The oxidation state of S in SO₃²⁻ is +4. Here's how to determine it:

* Oxygen's usual oxidation state: Oxygen typically has an oxidation state of -2, except in peroxides where it's -1.

* Overall charge: The overall charge of the sulfite ion (SO₃²⁻) is -2.

* Calculation: Let x be the oxidation state of sulfur.

* x + 3(-2) = -2

* x - 6 = -2

* x = +4

Therefore, the oxidation state of sulfur in SO₃²⁻ is +4.