two properties of iron or its compounds that show
that it is a transition element.
(b) You are provided with a solution of KMnO4 of known concentration in a burette. Outline how
you could use this solution to find out the concentration of Fe 2+ (aq) in a solution. You should
include relevant equations for any reactions you describe.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
479
(c) For each of the following equations, write the oxidation number of the element printed in bold
underneath its symbol, and balance the equation by adding appropriate numbers before each
species.
(i) ........ MnO4– + ........ SO2 + ........ H2O → ........ Mn2+ + ........ SO42- + …… H+
oxidation numbers:
........
........
........
........
(ii) ........ Cr2O72– + ........ NO2 + ........ H+ → ........ Cr3+ + ........ NO3– + ........ H2O
oxidation numbers:
........
........
........
........
(d) Outline the role that Fe3+ ions play in catalyzing the reaction between iodide ions and
peroxodisulphate (VI) ions.
2I– + S2O82-  I2 + 2SO429701/42/O/N/2009
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
480
Section 3
Organic
Chemistry
Introduction
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
481
Organic chemistry is the study of organic compounds. An organic compound is one that contains
C and H or
C, H and any other atom, for example O, N, S, Cl, Br.
An organic compound which contains C and H only is known as a hydrocarbon. The key element in organic
chemistry is carbon. Organic chemistry is essentially the study of carbon compounds. There are several
millions of known organic compounds, and every day new ones are either being discovered or synthesized.
Carbon compounds occur widely in living things and in the environment. Many synthetic compounds of
carbon exist, and these include drugs, dyes, plastics and fabrics.
Carbon has certain unique properties which allow it to form so many stable compounds.
 it can catenate. This means that it can form relatively long chains and rings by forming strong and
stable C-C bonds. These chains may also contain C to C double or triple bonds, which are also very
strong and stable. A wide variety of organic compounds is possible because the chains and rings
can have different sizes.
 It can form four bonds. This is important in the formation of side chains. There are many possible
ways in which this branching can take place, and once more, this increases the number of
compounds that carbon can form.
10.1
The structure of organic compounds
Before you study in depth the chemical and physical properties of organic compounds, you need to
understand their structure first. An organic compound is mainly made up of carbon and hydrogen. These
two elements are always present in any organic compound. However, many organic compounds also
contain other atoms such as O, N, S, Cl and so on. These atoms, which we may refer to as hetero atoms,
occur at regions of the organic molecule known as functional groups. Note, though, that a C to C double or
triple bond is also a functional group, though it contains no hetero atom. The concept of functional group
is very important in organic chemistry because it helps us to predict the chemical behaviour of an organic
compound.
A functional group is the reactive centre of an organic molecule. It is an atom or a group of atoms that
confers reactivity to the organic molecule.
A functional group is an atom or group of atoms that is responsible for the chemical reactions of
an organic compound.
An organic compound that lacks a functional group is known as an alkane. Such a compound is saturated
(it has no C to C double or triple bonds) and it contains carbon and hydrogen only. Table 10.1 below shows
the functional groups that you should study.
10.1.1
Classification of organic compounds
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
482
Organic compounds are classified into groups on the basis of the functional group present. Compounds
with the same functional group fall into the same group, which is known as a homologous series. Of
course, if the compound has more than one functional groups, then it cannot be classified into any one
group, for example, a compound could be an alcohol and a carboxylic acid at the same time. In Table 10.1
below we assume that the compounds have only one functional group (or none at all).
Table 10.1
Functional groups and representative compounds
Homologous
series
Functional
group
Alkanes
None
arenes
HYDROCARBONS
Alkenes
Examples
methane, CH4
butane , CH3CH2CH2CH3
C=C
Carbon to carbon double bond
ethene, CH2 = CH2
propene, CH3CH=CHCH2
Benzene (aromatic) ring
benzene
CH3
methylbenzene
C-OH
(OH not attached to benzene ring)
H
H
There are three classes, primary,
secondary and tertiary alcohols.
C
Primary alcohol: OH
group is bonded to a
C atom that carries
OH
two hydrogen atoms.
H
methanol
Alcohols
CH3
CH3
C OH
H
propan-2-ol
CH3
CH3
C OH
CH3
Secondary alcohol:
OH group is bonded
to a C atom that
carries one hydrogen
atoms.
Tertiary alcohol: OH
group is bonded to a
C atom that does not
carry any hydrogen.
atoms.
2-methylpropan-2-ol
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
483
Phenols
OH
phenol,
OH
CH3
OH
4- methylphenol
Halogenoalkanes
C-X
(C is non aromatic.
X = F, Cl, Br, I)
H H
chloroethane,
H C
C
Cl
H H
Halogenoarene
Aldehydes
CARBONYL COMPOUNDS
Ketones
C-X
(C is aromatic. X = F, Cl, Br)
Br
bromobenzene,
O
O
C
C
O
C
H
Both
aldehydes and
ketones
contain the
carbonyl
group,
CH3
C
CH3
propanone
O
CH3
C H
ethanal
Carboxylic acids
O
C
OH
O
CH3 C
OH
ethanoic acid
O
Esters
C
O
O
CH3
C
O
CH3
ethyl ethanoate
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
484
O
Acylchlorides
C
O
Cl
CH3
C
Cl
ethanoylchloride
Amines
H
C
N
There are three classes, primary,
secondary and tertiary amines
CH3 N
primary amine
H
methylamine
In a secondary amine, the N is bonded to
only one hydrogen atom. In a tertiary
amine, the N is not bonded to any H
atoms (see amides below).
O
CH3 C
H
O
Amides
C
H primary amide : N is bonded
to two H atoms
N
ethanamide
N
There are three classes, primary,
secondary and tertiary amides
O
H secondary amide : N is
bonded to one H
N
atom
CH3 C
CH3
N-methylacetamide
O
CH3 C
CH3
N
CH3
Tertiary amide : N is not
bonded to any H
atom
N, N-dimethylacetamide
Homologous series
A homologous series is a group of organic compounds that share the same functional group. The term is
used particularly for:
 straight chain organic molecules.
 organic molecules with only one type of functional group.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
485
Properties of a homologous series
1. Members of the same homologous group share the same functional group.
2. Members of the same homologous group have similar chemical reactions. This is expected, since the
functional group is the weak center in an organic molecule. This fact makes the study of organic
chemistry a light task, for example, instead of studying all alcohols individually, the student need only
study the chemical behavior of the alcohol functional group.
3. One member of a homologous group differs from the next by a single –CH₂- group.
4. Going down the series from one member to the next, the relative molecular mass increases by 14, which
corresponds to the mass of an – CH₂- group which is added on going down the series.
5. The increase in mass down the series is accompanied by an increase in boiling and melting
points. As the number of electrons increases, so does the strength of the Van der Waals forces. More
energy would be needed to break these forces down the series.
6. The increase in strength of Van der Waals forces results in a change in state from gas through liquid to
solid. As the forces of attraction become stronger, the molecules are held more tightly in fixed
positions, and their kinetic energy decreases, that is, the compounds begin to assume a liquid state, and
then finally a solid state. Compare methane, which is a gas, with octane (liquid) and candle wax
(C₃0H₆₂), which is a solid at room temperature. These three compounds belong to the same homologous
series, the alkanes.
10.1.2
Structure and bonding
Here are important guidelines about the structure and bonding in organic compounds.
 Carbon always makes four covalent bonds with other atoms. Take particular care here, especially if
you draw a structure in which a carbon atom is associated directly with a double or triple bond, or
is bonded to a functional group.
 Nitrogen in most organic compounds makes three bonds to other atoms. The fourth position on the
nitrogen atom is occupied by a lone pair, just as in ammonia.
 Oxygen in most of its compounds makes two bonds to other atoms. The other two positions are
occupied by lone pairs, just as in water.
Study the following illustrations .
x
H
H
H
C
C
O
H
H H
ethanol
Q
A
H
H
H
C
C
H
H
H
y
N
H
ethylamine
What are the bond angles marked x and y?
X is 104.50 (as in water).
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
486
Y is 107.50 (as in ammonia).
Hint:
Do not be misled by the way the compounds have been drawn. Remember the existence of two lone pairs
of electrons on the oxygen atom in ethanol and one lone pair on the nitrogen atom in ethylamine. These
lone pairs affect the bond angle, according to the VSEPR theory.
Q
A
Circle and name all functional groups present in the anticancer drug Taxol®. The ‘broken
wedges’ in the diagram show bonds that are pointing to the back of the paper, away from
you. The ‘solid wedges’ show bonds that are pointing out of the paper, towards you.
There are 7 different functional groups, numbered 1 to 7 in the following diagram.
1. benzene ring (phenyl group)
5. Ketone
2. amide
6. C to C double bond/alkene
Prepublication manuscript
3. (secondary) alcohol
4. ester
7. ether (not in your syllabus)
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
487
A
3
5
2
4
6
7
3
3
1
4
4
10.1.3
Formulae of organic compounds
You will encounter the following types of formulae:
 Molecular

Structural

Displayed

Skeletal
The molecular formula is rarely used in organic chemistry because it frequently gives rise to a case of
ambiguity known as isomerism.
Consider the molecular formula C2H6O. This compound cannot be named with certainty, because it could
be ethanol or ether (dimethyl ether).
C2H6O
H
H
H
C
C O
H
H
H
H
H
H
C
O C
H
H
ether
Ethanol
H
The difference between the two compounds in the
diagram above is clearly seen using displayed
formulae. A displayed formula shows (displays) all
the bonds and how atoms are arranged relative to
each other. The displayed formula therefore helps us
to see exactly which atom is bonded to which. One
other way in which the displayed formula becomes
handy is in checking the valencies of the atoms used
in bonding.
Check from the diagrams above that carbon has four bonds around it, oxygen and hydrogen have 2 and 1
respectively.
The structural formula
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
488
The illustration in Fig 10.1 shows the relationship between structural and displayed formulae. Note that the
groups of atoms shown in the displayed formula also appear in the structural formula, but this time bonds
are not shown. The structural formula is the most used type of formula in organic chemistry because it
overcomes the ambiguity associated with the molecular formula and at the same time it has greater
convenience than the displayed formula.
H
H H
displayed
structural
Fig 10.1
H
C
C O
H
H
H
H
C
H
CH3CH2OH
H
O
C
H
H
CH 3O CH3
Relationship between structural and displayed formula
The skeletal formula
This is a shorthand notation for showing the structure of organic molecules. A shorthand formula is very
useful in drawing structures of complex molecules. The diagram below shows the skeletal formula for
propane.
The rules for drawing a skeletal formula are straightforward.
 Carbon atoms are not shown.
 An ‘elbow’ in the structure represents the position of a carbon atom. This position represents a
–CH2 group, unless if there is an associated branch or multiple bond. In these two cases, the
number of hydrogen atoms is reduced from 2 to satisfy the valency of C.
 Hydrogen atoms are not shown unless if they are directly bonded to a non-carbon atom, for
example, O, N and S.
 A carbon chain is shown by a ‘zig- zag’ line, with each ‘elbow’ representing the position of a carbon
atom. However, if the carbon chain is interrupted by a non carbon atom, for example O as in esters
and ethers, the non carbon atom must be shown.
 The end of a line segment represents a –CH3 group unless if there is a functional group (such as
the carbon to carbon double bond) associated with this carbon atom.
Fig 10.2 illustrates the relationship between structural formula and skeletal formula of propane.
CH3 CH2 CH3
end of line segment
Prepublication manuscript
end of line segment
elbow
Fig 10.2
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
489
Now consider the following skeletal formula.
end of line : CH3
end of line : CH3
4
1
2
elbow:
CH2
3
O
O
5
6
elbow with one
branch : CH
elbow: C with no H
7
end of line : CH3
elbow:
CH2





Carbon 3 has only one H atom because there is a branch here. This gives the carbon atom a
correct valency of four.
Do not forget that the position marked 4 is a -CH3 group.
Carbon 5 has no H atom. Notice that at this position the carbon atom already has four bonds
around it because of the presence of the C to O double bond.
The molecule has 7 carbon atoms, which have been marked 1 to 7. Each of these carbon atoms
can have 3, 2, 1 or 0 H atoms.
Reading from carbon 1, the structural formula of the molecule is
CH3CH2CH(CH3)OCOCH2CH3 ... (i)
5 6 7
1
2 3 4


Note that branches are shown in brackets, in this case the CH 3 branch bonded to carbon 3.
The molecular formula of the compound is therefore C7H14O2. This compound is an ester.
The structural formula of the compound could have been written from the other direction, that is,
from carbon 7. This would give
CH3CH2COOCH(CH3)CH2CH3 ... (ii)
2
3
4
1
7
6 5
Structures (i) and (ii) are exactly the same.
Q
Limonene is an oily liquid that occurs naturally in the peels of citrus fruits. Its structure
is shown in Fig 10.3. Calculate the mass of 2 moles of this compound.
limonene
Fig 10.3
A
The molecular formula of limonene is C10H16. Its Mr is 136.
mass = n x Mr = 136 x 2 = 272 g
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
490
CH3
C
CH2
=
CH2
CH
H
CH2
C
limonene
C
CH3
CH2
Skeletal formulae that involve the benzene ring
The benzene ring is one of the functional groups whose chemistry you will study in detail later on. The
simplest compound that contains the benzene ring is benzene itself. The formula of benzene is C6H6 . It is
a cyclic compound whose structure is shown below.
H
C
H
C
C
H
=
H
C
C
H
C
This ring represents
delocalized electrons
skeletal form ula
H
display ed form ula
From the displayed formula it appears that each carbon has only three covalent bonds, instead of four.
This is because the other electrons that would have constituted the ‘missing’ bonds are delocalized within
the molecule.
In a substituted benzene, the number of hydrogen atoms is less than 6, depending on the number of
substituents. In such compounds, one or more hydrogen atoms have been replaced by other groups, for
example, the methyl group in methylbenzene. The structure of methylbenzene and other substituted
benzenes are shown in the following panel. The formula of methylbenzene can be written as C 6H5CH3.
Written like this, we can see that one of the six H atoms of the benzene ring has been replaced by a CH 3
group. However, if you were asked to write the molecular formula, it would be C 7H8.
Panel 10.1
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
491
m ethy lbenzene
CH 3
0
1 20
H
CH 3
C
C
H
C
=
=
H
C
C
H
skeletal formula
C
display ed formula
C6H5CH3
H
NB The position on which the -CH3 group is drawn is not important.
4- chlorom ethy lbenzene
2,4,6 - trichlorophenol (T CP)
= C6H2 OH(Cl)3
= C H Cl(CH )
6 4
3
[TCP is the active ingredient
in Dettol]
two hydrogen atoms replaced
Q
four hydrogen atoms replaced
Draw displayed formulae from the following skeletal formulae. Hence deduce the
molecular formula of each compound. The benzene ring is acceptable in a displayed
formula.
O
(a)
O
(b)
O
(c)
OH
H
H C
H H
A
C
(a)
C
C
H H
O
H
H
H
C
C
C
H
C
(c)
H
H
C
Prepublication manuscript
H
H
H
(b)
C
H
H
C
H
C
H
H C
O
C
H
H
H
H
C
C H
H
H H
H
C
H
C12H22O2
C
C
H
H
C12H14O
H
C
O
O
H
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
C4H8O
H
H H
C
(a)
492
O
H C
H
H
C
C
C
C12H14O
(b)
C
H
H
H
C
H
C
H
H H
O
H
C
C
C
H
O
(c)
C
H
H
H
C H
H
C
H
H H
H
C
H
C
H
H
C
H
H
C
C
H
O
H
C4H8O
H
C12H22O2
H
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
Exercise
10.1in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
All of the following compounds are naturally occurring. Work out their molecular formulae
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript
(b)
H
H
H
H C
C
H
C
H
C12H14O
H
H
C
H
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
H
H
C
H
H
493
10.1.4
Isomerism
This is a phenomenon in which different compounds share the same molecular formula.
An illustration has already been given using ether and ethanol.
ISOMERISM
Structural
Stereoisomerism
chain(skeletal) positional
isomerism
isomerism



functional
group
isomerism
cis-trans(geometric)
isomerism
optical
isomerism
There are two main forms of isomerism, structural and stereoisomerism.
There are two forms of stereoisomerism, namely optical and geometric isomerism.
Three forms of structural isomerism are chain (skeletal) isomerism, functional group isomerism
and positional isomerism
Definitions
Stereoisomers are compounds that share the same molecular formula but differ in the spatial arrangement of groups
(spatial means ‘… in space’)
Structural isomers: compounds that share the same molecular formula but differ in which atom is connected to
which.
Optical isomers: these are non-superimposable mirror images of each other. Optical isomerism is a form of
stereoisomerism. Optical isomers differ in the direction in which they rotate the plane of polarized light. One of the
isomers rotates to the right, and the other to the left.
Cis-trans isomers: also known as geometric isomers. These are stereoisomers that share the same molecular
formula but differ in the spatial arrangement of groups about a rigid bond, for example, a carbon to carbon double
bond. Cis-trans isomerism is a result of restricted rotation about a rigid bond.
Functional group isomers: these are compounds that share the same molecular formula but differ in which
functional group is present. Well known cases exist between esters and carboxylic acids, ethers and alcohols and
ketones and aldehydes.
Positional isomers: Positional isomers have the same molecular formula and functional group but differ in the
position of the functional group. Examples of positional isomers are but-1-ene and but-2-ene.
Chain (skeletal) isomers: These differ in the length of the main carbon chain. For example, pentane has three chain
isomers.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
494
1.4.1
Structural isomerism
Structural isomers differ in the exact way in which atoms are connected to each other. This type of
isomerism is very common among organic compounds due to the ability of carbon atoms to form chains of
variable length. The position of functional groups or chemical groups on these chains can also vary, giving
rise to a wide variety of structures that share the same molecular formula.
There are three forms of structural isomerism; chain isomerism, functional group isomerism and
positional isomerism.
Chain isomerism
Chain isomers, also known as skeletal isomers, differ in the length of carbon chains. They are a direct
consequence of the ability of carbon to catenate. One isomer, termed the normal chain isomer, has one long
continuous carbon chain. The other isomers will be branched, consisting of a main chain and groups
branching at some points along the main chain. Chain isomerism is significant in alkanes or compounds
that contain hydrocarbon chains. The smallest alkane that can have isomeric forms is C₄H₁0
The normal chain isomer is n- butane (which is usually just written as butane). The prefix n is used to
denote the unbranched isomer. This is often referred to as the straight chain isomer, but this term can be
misleading. Carbon chains are not straight (linear), but they tend to be jagged (as shown in skeletal
formulae) so that the geometry around each carbon atom minimizes repulsion between electron groups
(VSEPR theory).
Isomers of C₄H₁₀
H
H
H
H
H H
H
C C C C
H
H H H
n - butane
H
C
H
H
H
H
H H
1 2 3
H C C C H
2- methylpropane
Nomenclature
This refers to a systematic way of naming compounds. Isomers are named using IUPAC rules (IUPAC stands
for the ‘International Union of Pure and Applied Chemistry’).
Notice how the second isomer of C4H10 has been named. The longest continuous carbon chain is identified
and all of its carbon atoms are numbered. The longest continuous carbon chain gives the root of the name.
In this case the compound is a propane because the longest continuous carbon chain has three carbon
atoms. There is a methyl (CH3) branch at position 2 of the longest carbon chain. The compound is
therefore named 2- methylpropane.
Carbon compounds are named on the basis of the number of C atoms in the parent chain (longest continuous chain), side groups and functional groups present.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
495
Number of
carbon
atoms in the
parent chain
The prefixes given in Table 10.2 apply to all types of
organic compounds. Consider, for example, the
names of the compounds below.
Prefix
1
Meth-
2
Eth-
3
Prop-
4
But-
5
Pent-
6
Hex-
7
Hept-
8
Oct-
9
Non-
10
Dec-
11
Undec-
12
Duodec-
CH3CH3
CH2 = CH2
CH3CH2OH
CH3COOH
ethane
ethene
ethanol
ethanoic
acid
Note that all of the compounds contain two carbon
atoms, so they have the prefix eth. A suffix is then
added to identify the class to which the compound
belongs, for example:
Suffix
Class of compound
-ane
alkanes
-ene
alkenes
-anol
alcohols
-anoic acid
Carboxylic acids
Table 10.2 Prefixes used to name organic carbon
chains
Naming alkyl side groups
Chain isomers show different branching. An alkyl branch contains C and H only. Consider two
compounds below, which contain a methyl and ethyl branch respectively.
methyl side group
ethyl side group
CH3
H
H H
1 2 3 4
H C C C C H
H H H H
2 - methybutane
CH2CH3
H
H H
H H H
1 2 3 4
H C C C C C C C H
H H H H
H H
H
2- ethylhexane
The first compound has a methyl group at carbon 2 of a butane chain. The second compound has an ethyl
group at the second carbon of a heptanes chain. Note that the name of an alkyl contains a prefix (as in
Table 10.2) and the suffix -yl. Examples are shown in Table 10. 3 .
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
496
Alkyl group
methyl
CH3 -
ethyl
CH3CH2-
pentyl
Q
.
A
Formula
CH3CH2CH2CH2CH2-
Table 10.3
Show the structures of the isomers of C5H12 and name them.
From the given formula, we can infer that the compound is an alkane (general formula
CnH2n + 2). The first obvious isomer is n - pentane, which has 5 carbon atoms in one continuous
chain. We now draw the other isomers by successively reducing the number of carbon atoms in
the longest chain by one. That is, the second isomer is a butane because it has four C atoms in
the longest chain. The other isomer is a propane. The three chain isomers of C5H12 are shown
below.
I
H H H
H
H H
C C C C C H
H
H H H H
n-pentane
II
H H H
H
1 2 3 4
H C C C C H
H
H H
CH3
2 - methylbutane
CH3
1
H
H
H
H
CH3
2
3
H C C C H
2,2 - dimethylpropane
Structure I is 2 - methylbutane. The name tells us that there is a methyl group on a butane (4 carbon) chain.
Structure II is 2, 2 - dimethylpropane. The name shows that there are two methyl groups at carbon 2 of a
propane chain.
Look again at structure I. It could have been possible to number the main carbon chain from the other
direction. This would place the methyl group at position 3, and the compound would have a different
name, 3- methlypropane. This name would be wrong.
The carbon chain is numbered from that direction which would locate the branch at the smaller number.
The numbering in structure I is correct because it places the branch at the smaller number, 2. The
numbering in structure III below is wrong because it places the branch at the larger number, 3.
III
H H H
H
4 3 2 1
H C C C C H
H
Another point which should be stressed is that the choice of the parent chain
is not important in naming a compound. What is important is that the carbon
chain selected be the longest and it should be continuous. Structure I above
could have been numbered by choosing a ‘different chain’, as shown in
structure IV.
H H
CH3
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
497
IV
H H H
4
H C
H
3 2
C C C
H
methyl branch at carbon 2
H
H
1 H
CH3
longest continuous chain has
4 carbon atoms
Note that structure IV still has a methyl group at carbon 2 of a butane chain. Its name is therefore
2-methylbutane, that is, it is exactly the same compound as structure 1. If the name of the structures is the
same, they do not represent isomers. They are the same compound. This is useful in checking for
repetitions. You may draw two structures which, to your eyes, may look totally different. If you properly
assign names to them and the names are the same, then you know that you have not drawn two isomers,
you have simply drawn the same compound twice!
Consider the case of a student who draws the structures V and VI and claims that he has drawn two chain
isomers of C5H₁₂.
structure VI
H
structure V
H
C
H H H H
H
C C C C
H
H
H
H
H H
CH3
H C C
CH3
H
H
C
H
H
Notice that in both structures, the longest continuous carbon chain has four carbon atoms and that in both
cases there is a methyl group at position 2 of that chain. Both structures are identified by the name 2methylbutane. They represent the same compound.
Positional isomerism
Positional isomers differ in the position of the functional group.
As before, numbers are used to locate the position of the functional group. Consider the molecular formula
C₄H₈. This compound is an alkene (general formula CnH2n). But the formula does not tell us where the
double bond is located. There are two different alkenes which share this formula:
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
498
I
II
H H H H
2 3 4
1
H C C C C H
displayed formula
H
butene
H
H H H H
1 2 3 4
C C C C H
H
but-2-ene
H
H
skeletal formula
structural formula
CH2CHCH2CH3
CH3CHCHCH3
Compound II is but-2-ene. The prefix but- tells us that there are four carbon atoms in a continuous chain.
The suffix- ene tells us that the compound is an alkene. The locant (locator number) 2 shows that the double
bond begins at carbon 2. The locant is always where the double bond begins.
The rules for naming organic compounds given in the previous discussion apply here. The carbon chain
should be numbered from the direction that locates the double bond by the smaller number. The carbon
chain should therefore always be numbered from the end which is closer to the double bond, as has been
done for structure I above. If the double bond begins at carbon 1, it is not necessary to include a locator
number in the name of the alkene.
Structures III and IV represent the same compound, pent-2-ene, that is, the two structures do not
represent positional isomers. The carbon chains have been numbered from the end closest to the double
bond.
III
H
IV
H H H H H
1 2 3 4 5
C C C C C H
H
H
pent-2-ene
H
H H H H H
5 4 3 2 1
C C C C C H
H
H
H
pent-2-ene
Other classes of organic compounds may also exhibit positional isomerism. Consider the alcohols whose
molecular formula is C₃H₈O. There are two positional isomers, as shown below. They differ only on the
carbon to which the –OH group is attached. The names of the isomers tell us where to find the functional
group.
Propanol may also be written as propan-1-ol, but if
the name is written without a locator number then it
refers to the isomer in which the functional group is
at the first carbon. Note that the two structures refer
to different compounds, despite the fact that they are
both alcohols. For instance, the two compounds form
totally different esters.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
499
H
H C
C
C
H H
H
OH H
H
H H
OH
H C
C
C H
H
H
H
propanol
propan-2-ol
The two isomers also have different boiling points. Propan-1-ol has the higher boiling point because there
is less hindrance around the OH group by surrounding groups. Consequently, hydrogen bonding is strong
(hydrogen bonding requires the participation of an orbital carrying a lone pair of electrons. In propanol this
lone pair is freely available to participate in hydrogen bonding).
Functional group isomerism
A third type of structural isomerism is functional group isomerism.
Functional group isomers are compounds that share the same molecular formula but have different
functional groups.
This type of isomerism is common among



ethers and alcohols.
esters and carboxylic acids
ketones and aldehydes.
Functional group isomerism among esters and carboxylic acids
Both esters and carboxylic acids contain two oxygen atoms. Consider the formula C₃H₆O2. It could imply
propanoic acid, CH₃CH₂COOH, which is a carboxylic acid, or it could represent methyl ethanoate,
CH₃COOCH₃, which is an ester.
O
O
CH3 C
CH3 CH2 C
O
CH3
OH
propanoic acid
methyl ethanoate
Functional group isomerism among ketones and aldehydes
Both ketones and aldehydes contain the carbonyl group.
O
C carbonyl group
The difference between aldehydes and ketones is that in aldehydes there is a
hydrogen atom bonded to the carbonyl group. The presence of this hydrogen
atom in aldehydes and its absence in ketones affects the chemistry of these
compounds so significantly that they are classified into different classes.
Consider the molecular formula C₃H₆O. It could imply propanone, which is a ketone, or it could represent
propanal, which is an aldehyde.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
500
O
O
CH3 C
CH3 CH2 C
H
ethanal - aldehyde
propanone - ketone
O
O
skeletal formula
CH3
H
1.4.2
Stereoisomerism
Stereoisomerism is the phenomenon in which compounds of the same molecular formula and sharing the
same functional group differ in the spatial arrangement of atoms ( spatial arrangement= 3D arrangement
in space).
There are two types of stereoisomerism, cis - trans (geometric) and optical isomerism.
Cis- trans isomerism
This is also known as geometric isomerism. Cis - trans isomerism frequently arises in alkenes because of
restriction to rotation about the rigid double bond. It can also arise in cycloalkanes because their ring
structures also cause single C to C bonds to become rigid and resist rotation.
Consider the alkene whose molecular formula is C₄H₈. Recall that there are two positional isomers that
share this formula:
I
II
H H H H
2 3 4
1
H C C C C H
H
H
butene
H
H H H H
1 2 3 4
C C C C H
H
but-2-ene
H
Now, structure II also has two isomeric forms, which are cis-trans isomers. First, notice that structure II
can also be drawn as below.
1
2
CH3 C
A
H
B
3
C
4
CH3A
H
B
Each carbon atom of the double bond is bonded to two groups, which have been
labeled A and B. Now, recall that geometry at each carbon atom of the double
bond is trigonal planar (bond angle 1200). The following structures are drawn to
clearly show the geometry at carbon 2 and 3.
1200
III
1
4 A
A CH
CH3
3
2
C= C 3
B
Prepublication manuscript
H
B
cis but-2-ene
H
IV
BH
4 A
CH3
2 C = C3
1
H
A CH3
B
trans but-2-ene
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
501
Structures III and IV do not represent the same compound. These are different compounds, known as
cis-trans isomers. The groups labeled A and B are locked in their respective positions in space because it
is not possible to rotate the double bond and convert one structure into the other.
Structure III is the cis isomer (cis = adjacent). Note that similar groups are adjacent (on the same side of
the double bond) to each other. Structure (IV) represents the trans isomer (trans = opposite) . Note that
similar groups are diagonally opposite each other with respect to the double bond.
Important note
For cis - trans isomerism to be possible, each carbon atom of the double bond must be bonded to two
different groups. In the example above, there is an H and a CH₃ group on each carbon.
Consider two alkenes P and Q whose structures are shown below.
P
C2H5COHCOHC2H5
Q CH3CHCH2
First draw out the structures to show correct geometry at the double bond. You can then see that cis -trans
isomerism is possible in compound P because each atom of the double bond is bonded to two different
groups.
CH2CH3
CH2CH2
CH2CH3
OH
C
C
CH3CH2
C
OH
C
OH
OH
cis isomer
trans isomer
Cis - trans isomerism is not possible in compound Q because one of the carbon atoms bears the same
group, H.
H
H
C=C
CH3
H
cis - trans isomerism is not possible if one
or both C atoms of the double bond carries
identical groups.
Optical isomerism
This is the second type of stereoisomerism.
Optical isomerism is the phenomenon in which compounds with the same molecular formula, same
structural formula and possessing the same functional group(s) exist in two forms which are nonsuperimposable mirror images of each other.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
502
A compound can exhibit optical isomerism only if it possesses a carbon atom to which four different
groups are bonded. Such a carbon atom is said to be chiral (From German cheir, meaning ‘hand’: The left
hand and the right hand palms are non superimposable mirror images, just like optical isomers).
The structure of lactic acid (2-hydroxypropanoic acid) is shown.
OH
The carbon atom marked with an asterisk (*) is chiral. It is an asymmetric
centre that gives rise to a pair of non-superimposable images known as optical
isomers, or enantiomers, as shown in Fig 10.4.
CH3 C* COOH
H
mirror
line
CH3
CH3
1090
C
HO
H
C
COOH
HOOC
H
OH
The two optical isomers (entantiomers) of lactic acid.
Fig 10.4
Whenever you draw a pair of optical isomers, you must emphasize the three dimensional aspect by using
the solid wedge (
) to show a bond pointing out of the paper towards you and a broken wedge (
) or
dashed line to show a bond pointing to the back of the page, away from you. Ordinary lines are used for
bonds which are in the plane of the paper. You should also make sure that one structure is the exact mirror
image of the other. Note that it is not important which group you actually put at the end of each bond.
Optical isomerism is possible in compounds such as lactic acid due to the tetrahedral geometry at the chiral
carbon atom. If the bond angle was 90 0 (square planar), optical isomerism would not be possible, because
the mirror images obtained would be superimposable.
Optical isomers undergo the same chemical reactions. This is expected; since they have the same
functional group; they must have similar chemical properties. Optical isomers differ in the direction in
which they rotate the plane of polarized light. One isomer rotates the plane of polarized light clockwise (to
the right), and the other isomer rotates anticlockwise (to the left). Enantiomers are distinguished from each
other on the basis of the direction of rotation of the plane of polarized light using the notation (D) and (L).
The (D) enantiomer rotates the plane of polarized light to the right (D is short for ‘dextrorotatory’ which
means ‘rotating to the right’). The (L) enantiomer rotates the plane of polarized light in an anticlockwise
direction (L comes from ‘laevorotatory’, which means ‘rotating to the left).
Light from a light source is made up of mutually perpendicular electric and magnetic waves vibrating in all
directions. It is possible to filter light so that the emergent light vibrates in a single direction along a single
plane. Such filtered light is known as polarized light. This is done by passing the light through an optical
filter such as polaroid.
A molecule with a chiral carbon atom is able to rotate the plane of polarized light, to the right or to the left,
depending on whether it is the (D) enantiomer or the (L) enantiomer. Compounds which are able to rotate
the plane of polarized light are said to be optically active. (D) and (L) -Lactic acid are examples of optically
active compounds. 19 of the 20 naturally occurring amino acids are optically active. Glycine is the only
amino acid which has no optical isomers, because it has no chiral carbon atom. The structures of glycine
and alanine, which is optically active, are shown below.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
503
alanine
glycine
H
H2N
C
C
H2N
OH
H
O
CH3
O
*
C
H
C
OH
optically active since a chiral carbon atom
is present
Not optically active
Racemates
If lactic acid is synthesized using usual lab methods, it is found that the product does not rotate the plane
of polarized light. This is not surprising. The synthetic method used is not selective; the yield will contain
50% of the (D) enantiomer and 50% of the (L) enantiomer. Since the two forms rotate the plane of polarized
light in opposite directions, the net effect is a cancelling out of the rotation. An equimolar mixture of (D)
and (L) isomers of a compound is known as a racemate or racemic mixture and it is optically inactive.
N.B The notations D and L can be substituted by the notations (+) and (-) respectively.
Biological significance of optical isomers
Biological reactions are carried out by efficient biological catalysts known as enzymes. Enzymes are
specific, that is, in most cases they will recognize only one type of substrate (the reactant on which it acts).
Enzymes are able to recognize and distinguish between optical isomers. Often they can only work on one of
the isomers and not the other. An example is provided by the mould Penicillium glaucum. It can feed on
the (D) enantiomer of lactic acid, but it cannot use the (L) isomer. Geometry of molecules is important in
the action of enzymes. An enzyme has a catalytic region known as the active site. The active site has a shape
complimentary to the shape of the substrate, so the substrate can fit into the active site like a key fitting
into a lock. Only the correct molecule, or a molecule with the correct shape, will fit into the active site. In
the case of P. glaucum, (L)-lactic acid does not have the correct geometry, so it will not fit into the active
site (This is rather like trying to fit a left glove into the right hand palm).
Sometimes the isomer which is not metabolized acts as a poison. A good example is provided by the drug
thalidomide, which was the preferred sedative for pregnant women in the early ‘60s. The drug at that time
was produced as a racemate. One isomer had the required medical effect, but the other one proved to have
teratogenic effects (causes babies to be born with deformities such as shortened limbs). The structure of
thalidomide is shown below. The chiral carbon atom has been marked with an asterisk.
O O
N
*
Chiral carbon. Do not forget that there is
a hydrogen atom at this carbon.
O
O
Thalidomide
Q
What is the molecular formula of thalidomide?
A
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
504
C14H11O4N
Number of possible optical isomers
A compound such as lactic acid with only one chiral centre will give rise to two optical isomers. If a
molecule has two chiral centres, each center will give rise to a pair of optical isomers so there will be a total
of four. In general, the number of optical isomers possible for a molecule is given by 2n where n is the number of chiral centres.
Glucose is commonly represented by the structure below:
CHO
C* OH
H
*
HO
C
H
C
*
*
H C
There are four chiral C atoms, marked with asterisks. The
total number of possible optical isomers is therefore 2⁴ =
16.
H
H
OH
CH2OH
(D)- glucose
Q
(a) The structure of the drug cortisone is shown below. Mark out all the chiral
centres present.
(b)
State which functional groups are present in cortisone?
cortisone
O
O
OH
HO
O
A
It is important to be able to interpret such skeletal formulae. Always check the number of bonds
around each carbon atom (remember to treat a double bond as two bonds, and a triple bond as
three bonds). If you see only three bonds around a carbon atom, then the fourth position is
occupied by a hydrogen atom. Similarly, if you see only two bonds around a carbon atom, the
other two positions are occupied by two hydrogen atoms, giving a -CH2 group (‘elbow’).
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
505
OH
cortisone
O
HO
O
11
10
1
O
9
8
*
3
2
20
19
18
*
7
17
6
*
* 16
*
12
21
Carbons with two H atoms (CH2)
4, 5, 9, 10, 13, 14, 17, 21
*
15
Total H = 16
14
13
5
4
Carbons with one H atom
2, 6, 7 , 12
Total H = 4
Carbons with no H atom
1, 3, 8, 15, 16, 18, 20
Total H = 0
Analysis of the structure of cortisone
There are 21 C atoms , marked 1 to 21
The molecular formula is C21 H28O5
There are 6 chiral carbon atoms, marked*
Note that a carbon atom which is part of a
double bond can not be chiral (1, 2, 3, 18,
2o). The geometry at these carbon atoms is
1200, not 1090.
Carbons with 3 H atoms (CH3)
11, 19
Total H = 6
Other H atoms
OH
Total H = 2
grand total
28
Never forget that the end of a line segment (11,
19) represents a -CH3 group.
The functional groups present in cortisone are ketone, alkene/C to C double bond and alcohol.
Effect of symmetry
Now study the compound below.
Each of the highlighted carbon atoms has four bonds around it. However, both atoms
are not chiral (they are achiral). Note the perfect symmetry at each of the highlighted
carbon atom with respect to its right and left hand sides.
The groups to the left and to the right of these carbon atoms are therefore identical. The compound is
optically inactive. The molecule below is given to help clarify further the effect of symmetry.
1
This time both highlighted C atoms are chiral, because there is no symmetry around each,
with respect to the LHS and the RHS. For instance, if you look at C2, you will notice that
there are more –CH₂ groups (‘elbows’) going to its left than going to its right.
2
The LHS group is therefore different from the RHS group. However, the molecule is still
optically inactive, despite the presence of two chiral carbon atoms! The explanation is in the internal
symmetry exhibited by the molecule, as shown below.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
506
1
In this compound, there is internal compensation, whereby the two
carbon atoms rotate the plane of polarized light in opposite
directions. The effects of the two carbon atoms therefore cancel
each other out.
line of symmetry
2
The concept of internal compensation has been used to explain why
tartaric acid exists as three isomers, of which only two are optically active. The isomers are shown below.
COOH
H
HO
COOH
OH
HO
H
H
(+)- tartaric acid
(optically active)
OH
H
(-)- tartaric acid
(optically active)
enantiomers
A
H
COOH
COOH
Q
H
COOH
OH
OH
COOH
meso - tartaric acid
(optically inactive)
line of symmetry
Explain whether or not optical isomerism is possible in the compound shown
below.
Optical isomerism is possible due to the carbon atoms marked *. Both atoms are asymmetric.
Also, the whole molecule has no internal symmetry, so the effects of the two chiral carbon atoms
do not cancel each other out.
*
*
Summary: optical isomerism

A molecule is optically active if it has a chiral carbon atom, that is, a carbon atom to which are
bonded four different chemical groups. The geometry around a chiral carbon atom is always
tetrahedral, bond angle 109.5⁰. The term ‘optically active’ is used for a compound which is able to
rotate the plane of polarized light.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
507


Presence of a chiral centre gives rise to a pair of optical isomers, also known as enantiomers or
enantiomeric pair, designated (+) or (-) depending on the direction in which they rotate the plane
of polarized light. Alternatively, the notation D and L is used. Enantiomers differ in the direction
in which they rotate the plane of polarized light; otherwise they have similar chemical reactions. A
racemate (racemic mixture) is a 50%-50% mixture of the (+) and the (-) isomers. A racemate is
optically inactive.
The theoretical number of optical isomers possible for a molecule is given by 2 n, where n is the
number of chiral carbon atoms present.
Questions, solutions and discussions
Q1
Draw the displayed formula of two different compounds that share the molecular formula C 4H8O2.
Circle and name the functional group in each structure you have drawn.
Solution
ester group
H O
H
H H
C C O C C H
H
H H
ethyl ethanoate
carboxylic acid group
H H H
O
H C C C C
H H H
OH
butanoic acid
This is an example of functional group isomerism.
Tip: If an organic compound contains two oxygen atoms, it could be an ester or a carboxylic acid; otherwise
it is a compound containing two –OH groups, or two different functional groups that contain oxygen.
Q2
Lactic acid is one of the chief products of anaerobic respiration in animal tissues. It is responsible for
muscle cramps.
(a) Circle and name the group which is responsible for the acidity of lactic
HO
acid.
O
( b) Write down the molecular formula for lactic acid.
OH
( c) Name the functional groups present in lactic acid.
lactic acid
(d)
Prepublication manuscript
Calculate the mass of lactic acid required to react with exactly 25.00
cm3 of 0.1 moldm-3 NaOH.
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
508
Solutions
(a)
carboxylic acid group
HO
O
OH
(b)
C3H6O
(b) Carboxylic acid and secondary alcohol.
(c)
Left as an exercise to the reader.
Q3
(a) Explain the term optical isomerism.
(b)
State a structural feature which must be possessed by an organic compound for it to
show optical isomerism.
(c)
Draw the two optical isomers of lactic acid (see question 2).
Solutions
(a) Optical isomerism occurs when two different compounds share the same molecular and structural
formula but differ in the arrangement of groups in space.
(b) A chiral carbon atom should be present. Such a carbon is surrounded by four different groups.
(c)
CH3
CH3
C
C
HO
OH
H
H
COOH
COOH
mirror line
Prepublication
manuscript
Notes
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
Optical isomers, also known as enantiomers, are stereoisomers. Stereoisomers have the same molecular and structural
formula. The only difference is the three dimensional arrangement of groups in space. The notation used to draw optical
isomers (a mixture of normal lines, solid wedge and broken line) emphasizes the 3D orientation of groups in space
509
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise
10.2
text box.]
The structure of a derivative of one of the primary
sex hormones is shown below.
(b) State which carbon atoms
18 OH
17
6
1
HO
(a)
5
3
14
15
10
7
8
have trigonal planar geometry
(ii)
bear only one hydrogen atom
(iii) are tetrahedral but not chiral.
13 OH
11 12
9
4
2
16
(i)
(c)
(iv)
contain two hydrogen atoms
(v)
contain no hydrogen atoms
Name all the functional groups present in this
compound.
Mark out all chiral centres.
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
10.2
Mechanistic organic chemistry
We now discuss in detail the chemical reactions of functional groups. It would be worthwhile to invest
time and effort in this section to make the study of the individual classes of organic compounds easy and
interesting.
Too often students try to learn organic chemistry by memorizing individual reactions of compounds.
This usually turns out to be a difficult task. There are several millions of known organic compounds, and
every day new ones are being discovered or synthesized. The brain finds it very difficult to organize and
store information which is not systematically presented to it. It must see an order and a pattern if it is to
process and store information efficiently. In organic chemistry, the ordering of information can be done by
understanding a few reaction mechanisms thoroughly. The reactions of organic compounds fall
into one or more of these mechanisms. By understanding reaction mechanisms, you do not have to be
intimidated by molecules that look complex on paper, for example, the anti-cancer drug Taxol® (Fig 10.5).
In fact, chances are that in your final exam, you will encounter organic compounds you have never heard of
or learnt about. All you need to do is retrieve information from your brain about how certain reactive
centres in an organic molecule react. These reactive centres are known as functional groups.
A functional group is a group of atoms that is responsible for the chemical behavior
of an organic molecule.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
510
Fig 10.5
The structure of Taxol®, an anticancer drug.
Name the functional groups
present and identify any chiral
centres.
A functional group is the reactive center of an organic molecule. No matter how complex an organic
molecule is, it will always react in a predictable way. This is because its functional group(s) will always react
in a specific way with the specified reagent and under the specified conditions.
A reaction mechanism can be defined as the specific method by which a reaction occurs. In a more strict
sense, a reaction mechanism is the detailed sequence of steps leading from reactants to products. In this
section we will focus on six reaction mechanisms that should be understood thoroughly in ‘A’ Level
chemistry.
10.2.1 Addition versus substitution reactions
During a substitution reaction, an atom or group of atoms replaces another atom or group of atoms
from the organic reactant. Reaction I is an example of a substitution reaction. In this reaction, the OH
group replaces the Cl group in chloroethane.
CH3CH2Cl + NaOH (aq)  CH3CH2OH + NaCl ... I
Note that a substitution reaction results in the formation of two products.
During an addition reaction, two reactants combine to form one product. An example is the addition of
bromine to the double bond of an alkene.
H
H
H
C
C
ethene
+
H
Br2
Br
H C
H
H
C H
Br
1,2- dibromoethane
In some cases it may appear as if an addition reaction has formed two or more products. This happens when
the product of the addition reaction further reacts by a different mechanism to form other substances.
10.2.2
Homolytic versus heterolytic bond fission
Bond fission (breaking) may take place by one of two possible methods, homolysis or heterolysis.
During homolysis (homolytic bond fission), the bond breaks in such a way that each atom of
the bond leaves with a single electron of that bond. This results in the formation of two free
radicals.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
511
odd electron
A
B
A
+ B
Homolysis of bond A-B results in the formation of two free
radicals A and B. Notice that we show homolysis by a halfheaded arrow. This notation shows the movement of a single
electron.
free radicals
Free radicals are neutral particles which are very reactive by reason of them containing an odd
number of electrons in the outer-most shell.
One electron in the outer-shell is thus unpaired (it is an odd electron). The formula of a free radical is usually
written to emphasize the odd electron, ignoring the paired electrons, as illustrated for a chlorine atom,
which is a well known example of a free radical.
odd electron
lone pair
Cl
Cl
During heterolysis, the bond breaks so that one atom leaves with both electrons of the bond.
The leaving group takes both electrons of the bond and so becomes a negatively charged ion. In the following
illustration , B is the leaving group. The remaining group may or may not become a cation.
Ax B
+
A + Bx
For example, the C-Cl bond in 2 -chloro-2-methylpropane undergoes heterolysis as shown below.
CH3
CH3 C x Cl
CH3
CH3
CH3
+
C
+
Cl x
CH3
2-choro-2-methylpropane
Heterolysis of a bond in intermediates of some reactions generates an anion, but does not generate a cation,
for example.
H
H
HO
C x Cl
H
H
H C
OH + Cl x
-
H
intermediate
(transition state)
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
512
Take note of the use of the arrow notation. An arrow is used to show the movement of an electron or a pair
of electrons. A half headed arrow shows the movement of a single electron, for example, during homolysis.
A full headed arrow shows the movement of a pair of electrons, for example during heterolysis.
10.2.3
The importance of reaction conditions
Organic reactions frequently involve the breaking of strong covalent bonds. The reactions therefore have
high activation energies, and relatively harsh conditions must be used to overcome this energy barrier to
reaction. Harsh conditions may be provided by using high temperatures and pressures. A special technique
known as refluxing ensures that reactants remain in contact and at high temperatures for a long time. This
increases the yield of the reaction. Refluxing is carried out in a refluxing apparatus (Fig 2.1), which is in
fact a reaction vessel attached to a condenser
During refluxing, the reaction mixture is heated to its
boiling point for a long time. Reactants rise as vapour
and are cooled in the condenser, allowing them to fall
back into the reaction vessel. In this way, the
reactants remain in contact for a long time.
One other reason why organic reactions are
sometimes slower than inorganic reactions is the
orientation factor. According to the collision theory of
reaction kinetics, reactant particles must first collide
with sufficient energy and with the correct orientation
for them to react. The orientation factor is not much
of a problem when the reacting particles are small.
However, in organic reactions, relatively large
molecules are often involved, and this decreases the
chances of the molecules colliding with the correct
orientation. Consider the reaction between ethanoic
acid and methanol to form methyl ethanoate.
Fig 10.6
A reflux apparatus
If a collision is to be successful, the two molecules should approach each other in such a way that the –OH
group on ethanol collides with the –COOH group on ethanoic acid. Collisions which do not occur in this
way will not result in a reaction, as shown in the following illustration.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
513
The need to specify reaction conditions
The product of an organic reaction is affected by the reaction conditions employed. Changing the
conditions of the reaction usually results in the formation of different products. For this reason, the
student of organic chemistry needs to be aware of the reaction conditions that must be used for specific
reactions. The term condition here refers to
 temperature
 whether or not refluxing is necessary
 pressure
 catalyst
 whether reagent should be used in excess or in limited amounts.
10.3
Organic reaction mechanisms
We are now ready to explore reaction mechanisms that are encountered in ‘A’ level chemistry. The student
is reminded once more: it is better to spend a large amount of time trying to understand these mechanisms
than rushing to memorize reactions of individual compounds. A reaction mechanism in organic chemistry
is the method by which a particular functional group reacts. Once you understand this method, you can
apply the knowledge gained to predict the reactions of any other compound that bears the same functional
group.
Six reaction mechanisms
1. Nucleophilic reactions
4. Free radical substitution reactions
2. Electrophilic reactions
5. Organic redox reactions
3. Elimination reactions
6. Organic neutralization reactions
.
10.3.1
Nucleophilic reactions
A nucleophilic reaction involves the participation of a particle known as a nucleophile.
Nucleophilic reactions can occur by substitution or by addition.
A nucleophile is an electron rich species which is therefore attracted to a relatively
positively charged carbon atom in an organic molecule, to which it makes a dative bond.


Nucleophiles are either neutral or negatively charged.
A nucleophile, whether it is neutral or negatively charged, contains a lone pair which it uses
to attack an electron deficient (relatively positively charged) carbon atom. This carbon atom
is denoted, C
Prepublication manuscript

©L.Mwanawenyu. Please do not photocopy or reproduce in any way
514
The carbon atom attacked by a nucleophile is relatively positively charged by reason of it being bonded to
a more electronegative atom, for example, the halogen or oxygen atom.
this bond could be single or
multiple
C Y

more electronegative atom
nu
Some of the functional groups which contain an electron deficient carbon and are therefore likely to be
attacked by nucleophiles include
OH
O

C
 C

 C
carbonyl(aldehydes
and ketones)
alcohol
 
O 
O
C
 

O
 
 
  C
N 
amide

 
Cl
acyl chloride

 
O
C
 
Hal

ester
halogenoalkane
Whether nucleophilic addition or substitution occurs depends on the nature of the C-Y bond. If the bond is
relatively weak, the likely reaction is substitution, because the bond can easily break to make room for the
incoming nucleophile. If the bond is double (particularly the C to O double bond), substitution does not
take place because the bond does not easily break. In this case nucleophilic addition occurs.
The following panel shows the nucleophiles commonly encountered in organic chemistry.
Panel 10. 2
Examples of neutral nucleophiles

H2O

NH3

Amines, e.g. CH3CH2NH2 (ethylamine)

Alcohols, e.g. ethanol
Prepublication manuscript
Examples of negatively charged nucleophiles

Cyanide, CNʒ

Hydroxide, OHʒ

Halide, e.g. Clʒ

Hydride, H-
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
515
3.1.1 Nucleophilic substitution
This is a reaction in which an electron rich species, known as a nucleophile, replaces an atom or a group of
atoms in an organic molecule that has an electron deficient (relatively positively charged) carbon atom.
The atom or group of atoms replaced is known as a leaving group and it is itself nucleophilic in nature. A
nucleophilic reaction therefore involves replacement of one nucleophile by another, e.g.
CH₃Cl (aq) + OH⁻ (aq) → CH₃OH (aq) + Cl⁻ (aq)
… (i)
The nucleophile is the OH⁻ ion from a reagent such as NaOH. The leaving group is the chloride ion. Notice
that the leaving group is also nucleophilic in nature since it is electron rich. This raises an important
question. Why should the nucleophile replace the leaving group if both are nucleophilic in nature? Is it not
likely that once freely available in the reaction mixture, the leaving group, Cl⁻ in this case , will attack the
product, forming the reactants once more (both the reactant and the product carries an electron deficient
centre and so both are susceptible to attack by a nucleophile). This is a very valid argument. We should turn
our attention to bond strengths to understand why one nucleophile can replace another. In the example
above, a weaker C-Cl bond is replaced by a stronger C-O bond. The product is therefore more energetically
stable, so the forward reaction is favoured. However, it is important to know that the reverse reaction can
still happen under certain conditions. If the concentration of Cl⁻ ions in the reaction mixture is increased ,
for example, by adding concentrated HCl, followed by heating, then the reverse reaction will take place,
with the Cl⁻ now being the nucleophile and the OH⁻ the leaving group.
Nucleophilic substitution bimolecular (SN2)
Reaction (i) is an example of an SN2 reaction. The mechanism is as follows.
H

C
HO
H
Cl
H
H
HO
H

C

Cl
H
Prepublication manuscript
The C-Cl bond is polarized since chlorine has a greater electronegativity than
carbon. The carbon atom of this bond is therefore partially positively charged
since chlorine takes a greater share of electrons of the bond.
The nucleophile, being electron rich, is attracted to the Cδ+ of the C-Cl bond,
resulting in the formation of a new C-O bond. Note that the new bond formed
is dative since it makes use of a pair of electrons from only one atom, the
oxygen of the -OH group. As the C-O bond forms, the C-Cl bond becomes
longer and weaker as it prepares to break.
A transition state (very unstable intermediate) is formed. Instability of this
transition state is caused by the presence of too many bonds around the
central carbon atom. Carbon, in Period 2, is not able to expand its octet, that
is, it can only have four bonds (eight electrons) around it in its compounds.
The transition state is said to be pentavalent, since the carbon atom has five
bonds (10 electrons) around it. Interelectronic repulsions between these
bonds is responsible for the instability of the transition state. To relieve the
excess electron density around the carbon atom, the C-Cl bond undergoes
heterolysis. The chlorine atom leaves with both electrons of the bond, and so
becomes an anion. Heterolysis of the C-Cl bond results in the formation of a
product in which the carbon atom has the correct number of bonds (four)
around it.
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
516
The final product of the reaction, methanol, is more stable than the
reactant, chloromethane. This is because the new C-O bond in the
product is more stable than the C-Cl bond broken in the reactant.
H
C
H
H
+ Cl-
OH
The kinetics of the reaction
The reaction mechanism is said to be concerted. In other words, the intermediate (transition state) is so
unstable and short lived that the reaction may be considered as taking place in one step
CH₃Cl (aq) + OH⁻ (aq) → CH₃OH (aq) + Cl⁻ (aq)
The rate law is written as
R = K [CH₃Cl]m[OH⁻]n
Now, the values of both m and n have been experimentally determined to be 1. That is, the reaction is first
order with respect to both reactants. Overally, the reaction is second order. It can also be said that the
reaction is bimolecular, that is, the rate determining step involves the collision of two particles. The
mechanism of the reaction between a primary halogenoalkane such CH 3Cl and NaOH has been
appropriately termed SN2.
substitution
SN2
second order (bimolecular)
nucleophilic
You can split the rate law above so that it reads
R = K [CH₃Cl]
R=K
[OH-]
first order kinetics with respect to chloromethane
first order kinetics with respect to NaOH
The rate/concentration graph is therefore linear for both reactants. That is, changing concentration of
either reactant results in a proportional change in the rate of reaction.
Rate
The energy profile of the reaction is shown in Fig 10.6


reactant
concentration
Prepublication manuscript
The value of ∆H is negative, that is, the product, CH3OH is lower
in energy than the reactant, CH3Cl, as already explained.
The reaction is not spontaneous since Ea is quite high. This is
usually the case with organic reactions. Activation energy is the
minimum energy reactant molecules must possess when they
collide so that a reaction can take place. In this reaction, the
collisions between CH3Cl and must be energetic enough to
initiate breaking of the C-Cl bond.
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
517
transition state
Energy
Ea
H
progress of reaction
Fig 10.6 Energy profile of an
SN2 reaction. The reaction has
only one ‘hump’ since it is
concerted.
Other SN2 reactions involving halogenoalkanes
In the reaction we have just studied, the OH - nucleophile replaces the Cl atom from the organic reactant.
Other nucleophiles can also replace the chloride atom, and these include CN - and NH3. The same reactions
are also possible in other halogenoalkanes, for example, bromoethane.
Some reactions of bromoethane which occur by nucleophilic substitution second order (S N2) are shown
below.
CH3CH2OH
(i)
CH3CH2Br
(ii)
CH3CH2CN
(iii)
CH3CH2NH2
Reaction (ii) : formation of a nitrile
It is clear that in this reaction the nucleophile is the cyanide ion, which replaces the Br atom from the
reactant. The result is that a halogenoalkane is converted to a nitrile. Now, here we need to make a clear
distinction between a nucleophile and a nucleophilic reagent. The nucleophile is the CN- ion; but it does not
exist on its own, rather, it is found in some reagents such as potassium cyanide, KCN. If you wish to carry
out this reaction, you do not look for cyanide ions; you look for a suitable reagent which can easily dissolve
to release the CN- nucleophile.
CH3CH2Br + KCN





ethanol
reflux
CH3CH2CN + KBr
KCN is not the nucleophile. It is simply a reagent that supplies the nucleophile, CN -.
Ethanol serves as a solvent. Recall that organic reactants do not dissolve well in water.
The reagent used should be a salt that can easily release the CN - ions. HCN should not be used.
It is a weak acid and liberates only a very small amount of CN- ions.
Nucleophilic substitution by CN- is very important in synthesis as it increases the length of the
carbon chain by one atom. If a synthesis involves conversion of an organic reactant to a product
that has one more carbon atom in the carbon chain, you may suspect that CN - was used as a
nucleophile.
Note that the organic product in the reaction above contains a new C-C bond.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
518
H
H
H
C
C
C
H
H
new bond
N
This information is important because it allows us to see that during the reaction, the cyanide
ion attacks the Cδ+ centre of the organic reactant using the carbon atom, not the nitrogen atom.
In other words, the negative charge on the CN- ion is on the carbon atom, not on the nitrogen
atom! The cyanide ion is usually written with the negative charge on the nitrogen atom, but this
is only for convenience. From this point on, we will show the negative charge on the carbon
atom, as shown on the dot - and - cross diagram below.
x
you study the dot-and-cross diagram, you will notice that carbon
C . . . N.. Ifappears
to have used 5 electrons in bonding, and yet it has only four
xxx
valence electrons in its outer shell. The extra electron therefore
originally came from another atom, for example H in HCN. When
HCN dissociates, the H-C bond breaks and the H atom leaves
without any electrons, so the remaining cyanide ion
will have an extra electron on its carbon atom, as illustrated below.
extra electron on C
= charge of -1
HCN
H x C . . . N..
xxx
x electron from C
.
dissociation
x
extra electron on C,
left behind by H
electron from N
electron from H
C . . . N.. + H
xxx
+
cyanide ion
Reaction (iii): formation of an amine
CH3CH2Br +


NH3
heat in bomb
CH3CH2NH2 +
HBr
The reaction is carried out in a sealed vessel known (bomb). This allows pressure to build up. In
turn, this speeds up the reaction.
The reaction results in the formation of an amine. The mechanism is outlined below.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
519
H
H
H
N
..
CH3
C
H
Br
H
H
H
+
N
H
H
.. N
H
 
C
CH3
H
Br
H
H
C
CH3 + HBr
H
Remember that N in its neutral organic compounds usually forms three bonds, remaining with a lone pair
of electrons. A nitrogen atom bonded to four groups is positively charged, as in the intermediate above.
Notice how one N-H bond breaks by heterolysis, supplying electrons to the N atom and neutralizing the
positive charge.
Nucleophilic substitution first order (SN1) reactions
Consider the reaction below which also occur by nucleophilic substitution, but the mechanism is different.
The difference in the details of the reaction can be traced to the nature of the organic reactant.
(CH3)3CCl + OH -
reflux
H2O
(CH3)3COH + Cl-
In this reaction, the OH⁻ nucleophile replaces the Cl⁻ as in the previous example. The mechanism of the
reaction consists of two steps. In the first step, the organic reactant dissociates into positive and negative
ions, just as if it were an ionic compound. The organic ion formed is positively charged and is known as a
carbocation. In the second step, the nucleophile, being electron rich, is attracted electrostatically to the
carbocation, to which it forms a dative bond, forming the product.
( CH₃)₃C-Cl → ( CH₃)₃C⁺ + Cl⁻
…
step 1: slow
( CH₃)₃C⁺ + OH⁻→ ( CH₃)₃COH
…
step 2: fast
Step 1 is the slower step because it involves heterolysis of a strong covalent bond. Moreover, a very unstable
intermediate (the carbocation) is formed.
Step 2 is faster because it involves the attraction of two ions of opposite charge. Step 1 is therefore the rate
determining step, and we use it to write the rate law.
R = k[( CH₃)₃CCl]
The order of reaction has been experimentally shown to be 1. This reaction is termed SN1 (nucleophilic
substitution first order). Such a reaction involves only one reactant in the slow rate determining step.
The mechanism of reaction was proposed after it was noticed that the rate of reaction varies directly as the
concentration of (CH₃)₃CCl but is not affected by changing the concentration of sodium hydroxide. The only
explanation is that OH- does not appear in the rate law because it is involved in a fast step
(non-rate determining).
The rate/concentration graphs with respect to both reactants are shown in Fig 10.7.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
520
Rate
Rate
first order
zeroth order
[NaOH]
[(CH3)3CCl]
Fig 10.7
The energy profile of the reaction is shown below.
CH3
H3C
Energy
+
C
intermediate
CH3
Ea1
Ea1
Ea2

progress of reaction



The reaction is non-concerted. There are two distinct steps leading from reactants to products.
This is a result of the relative stability associated with the intermediate. An intermediate is more
stable than a transition state. It can exist for a relatively longer period, so that there is a clear
separation in time between the first and the second step of the reaction.
Since the reaction has two steps, the energy profile also has two activation energies (two ‘humps’).
In this case, the first step has the higher activation energy, since it requires the breaking of a
relatively strong covalent bond. The second step has the lower activation energy because it
involves conversion of a very unstable intermediate (carbocation) into a stable
product.
Overally, the reaction has a negative enthalpy (exothermic) because it involves replacement of a
relatively weaker bond (C-Cl) by one which is stronger(C-O).
SN1 or SN2?
Reaction (i) below proceeds by SN2. Reaction (ii) proceeds by SN1.
CH₃Cl (aq) + OH⁻(aq) → CH₃OH(aq) + Cl⁻(aq)
… (i)
(CH₃)₃CCl (aq) + OH- (aq) → ( CH₃)₃COH(aq) + Cl⁻(aq)
... (ii)
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
521
But why should the reactions occur by different mechanisms? Both reactants are halogenoalkanes and they
are reacting with the same reagent. The answer is in relative stabilities of the carbocation intermediates that
could be formed from each of the halogenoalkanes.
The halogenoalkane in reaction (ii) is tertiary, that is, the carbon atom that bears the halogen atom is
bonded to 3 alkyl groups, as shown below.
H3C
H3C
H3C
H3C
H
H
C
The carbocation formed from this molecule by heterolysis of the C-Cl bond has a
positively charged C atom which is bonded directly to three alky (methyl groups),
which have an electron pushing (positive inductive) effect, as shown by the arrows in
the structure shown below.
The effect of this positive inductive effect is to partially neutralize the positive charge
and make the carbocation relatively stable. Formation of this carbocation is
therefore favourable, explaining why (CH₃)₃CCl reacts via the carbocation (SN1).
Chloromethane (CH₃Cl) can not react via a carbocation. If this were to happen, the
carbocation intermediate formed would be as shown below. The carbocation has a
positively charged C atom to which is bonded three H atoms. Hydrogen atoms do not
have the electron pushing ability associated with alkyl groups. The positive charge on
this carbon is well pronounced, making the intermediate very unstable, so unstable
that it is not formed at all.
Cl
CH3
+
C
CH3
+
C
H The reaction route chosen avoids a carbocation and proceeds via a pentavalent interme-
diate (SN2 ), which is more stable than the carbocation that could have been formed had the reaction
proceeded by SN1.
In general, SN1 would be the favoured mechanism if there are three alkyl groups bonded to the carbocation.
The stability of the carbocation depends on this. The length of the alky groups is also important. The longer
an alkyl group is, the stronger its electron pushing effect and the more stable the carbocation would be.
Compare the three carbocations shown in fig below:
H
H
+
C
H
(I)
H3C
H3C
+
+
C
CH3
H3C
(II)
C
CH2CH2CH3
CH2CH3
(III)
Carbocation (I) would be the least stable and its formation is unlikely. Both carbocation (II) and (III) are
stable enough to be formed as intermediates during nucleophilic substitution. However, carbocation (III)
is more stable since it has longer alkyl groups which have a stronger electron pushing effect.
Water as a nucleophile: hydrolysis reactions
Water can act as a nucleophilic reagent since it possesses lone pairs of electrons. Nucleophilic substitution
of a halogen atom from a halogenoalkane can take place, for example
CH3CH2Br + H2O → CH3CH2OH + HBr ... (i)
In this reaction, the nucleophile is actually the hydroxide ion which is derived from water. This is therefore
an example of a hydrolysis reaction, since it involves splitting of a water molecule.
In a hydrolysis reaction, water acts as a reactant and it is split into OH- and H+
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
522
Reaction (i) is very slow and an insignificantly small amount of product is formed. This is because the
O-H bond in water is so strong that it does not easily dissociate to release the OH - nucleophiles. Another
way to explain this is to say that the autodissociation of water is so slight (K w is a very small number =
10 -14 mol2dm-6) that the concentration of OH - ions is very small. The probability of collisions between the
halogenoalkane and OH- ions is therefore very small. The rate of formation of the product is increased by
boiling or refluxing the halogenocompound with an alkali such as NaOH, as already explained. The reaction
taking place is then referred to as alkaline hydrolysis. The mechanism of the hydrolysis reaction between a
halogenoalkane and water occurs by SN2 in primary halogenoalkanes and by SN1 in tertiary
halogenoalkanes. This is illustrated for a primary halogenoalkane below. In a primary halogenoalkane, the
carbon atom that carries the halogen is bonded to two hydrogen atoms.
C
H


C
Br
H
H
H
.. H
O..
H
H
H
H
C
C
H
.
O.
H
Br
H
H + H
H
H2O attacks electron deficient
carbon atom. New C-O bond
begins to form. At the same
time, C-Br bonds begin to
break.




H C
H
C OH
H H
+ HBr
unstable pentavalent transition
state.
Note positive charge on the bonded
'H2O' group. Heterolysis of C-Br
bond continues.
If the halogenoalkane contains more than one carbon atoms, it is the carbon atom which is directly
bonded to the halogen atom which is attacked by a nucleophile. This carbon is partially positive.
The mechanism of the reaction is concerted, as already explained. Bond formation and bond
breaking occur simultaneously.
In the transition state, the O atom from water has three bonds, not two as is the usual case. The
oxygen atom is therefore positively charged. An O-H bond then undergoes heterolysis, and the
electrons from this bond are shifted onto oxygen, neutralizing the positive charge. The hydrogen
ion produced from this heterolysis then combines with the Br - ion from the heterolysis of the
C-Br bond to form HBr.
The mechanism of the reaction is SN2 because the slower rate determining step involves collision
of two reactant particles, H2O and the primary halogenocompound. If the halogenocompound
was tertiary, the reaction would be SN1.
Table 10.3 Differences between SN2 and SN1 reactions in halogenoalkanes
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
523
SN2
SN1
Concerted mechanism. Bond formation and
breaking of the C-Hal bond is simultaneous.
Non- concerted mechanism. Breaking of the C-Hal
bond and formation of a new bond with the
nucleophile occurs in two distinct steps.
Nucleophile attacks in ‘first step’. However, in a
strict sense, it is difficult to really distinguish
between one step and the next.
Nucleophile participates in the second step.
Nucleophile directly attacks the reactant
(halogenoalkane).
Nucleophile does not attack the organic reactant.
Rather, it attacks the intermediate formed in the
first step.
A very unstable and extremely short-lived
pentavalent transition state is formed.
A relatively stable carbocation intermediate is
formed.
The energy profile of the reaction has only one
‘hump’, that is, one activation energy, since the
reaction essentially takes place in one step.
The energy profile of the reaction has two ‘humps’,
that is, two activation energies, since the reaction
essentially takes place in two steps.
Takes place in primary halogenoalkanes.
Takes place in tertiary halogenoalkanes and some
secondary halogenoalkanes.
Example : CH3CH2Br + OH-  CH3CH2OH + Br-
(CH3)3CBr + OH-  (CH3)3COH + Br-
Conditions required for nucleophilic substitution
From the foregoing discussion, it has probably become clear that the following conditions are required for
a nucleophilic substitution reaction to take place:
 Presence of a nucleophile.
 An organic reactant with an electron deficient (relatively positively charged) C atom, as in halogenoalkanes.
 A relatively weak bond in the organic reactant (such as the C-Hal bond) which can break to make
room for the incoming nucleophile.
This condition is important in deciding whether nucleophilic substitution or addition takes place.
If the first two conditions are satisfied, and yet there is no bond that can easily break to allow
substitution, then the reaction likely to take place would be nucleophilic addition.
Effect of C-Hal bond strength on the rate of nucleophilic substitution
Which of the following halogenoalkanes will react the fastest with OH - ions?
CH3CH2Cl
CH3CH2Br
CH3CH2I
There are two things to consider
 the ability of the Cδ+ in the organic reactant to attract the nucleophile. This depends on the polarity
of the C-Hal bond.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
524

the strength of the C-Hal bond. Nucleophilic substitution depends on the ability of the C-Hal to
undergo heterolysis. The weaker the bond, the faster the rate of substitution of the halogen atom
by a nucleophile.
If we consider the first point, then we expect CH3CH2Cl to react the fastest with nucleophiles. This is because
the C-Cl bond is the most polar. CH3CH2I will react the slowest because the C-I bond has the least polarity.
It is not so easy for the carbon atom of the C-I bond to attract a nucleophile.
However, from a consideration of bond strength, we would expect CH 3CH2Cl to have the smallest rate of
reaction, since the C-Hal bond is the strongest. It does not easily break to make room for the incoming
nucleophile. CH3CH2I would react the fastest since the C-I bond is the weakest. Bond polarity and bond
strength therefore have opposite effects on the rate of substitution with nucleophiles. However, bond
strength has the greater effect, so reactivity of the halogenoalkanes is in this order :
CH3CH2I > CH3CH2Br > CH3CH2Cl
least reactive
most reactive
weakest
C - Hal bond
strongest
C - Hal bond
Summary: nucleophilic substitution



In these reactions, a nucleophile replaces an atom or group of atoms from the organic reactant. A
nucleophile is an electron rich species (neutral or negatively charged) which is therefore attracted
to an electron deficient carbon in the organic reactant.
Nucleophilic substitution can occur by SN1 or by SN2, as summarized in Table 10.4.
Nucleophilic substitution is well known in, but not limited to halogenoalkanes. The product of the
reaction between a halogenoalkane and a nucleophile depends on which nucleophile has been used.
Let the nucleophile be Nu:, then the substitution reaction can be written as
R-Hal + Nu: → R-Nu + Hal-
nucleophile
Product formed
Conditions
OH- (e.g. from KOH)
alcohol
Boil or reflux in water.
alcohol
Yield is very poor even when
the reaction mixture is boiled.
Nitrile (need to reflux
in alcohol solvent)
Reflux in ethanol (solvent)
Amine
Heat in a sealed bomb.
H2O
CN- (e.g. from KCN)
NH3


3.1.2
All of the reactions above require refluxing or strong heating because they involve breaking of a
relatively strong covalent C-Hal bond.
Nucleophilic substitution with water as the nucleophile is known as hydrolysis. The reaction is very
slow because water does not easily release OH- nucleophiles.
Nucleophilic addition
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
525
First recall the difference between a substitution and an addition reaction. In a substitution reaction, an
atom or group of atoms replaces another in the reactant, resulting in the formation of two products, e.g.
CH₃Cl (aq) + OH⁻ (aq) →CH₃OH (aq) + Cl⁻ (aq) … 1
In an addition reaction, atoms from one reactant are incorporated into another reactant, resulting in the
formation of one product, e.g.
BCl₃ + NH₃ → BCl₃.NH₃
Nucleophilic addition is a reaction in which atoms of a nucleophilic reagent are incorporated into
an organic reactant which has a relatively positively charged C atom.
Nucleophilic addition reactions involve compounds that possess the carbonyl group.

carbonyl group C
. . 
.
O.
Addition of a nucleophile to a molecule would result in the increase of electron density around the
electron deficient carbon of the organic reactant. This leads to instability. There are two possible ways of
removing this instability: either a bond breaks by heterolysis, resulting in an atom or group of atoms moving
away with the extra electron density. This happens in nucleophilic substitution reactions in which there is
available a bond weak enough to be broken and make room for the incoming nucleophile.
The second way of redistributing electron density to accommodate the incoming nucleophile involves
shifting the extra electron density from carbon and accommodating it on an atom that has the ability to do
so. The carbonyl group is unique in this sense. The oxygen atom of the carbonyl group is able to accept and
stabilize an electron pair to create room for the incoming nucleophile. Also the carbonyl bond, being
double, is strong and cannot be easily broken as in nucleophilic substitution.
Here is a well known example of a nucleophilic addition reaction.
atoms from HCN
CH3COCH3 + HCN
propanone
(ketone)
trace NaCN/
NaOH
CH3C(OH)CH3CN
2-hydroxybutanenitrile
( cyanohydrin)
Notice why this is an addition reaction; atoms from the nucleophilic reagent (HCN) have been incorporated
into the organic reactant, to form one product. The reaction is very slow without a catalyst. NaCN or NaOH
are effective catalysts. The Mechanism of the reaction is as follows.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
526
H3C
 
C
O 

CH3
propanone
-.
.CN
Carbonyl compounds are trigonal planar (flat) at the carbonyl group
(bond angle, 1200).The C=O bond is polarized, since O is more electronegative than C. The C atom of the carbonyl group is relatively positively
charged and is therefore prone to attack by a nucleophile. The nucleophile (-CN in this case) approaches and attacks perpendicularly from
below or above the plane of the molecule to minimize repulsion from the
lone pair of electrons on the O atom. The nucleophile is generated from
the dissociation of HCN, which is a weak acid.
HCN Ý H+ + -CN
As the nucleophile bonds to the electron deficient C, there is an increase
in electron density around the central carbon atom, leading to instability.
The excess electron density brought in by the nucleophile is shifted on to
the O atom. This involves breaking of the pi bond of the carbonyl group (a
π bond is weaker than a σ bond). The electron pair from the pi bond is
accommodated by O. It is the ability of O to accept and stabilize a pair of
electrons which makes nucleophilic addition possible.
H+ The intermediate is a negatively charged ion. Note that this ion has a
H3C
C
H3C
-
....
O ..
tetrahedral geometry, bond angle 1090. The extra electron density that
came in with the nucleophile is now accommodated on the O atom of the
intermediate. The negative charge on the O atom is now neutralized by the
H+ ion from the dissociation of HCN. A stable product is thus formed.
CN
negatively charged
intermediate
H3C
H3C
C
OH
The product of this reaction is a compound known as a cyanohydrin or a
2-hydroxynitrile.
-
CN
cyanohydrin
(2-hydroxynitrile)
The reaction described above gives rise to a cyanohydrin. On their own, these compounds have limited
uses, but they are extremely important as intermediates in the synthesis of more useful compounds. These
synthesis reactions rely on the conversion of the nitrile group, particularly to an amine (reduction) and
carboxylic acid (hydrolysis).
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
527
OH
C
H3C
CH3
C N
LiAlH4 (lithium tetrahydrido aluminate) is a
very powerful reducing agent used to carry out
reductions where many other reducing agents
cannot. Its major disadvantage is that it is very
sensitive towards water, in which it explodes to
give a mixture of products. This reagent adds H
atoms to the nitrile group.
cyanohydrin
LiAlH4 /ether
H2SO4(aq) / reflux
(reduction)
(hydrolysis)
OH
OH
C
H3C
CH3
The nitrile group can be converted to a
carboxylic acid by hydrolysis. The reactant in
all hydrolysis reactions is water, but an acid
catalyst is often required. In this case, H2SO4
supplies H+ ions which catalyze the reaction.
C
H3C
CH2NH2
CH3
COOH
Catalytic role of NaOH or NaCN
The reaction between propanone and HCN is very slow. This is because HCN is a weak acid; its dissociation
in water is only slight, so the concentration of CN - nucleophiles is very small. The reaction can be made
faster by adding trace amounts of NaCN or NaOH.
Catalysis by NaOH
The following equilibrium exists in the reaction vessel
HCN Ý H+ + CN-(aq) ... (i)
Upon addition of NaOH (aq), H+ ions are removed from the equilibrium
H+ (aq) + OH-(aq) → H2O
This forces equilibrium (i) to the right, releasing more CN - nucleophiles (Le Chatelier’s principle). The
increase in the concentration of cyanide ions increases the probability of collisions with propanone
molecules, and this in turn increases the rate of reaction.
Catalysis by NaCN
Being a salt of high solubility, NaCN dissociates in water to release CN- nucleophiles. These are the ligands
which participate in the first step of the reaction, resulting in the formation of a negatively charged intermediate. The intermediate has a high affinity for hydrogen ions, so it rapidly removes H+ from HCN,
generating - CN ions (the catalyst is thus regenerated). Once the first anion intermediates are formed, the
reaction becomes rapid as these anions rapidly remove H+ ions from HCN, thus generating more CN- ions.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
528
H
H3C
 
C
H3C
O  
H3C
CH3
propanone
C
-
CN
.. ..
O ..
CN
-.
. CN from NaCN, not HCN
anion intermediate
H3C
H3C
C
-
CN
..
..
O ..H + CN
catalyst regenerated
Conditions for nucleophilic addition

A nucleophilic reagent must be present.

A carbonyl group should be present in the organic reactant. This group has an electron deficient
carbon atom which can be attacked by the nucleophile.
Questions, solutions and discussions
Q1
The kinetics of the reaction between (CH3)3CCl and NaOH has been studied.
(a) Construct a balanced equation for this reaction.
(b)
State the conditions required for the two substances to react.
(c)
The rate of the reaction changes proportionally when the concentration of (CH 3)3CCl
is changed. Changing the concentration of NaOH does not affect the rate of reaction.
Comment on these observations as fully as possible.
Solutions
(a)
(CH3)3CCl + NaOH  (CH3)3COH + NaCl
(b)
Reflux in aqueous solution.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
529
(c)
From the given facts, we conclude that the reaction is first order with respect to
(CH3)3CCl and zeroth order with respect to NaOH. NaOH is therefore involved in a fast
non-rate determining step.
(CH3)3CCl (CH3)3C+ + Cl-
…I
slow
(CH3)3C+ + OH-  (CH3)3COH
… II
fast
Step I is the slower rate determining step because it involves the breaking of a relatively
strong C-Cl bond and the formation of a very unstable intermediate (a carbocation).
The rate law is written using the slower rate determining step,
R = K [(CH3)3CCl]
That is, the reaction is first order with respect to the halogenoalkane. A change in the
concentration of the halogenoalkane results in a proportional change in the speed
of reaction. Step II, involving NaOH, is the faster non-rate determining step. Changing
concentration of NaOH therefore has no effect on the rate of reaction.
The reaction is said to proceed via an S N1 mechanism since the slow rate determining step
involves only one reactant molecule.
Q2
1-(4-hydroxyphenyl)butan-2-one, A is responsible for the aroma of raspberries.
A
O
H
O
1 -(4-hy drox y pheny l)butan-2-one
(a) Circle and name the functional groups in A.
(b) How would you expect A to react with
(i) dilute HNO3
(ii) HCN in the presence of a little NaCN?
Draw the structure of the product formed in each case.
Solutions
(a)
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
530
(b)(i) Electrophilic substitution (nitration) in the benzene ring.
[ no catalyst is needed since the benzene ring is activated by the -OH group]
The product formed is
(ii)
Nucleophilic addition to the carbonyl group, to form
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere
in the document. Use the Text Box Tools tab to change the formatting of the pull quote
Exercise
10.5
text box.]
1.
The cyanide ion reacts with both chloropropane and propanone.
(a) Name the type of reaction which takes place in each case
(b) What are the reagents and conditions required for each reaction?
(c) Outline the reaction mechanism for each reaction.
(d) Suggest why chlorobenzene has no reaction with cyanide under ordinary lab conditions.
2.
Draw the structure of the product formed when benzaldehyde reacts with HCN in the presence of a
trace amount of NaOH(aq).
(a) What type of reaction takes place?
(b) Why is the reaction very slow in the absence of OH- ions?
(c)
What other substance can be used to catalyze the reaction?
[Type a quote from the document or the summary of an interesting point. You can position the text
box anywhere in the document. Use the Text Box Tools tab to change the formatting of the pull quote
text box.]
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
531
10.3.2
Electrophilic reactions
Electrophilic reactions involve electrophilic reagents.
An electrophile is an electron deficient species, usually positively charged, that therefore seeks
an electron rich centre.
(Electrophile literally means ‘electron lover’. To ‘love’ electrons, a species must be relatively positively charged).
Most electrophiles are positive ions generated insitu, that is, they do not have an independent existence but
must be generated during the reaction in which they participate.
Electrophiles are attracted to electron rich centres, notably the C to C double bond and the benzene ring.
Examples of electrophiles include Br⁺ (generated insitu from bromine), H⁺ from a mineral acid such as HCl,
and the nitronium ion, NO2⁺, generated insitu from the reaction between concentrated sulphuric and nitric
acids.
There are two types of electrophilic reactions, addition and substitution. Electrophilic addition occurs to
the carbon to carbon double bonds in alkenes, whilst electrophilic substitution occurs in the benzene ring.
Elecctrophilic reactions
Addition
Occurs in C to C double bond
3.2.1
substitution
occurs in benzene ring
Electrophilic addition
This is a reaction in which atoms of an electrophilic reagent are incorporated to the electron rich centre
of the organic reactant.
These reactions occur at the C to C double bond of alkenes.
Halogenation of ethene
You will probably remember that alkenes decolourize cold bromine in the dark. This is a reaction of the C
to C double bond, for example
CH₂ = CH₂ (g) + Br₂ (l) → CH2BrCH2Br (l)
In this reaction, ethene is converted to 1, 2- dibromoethane. The numbers 1 and 2 emphasize that the Br
atoms are added across the double bond, rather than on the same carbon atom. The reaction above does
not reveal the electrophile. This is because the electrophile, Br⁺, is generated insitu from bromine.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
532
The mechanism
H
H
π electrons of the double bond repel the lone pairs of electrons in bromine.
This polarizes bromine, that is, one end of the molecule becomes richer in
electrons and the other becomes deficient in electrons. The electron deficient
pole of the molecule is relatively positively charged, so it is attracted to the
electron rich double bond, and a C-Br bond begins to form. At the same time,
the Br-Br begins to break. Heterolysis of the Br-Br bond produces the bromonium ion, Br+, which is the electrophile in this reaction. However, the Br + ion
is not released in solution but it immediately forms a bond to one carbon atom
of the double bond. This requires breaking of the π bond of the C to C double
bond; it is the electron pair from this bond which is used to form the new CBr bond. The first step in the reaction is very slow because it involves breaking
of a strong Br-Br bond.
H
C  C

H
.
Br x Br
H
H
Br
H
C
Breaking of the π bond in ethene leaves one carbon atom with three bonds
(electron deficient) instead of four. The intermediate of this reaction is therefore a carbocation.
C+
H
carocation intermediate
H
H
Br
H
H
C
C+
Br .. -
H
H
C
C Br
H
Br
H
1,2-dibromoethane

The intermediate is very unstable and it immediately reacts with the bromide ion from the heterolysis of bromine. In this reaction, the bromide ion
actually behaves as a nucleophile. However, this step is very fast (it involves
attraction of oppositely charged ions) and so it is non rate determining. The
reaction is termed electrophilic addition because the slower rate determining step involves the participation of an electrophile.
The product of this reaction is an oily colourless liquid known as 1,2-dibromoethane. The reaction of an alkene with bromine is a test for unsaturation
(presence of a C to C double bond). Alkenes decolourize bromine liquid in
the cold.
What evidence is there that the reaction proceeds via a carbocation?
If the reaction is carried out using pure bromine liquid, only one product, 1, 2-dibromoethane is formed.
However, when bromine water (bromine in water) is used, a mixture of products is formed, 1,2dibromoethane and 1-bromoethanol, whose structure is shown below. This shows that the reaction
proceeds via a relatively stable cation which then reacts indiscriminately with whatever nucleophile is
present. When pure bromine is used, the only nucleophile available to react with the carbocation is Br -.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
533
However, when bromine water is used, there are also OH - ions from the autodissociation of water. These
also react with some of the carbocations, to form a different product, 1-bromoethanol.
H
H
1
C+
C
H
H
Br
carbocation intermediate
Br- from the heterolysis of
Br2 attacks C1
H
1
C
C
H
H
H
Br
H
Br
H
Br
OH- from the dossociation of
water attacks C1
1,2-dibromoethane
C
1
C
H
OH
H
1-bromoethanol
If other negatively charged ions are added, for example, NO3- and Cl-, corresponding products are
formed, in addition to 1, 2 - dibromoethane, as shown below.
CH2Br
CH2NO3
NaNO3(aq)
CH2Br
CH2
NaCl(aq)
CH2Br

CH2Cl
Why does electrophilic addition to the double bond in alkenes involve breaking of a
pi bond, not a sigma bond?
A pi bond is weaker. This is because on average, pi electrons are a larger distance from the nuclei of both
atoms of the C to C bond.
Addition of HX to the double bond (X = Cl, Br, I)
Hydrogen halides can also be added to the double bond via electrophilic addition. The electrophile in this
case is the H⁺ ion generated insitu from the hydrogen halide. Consider the addition of HCl to ethene.
CH₂CH₂ + HCl → CH₃CH₂Cl
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
534
In this reaction, an alkene is converted to a halogenoalkane. The mechanism is similar to that of the
addition of bromine to the double bond. If a pure product is required, then the reaction should not be carried
out in water. As already explained, the reaction proceeds via a carbocation and if water is present, the
carbocation may react with OH- ions from water. The product, chloroethane, will be contaminated with
ethanol. A suitable organic solvent such as hexane can be used.
H
H
H
H
C  C

H
H
C
H
1
H
Cl-
C+
2
H
H
H
H
C
H
Cl
C
H
carocation intermediate
H
Cl
Once more, note that atoms of the electrophilic reagent, HCl, are added across the double bond. The first
atom of the double bond (1) accepts the electrophile and the C to C pi bond breaks. The other atom of the
double bond (2) acquires a positive charge, which attracts a negative ion (nucleophile) generated from the
electrophilic reagent.
If water is used as the solvent, contamination of the product by ethanol cannot be avoided. The OH⁻ ions
from the auto- dissociation of water can also react with the carbocation intermediate, resulting in the formation of ethanol. The amount of ethanol formed can be kept at a minimum by using concentrated HCl.
Markownikoff’s rule
Addition of HX to the double bond can give rise to a mixture two isomeric products.
Consider the reaction of propene with HCl. The electrophile can bond either to carbon 1 or to carbon 2. This
affects the carbon at which the positive charge develops, and in turn this affects the position where the
negative ion from HX bonds. Markownikoff’s rule helps us to predict the isomer which is formed in greater
amounts.
Markownikoff’s rule says:
During the reaction of an alkene with a hydrogen halide HX, the H+ electrophile from HX attacks that
carbon of the double bond which already contains the larger number of hydrogen atoms.
The following illustration pertains to the reaction of propene with HCl. Two possibilities are shown

Markownikoff addition in which the H+ ion attacks the carbon of the C=C bond that bears the
larger number of hydrogen atoms.
 AntiMarkownikoff addition in which the H+ ion attacks the carbon of the C=C bond which
bears the smaller number of hydrogen atoms.
AntiMarkownikoff addition is not favoured because t results in the formation of an unstable intermediate.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
535
1
H
2
H
C
C C
H
H H
H
If electrophile (H+) attacks
Carbon 2
Carbon 1
positive charge develops at C 2
positive charge develops at C 1
H H
+1 2
H C C C
Cl-
H
H
H
H
H H
C
C C
H
H H
H
H H
C
C C
H
H H
H
H
H
A larger proportion of the carbocation
intermediates formed have a positive
charge on carbon 2. This is because the
positive charge would then be stabilized
by the electron pushing effect of two
alkyl (methyl) groups. This is the
reasoning behind Markownikoff’s rule.
H
2-chloropropane
H
H
C
+
C
H
H
H
C
H
H
What would be the product or products formed when but-2-ene reacts with HBr in
a suitable organic solvent?
H
H
H
C
H
A
+2
C C
H Cl H
chloropropane
Q
Cl-
H
C
1
Cl H H
H
When propene reacts with HCl, two products
are formed, chloropropane and 2chloropropane. The major product is 2chloropropane, as predicted by
Markownikoff’s rule. H+ ions mostly attack
carbon 1 since it is the one that already has the
larger number of hydrogen atoms.
A large number of the carbocation
intermediates will therefore have a positive
charge on carbon 2.
C
C
C
H
H H H
but-2-ene
Only one product, 2-chlorobutane is formed. This is because the double bond in but-2ene is symmetrically positioned. It does not matter which carbon is attacked by the
electrophile, the same product will be produced.
Prepublication manuscript
©L.Mwanawenyu. Please do not photocopy or reproduce in any way
536
Summary: electrophilic addition






Occurs to the C to C double bond in alkenes
During the reaction, atoms from the electrophilic reagent are added across the double bond, to form
only one product.
The reaction is initiated by polarization of the electrophilic reagent by π electrons of the C to C
double bond. The π electrons have the ability to cause this polarization because they occupy a
diffuse (spread out) π orbital which is therefore only weakly attracted and stabilized by the nucleus.
Examples of electrophilic reagents which can add to the double bond in alkenes are the halogens,
X2 and hydrogen halides, HX. For the halogens, the electrophile is the ion X + which is generated
insitu from heterolysis of the X-X bond. For HX, the electrophile is H+.
The reaction proceeds via a carbocation intermediate. This explains why it is possible to form a
mixture of products when two or more nucleophilic particles (e.g. OH - from water) are present in
the reaction mixture.
The