Finding Oxidation Numbers
Example 3 Here, we’ll go through another example of finding the oxidation number of each element in a polyatomic ion.
We’re asked to find the oxidation number of each element in the ion with the formula C5H5 minus
Find the oxidation number of each element in the ion: .55C H
We’ll start by writing the formula up here
Find the oxidation number of each element in the ion: .
Element OxidationNumber
K +1
S +7
O –2
55C H
55C H
And a table here for the oxidation number of each element
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Carbon has variable oxidation numbers, so the oxidation number (click) of carbon in this ion is unknown. Therefore, we’ll call it (click) x
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
x
x
x
x
x
5 5C H
Find the oxidation number of each element in the ion: .55C H
Hydrogen’s symbol is written (click) to the right of the symbol for C, but carbon is not a metal, so this is not a metallic hydride, therefore the oxidation number of hydrogen is (click) the normal positive 1
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
x
x
x
x
x
5 5C H
Find the oxidation number of each element in the ion: .55C H
We’ve called the oxidation number of carbon x,
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
So the total charge on 5 carbon atoms is (click) 5x
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
The oxidation number of a hydrogen atom is plus 1
Element OxidationNumber
C x
H +1
5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
So the total charge on 5 hydrogen atoms is (click) 5 times positive 1
Element OxidationNumber
C x
H +1
5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
The net charge of this ion, shown on the top right (click) of the formula is negative 1, so the charges on all the atoms add up to (click) negative 1
Element OxidationNumber
C x
H +1
1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
5 5 1
x
x
x
x
x
5 5C H
Find the oxidation number of each element in the ion: .55C H
We can solve for x in this equation to find the oxidation number of carbon
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 1 or 1
5 5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
So we write 5x
Element OxidationNumber
C x
H +1
5 5 1
5 1 5
5 6
6 1 or
1 1
5 5
5
1
5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Plus 5
Element OxidationNumber
C x
H +1
5 1
5 1 5
5 6
6 1 or 1
5 5
5 1
5
15
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Equals negative 1
Element OxidationNumber
C x
H +1
1
5 1 5
5 6
6 1 or 1
1
5
5
5
5
5
5 1
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Subtracting 5 from both sides, gives us 5x = negative 1 minus 5
Element OxidationNumber
C x
H +1
5 1 5
5 6
6 1 or 1
5 5
5 5 1 1
5 5 1
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
which equals negative 6
Element OxidationNumber
C x
H +1
5 6
6 1
5 5 1 1
5 5 1
or 15 5
1 55
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Dividing both sides by 5 gives us…
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
6
5
1 or 1
5
6
5
5
1 5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
X equals negative 6 5th’s
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
6
5
1 or 1
5
6
5
1 5
5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Or as a mixed number, negative 1 and 1 fifth
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
5 1 5
5 6
6 or 1
5
5
1
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
So we can say the the oxidation number of carbon in this ion is
Element OxidationNumber
C x
H +1
5 5 1 1
5 5 1
6
5
1 or 1
5
6
5
1 5
5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
Negative 6 fifths or negative 1 and 1 fifth. Even though non-integer oxidation numbers are not as common as integer ones, we see that they are possible, so don’t be alarmed if you occasionally get a fraction for an answer.
Element OxidationNumber
C
H +1
5 5 1 1
5 5 1
6
5
1 or 1
5
6
5
1 5
5
x
x
x
x
x
55C H
Find the oxidation number of each element in the ion: .55C H
6 1 or 1
5 5
So we can summarize (click) by saying that the oxidation number of carbon in this ion is negative 6 fifths or negative 1 and 1 fifth.
Element OxidationNumber
C
H +1
55C H
Find the oxidation number of each element in the ion: .55C H
6 1 or 1
5 5
And the oxidation number of hydrogen is positive 1.
Element OxidationNumber
C
H +1
55C H
Find the oxidation number of each element in the ion: .55C H
6 1 or 1
5 5