(i) The structure of compound containing sulphur in $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ is :

Let the oxidation state of sulpher be $x$ :

$$ \begin{gathered} 4 \times x+2 \times(+1)+6 \times(-2)=0 \\ 4 x=12-2=0 \\ 4 x=10 \\ x=\frac{5}{2}=2.5 \end{gathered} $$

(ii) Each of the corner sulphurs utilises five valence electrons to form bond with oxygen atoms.

Their oxidation state is +5.

(iii) Oxidation state of central sulphur atom is zero (0).

Difference between two types of sulphur = 5 – 0 = 5.