Question

Give the oxidation state of underlined.
$N{a}_2\underset{–}{S_4}{O}_6$.

Answer 1

Second and third sulphur atoms are joined to two sulphur atoms so their oxidation number is 0. Two sulphur atoms which are joined to 3 oxygen atoms in which one oxygen is attached to $Na$ has -1 oxidation state and other two oxygen atoms has -2 oxidation state so sulphur has +5 oxidation state.

Oxidation state of sulphur is $\frac{(0+0+5+5)}{4}$ $=2.5$.