Oxidation numbers are given to an element in a compound, such that the sum of oxidation numbers of the compound’s constituent elements is equal to the charge on the compound.

Element in the compound Oxidation Number
Group 1 metals+1
Group 2 metals+2
Oxygen -2
Hydrogen +1
Fluorine-1
Chlorine-1

Example: 
Let us consider the following compound:

  • H₂SO₄

Find the oxidation state (number) of sulphur.

Hydrogen usually has oxidation number (+1)
There are two hydrogens atoms so we have (+1 x 2) = +2

Oxygen usually has oxidation number (-2)
There are four oxygen atoms, so we have (-2 x 4) = -8

The charge on the compound is 0.

So we must satisfy the equation: 

Ox no. of H + Ox no. of O + Ox no. of S = 0
2 + (-8) + Ox no. of S = 0
-6 + Ox no. of S = 0
Ox no. of S = +6


Example: 
Let us consider the following compound:

  • MnO₄⁻¹

Find the oxidation state (number) of manganese in the above compound. 

Oxygen usually has oxidation number (-2)
There are four oxygen atoms, so we have (-2 x 4) = -8

The charge on the compound is -1.

So we must satisfy the equation: 

Ox no. of O + Ox no. of Mn = -1
(-8) + Ox no. of Mn = -1
-8 + Ox no. of Mn = -1
Ox no. of Mn = +8 – 1
Ox no. of Mn = +7


Oxidation number of an element on its own is always zero (0). 

Example:
Oxidation number of ‘Na’ in Na = 0
Oxidation number of ‘O’ in O₂ = 0
Oxidation number of ‘H’ in H₂ = 0