The average oxidation state of sulphur in sodium tetrathionate $\left({\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6\right)$ is:

The average oxidation state of Sulphur in sodium tetrathionate  $\left({\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6\right)$ is 2.5.

Let x be the oxidation state of sulfur in sodium tetrathionate $\left({\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6\right)$.

The oxidation states of sodium and oxygen are +1 and -2 respectively.
In a neutral molecule, the sum of all the oxidation states should be zero.
Hence

$\begin{array}{l}2\left(1\right)+4\left(\mathrm{x}\right)+6(-2)=0\\ x=2.5\end{array}$

Hence, option C is correct.