In an ion, the sum of the oxidation states is equal to the overall ionic charge. Note that the sign of the oxidation states and the number of atoms associated with each oxidation state must be considered. In OH-, for example, the oxygen atom has an oxidation state of - 2 and the hydrogen atom has an oxidation state of + 1, for a total of (-2) + (+1) = -1. What is the oxidation state of an individual sulfur atom in SO32-?

Answer: the oxidation state of an individual sulfur atom in SO₃⁻² is + 4.

Explanation:

1) Fundamental considerations: There are important rules that you need to know to find the oxidation states of neutral compounds or ions.

In a neutral compound the sum of the oxidation states is equal to zero.

In an ion, the sum of the oxidation states is equal to the overall ionic charge.

Except in peroxides, oxygen (O) atoms bonded to other atoms have oxidation state - 2.

Those are the relevant rules for this question.

2) Procedure:

The ion for which you have to fiind the oxidation state of an indivitual sulfur atom is SO₃⁻².

Let x be the oxidation state of the sulfur (S) atom.

Oxidation state of O = - 2

Total oxidation number for 3 atoms of O: 3 * (-2) = - 6

Overall ionic charge of SO₃⁻²: - 2

Sum of the oxidation states: x + (-6) = overal charge

⇒ x + (-6) = - 2

Solve for x: x = - 2 + 6 = + 4

4) Conclusion: the oxidation state of an individual sulfur atom in SO₃⁻² is + 4.