Question

What is the difference in the oxidation numbers of the two types of sulphur atoms in $N{a}_2{S}_4{O}_6$?

Answer 1

The structure of $N{a}_2{S}_4{O}_6$ is.
$N{a}^{+}\stackrel{-}{O}-\stackrel{\stackrel{O}{\uparrow }}{\underset{\underset{O}{\downarrow }}{S}}-S-S-\stackrel{\stackrel{O}{\uparrow }}{\underset{\underset{O}{\downarrow }}{S}}-\stackrel{-}{O}N{a}^{+}$
Oxidation number of sulphur atom involved in coordinate bond formation is $(+5)$ and that of middle sulphur atom is zero. Hence, the difference in oxidation number of two types of sulphur atoms will be $(+5)$.