Multiple Oxidation Elements
Elements that form
more than one ion.
• I CAN write
formulas for and name
compounds containing Multiple
Oxidation elements.
What is a MultiOx element?
• Certain elements, particularly from the
TRANSITION ELEMENTS and a few HEAVIER
AMN ELEMENTS tend to lose both VALENCE
electrons as well as one or more from the
NEXT ENERGY LEVEL below valence.
• These are known as MULTIPLE OXIDATION
STATE ELEMENTS or MULTIOX for short.
• Since these come from the METALS of the
periodic table, they are always CATIONS (+).
• There are quite a few elements that exhibit
multiple oxidation. REFER TO THE HANDOUT
PROVIDED.
Compounds with MultiOx Elements
• Its easy to recognize a MultiOx element in the
NAME OF A COMPOUND.
– Following the name of the POSITIVE ELEMENT
there will be a PARENTHESIS WITH A ROMAN
NUMERAL.
– The ROMAN NUMERAL indicates the POSITIVE
CHARGE of the element IN FRONT OF IT.
• EXAMPLE Iron (III) Sulfide The roman numeral
indicates the IRON has a +3 charge in this compound.
– This makes it easy to determine the formula for a
compound containing a MultiOx element.
Practice Problems
• Write the formula for these compounds:
+3
+2
• Copper (II) Phosphide
Iron (III) Oxide
• 3 (+2) + 2 (-3) = 0
•
Cu
P
•
Cu3P2
2 (+3) + 3 (-2) = 0
Fe
O
Fe2O3
Naming Compounds with MultiOx
elements from their formulas.
• Naming a compound with a MultiOx element
from its FORMULA requires a bit of working
backwards.
• 1. Look at the formula and refer to your
MultiOx list if needed to see if there is a
MultiOx element in the compound.
• 2. Work BACKWARDS to determine the name
of the compound.
Practice Problems
• What is the name of the compound PbCl4 ?
• 1. Check for a MultiOx element: LEAD
• 2. Break the formula apart:
•
( ? ) + 4 ( -1) = 0
•
Pb
Cl
• 3. Use the NEGATIVE ION to determine the charge of
the MultiOx ion:
•
+4
-4
•
( ? ) + 4 ( -1) = 0
•
Pb
Cl
•
•
•
•
•
+4
-4
( +4 ) + 4 ( -1) = 0
Pb
Cl
Therefore this is LEAD (IV) so the name of
the compound is:
Lead (IV) Chloride
Another Practice Problem
• What would the name of SnI4 be?
•
+4
-4
•
( ? ) + 4 (-1) = 0
•
Sn
I
• Therefore this is Tin (IV) so the compound
would be:
•
Tin (IV) Iodide