Find the average oxidation number of sulphur in tetrathionate ion.

Step-by-step solution

The chemical formula of the tetrathionate ion is S₄O₆²⁻ . Let the average oxidation number of sulphur be x . The oxidation number of oxygen is -2 . The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the net charge of the ion. 4x + 6(-2) = -2 4x - 12 = -2 4x = 10 x = +2.5 Alternatively, from the structure of the tetrathionate ion ( ^-O₃S-S-S-SO₃^- ), the two terminal sulphur atoms have an oxidation state of +5 each, and the two central sulphur atoms have an oxidation state of 0 each. Avera