What is the average oxidation state of sulphur in ${\mathrm{Na}}_2{ \mathrm{S}}_4{\mathrm{O}}_6$ ?

Given: It is required to find the average oxidation state of sulphur in ${\mathrm{Na}}_2{ \mathrm{S}}_4{\mathrm{O}}_6$.

Concept Used: Oxidation State

Detailed Solution:

The structure of ${\mathrm{Na}}_2{ \mathrm{S}}_4{\mathrm{O}}_6$ will be,

The average oxidation state will be,

$\Rightarrow \frac{2\times 5+2\times 0}{4}$

$\Rightarrow \frac{10}{4}$

$\Rightarrow 2.5$

Hence , the average oxidation state of sulphur in ${\mathrm{Na}}_2{ \mathrm{S}}_4{\mathrm{O}}_6$  is 2.5