The oxidation state of sulphur in $\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$ is

The oxidation state of sulphur in $\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$ is
  1. 6
  2. $\frac{+3}{2}$
  3. $\frac{+5}{2}$
  4. -2

Solution

$\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$ $2+4 x-12=0$ $4 x-10=0$ $x=\frac{10}{4}=\frac{+5}{2}$ Oxidation state of $\mathrm{S}$ is $=\frac{+5}{2}$

Asked in: BITSAT 2013

Practice more Redox Reactions questions on Aicharya