# How would you find the oxidation number for SO4^2-?

Oxidation of sulfur in sulfate is $V {I}^{+}$
Oxidation number is the charge left on the central atom when all the bonding pairs of electrons are removed, with the charge going to the most electronegative atoms. If I do this for sulfate I get ${S}^{6 +}$ and $4$ $\times$ ${O}^{2 -}$.
What about thiosulfate anion, ${S}_{2} {O}_{3}^{2 -}$; here oxygen retains the $- I I$ oxidation state. What about the sulfurs?