Here, we’ll go through an example of finding the oxidation number of each element
in a polyatomic ion.
Find the oxidation number of each element in the
ion: H 2PO2 .
We’re asked to find the oxidation number of each element in the ion with the
formula H2PO2 minus

H 2PO2
Element
Oxidation
Number
K
S
O
+1
+7
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
We’ll start by writing the formula up here

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
And a table here for the oxidation number of each element

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
The assigned (click) oxidation number of hydrogen in a compound or polyatomic
ion is (click) positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
This is not a peroxide compound or molecular oxygen (click) so the oxidation
number of oxygen is (click) negative –2

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
We don’t (click) know what the oxidation number of phosphorus is so we call it
(click) x

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
The oxidation number of a hydrogen atom is positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
So 2 hydrogen atoms have a total charge of (click) 2 times positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
The oxidation number of a phosphorus atom we’ve called x

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
So below phosphorus, we write (click) plus x

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
The oxidation number of an oxygen atom is negative 2

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
So the total charge on 2 oxygen atoms is (click) 2 times negative 2

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
The net charge on this ion, (click) shown on the top right of the formula is minus 1,
so the charges on all the atoms add up to (click) negative 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
Now we solve this equation for x, to find the oxidation number of phosphorus in
this ion

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2   2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
2 times positive 1 is equal to (click) 2

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2   1   x  2   2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
We write x here

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
And 2 times negative 2 is negative 4, so we write (click) minus 4 here

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
And these add up to negative 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
Positive 2 and negative 4 add up to negative 2

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
So we can say that x minus 2 is equal to negative 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
Adding 2 to both sides of the equation gives us x equals negative 1 plus 2

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
Which is equal to positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
x
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
So we can say that the oxidation number of phosphorus in this ion

H 2PO2
Element
Oxidation
Number
H
P
O
+1
+1
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
Is equal to positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
+1
–2
2  1   x  2  2   1
2  x  4  1
x  2  1
x  1  2
x  1
Find the oxidation number of each element in
the ion: H2PO2 .
If you want a copy of the whole solution to this problem, you can pause the video,
take a screen shot, and print it.

H 2PO2
Element
Oxidation
Number
H
P
O
+1
+1
–2
Find the oxidation number of each element in
the ion: H2PO2 .
So we can summarize by saying (click) the oxidation number of hydrogen in this ion
is positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
+1
–2
Find the oxidation number of each element in
the ion: H2PO2 .
The oxidation number of phosphorus is also positive 1

H 2PO2
Element
Oxidation
Number
H
P
O
+1
+1
–2
Find the oxidation number of each element in
the ion: H2PO2 .
And the oxidation number of oxygen is negative 2