Oxidation number

Oxidation number is assigned to each element in a compound to keep track of electrons in oxidation and reduction processes.

Oxidation numbers can be zero, positive or negative numbers, and integers or non-integers.

Simple ions

For the simple ions in the compound, oxidation numbers bear no real physical meaning.

Polyatomic ions

In polyatomic compounds which contain polyatomic ions, elements in polyatomic ions do not have measurable ionic charge. For example, in sodium nitrate, NaNO₃,

• the sodium ion carries a +1 charge
• the nitrate carries a -1 charge
• the nitrogen in the nitrate ion does not have any ionic charge since it is covalently bonded to oxygen atoms. However, we can determine an apparent charge that is associated with nitrogen.

Since we know that a compound must be neutral in charge, we follow a set of rules to assign the oxidation number to nitrogen.

Rule #6: Oxygen is usually assigned an oxidation number of -2 for oxides. In this case, we have three oxygen atoms in NO₃^-.

Rule #9: The sum of oxidation numbers of an ion or complex ion is the same as the charge on that ion.

Let the oxidation number of nitrogen be represented as N. Set up the equation and solve for N.

N + 3 (-2) = -1

N = +5

We would determine that the apparent charge on nitrogen is +5.

Assigning oxidation numbers

Example 1: What is the oxidation number of sulfur in SO₂?

• The oxidation number of each oxygen is -2 (according to rule 6). There are two oxygen atoms.
• The overall charge of the compound is zero, so the sum of the oxidation numbers of all the atoms in the formula is zero (according to rule 8).

Let the oxidation number of sulfur be represented as S. Set up the equation and solve for S.

S + 2 (-2) = 0

S = +4

The oxidation number of sulfur in SO₂ is +4.

Example 2: What is the oxidation number of sulfur in H₂SO₄?

• The oxidation number of hydrogen is +1 (according to rule 3).

There are two hydrogen atoms.

• The oxidation number of oxygen is -2 (according to rule 6).

There are four oxygen atoms.

• The overall charge of the compound is zero, so the sum of the oxidation numbers of all the atoms in the formula is zero (according to rule 8).

Let the oxidation number of sulfur be represented as S. Set up the equation and solve for S.

2(+1) + 4(-2) + S = 0

-6 + S = 0

S = +6

Or, you can just consider the SO₄^2- ion:

Then, you must apply according to rule 9

Rule 9: The sum of oxidation numbers of an ion or complex ion is the same as the charge on that ion.

Let the oxidation number of sulfur be represented as S. Set up the equation and solve for S.

4(-2) + S = -2

-8 + S = -2

S = +6

Either way, you should still come out with an oxidation number of +6 for sulfur.