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