Here, we’ll go through another example of finding the oxidation number of each element in a polyatomic ion

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


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


Carbon has variable oxidation numbers, so the oxidation number (click) of carbon in this ion is unknown. Therefore, we’ll call it (click) x


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


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


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


We’ve called the oxidation number of carbon x,


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


So the total charge on 5 carbon atoms is (click) 5x


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


The oxidation number of a hydrogen atom is plus 1


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


So the total charge on 5 hydrogen atoms is (click) 5 times positive 1


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


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


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


We can solve for x in this equation to find the oxidation number of carbon


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


So we write 5x


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


Plus 5


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


Equals negative 1


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


Subtracting 5 from both sides, gives us 5x = negative 1 minus 5


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


which equals negative 6


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


Dividing both sides by 5 gives us…


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


X equals negative 6 5th’s


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


Or as a mixed number, negative 1 and 1 fifth


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


So we can say the the oxidation number of carbon in this ion is


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


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.


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


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.


C

H +1

55C H


6 1 or 1

5 5

And the oxidation number of hydrogen is positive 1.


C

H +1

55C H


6 1 or 1

5 5

Documents

Here, we’ll go through another example of finding the oxidation number of each element in a polyatomic ion