Find the average oxidation number of sulphur in tetrathionate ion.

The chemical formula of the tetrathionate ion is $S_4O_6^{2-}$.

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 ($^-O_3S-S-S-SO_3^-$), 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.

Average oxidation number $= \dfrac{5 + 0 + 0 + 5}{4} = +2.5$

Answer: $+2.5$