Question

Calculate the oxidation number of each sulphur atom in the following compounds:

A) $N{a}_2{S}_2{O}_3$

B) $N{a}_2{S}_4{O}_6$

C) $N{a}_{2}S{O}_3$

D) $N{a}_{2}S{O}_4$

Answer 1

A)
Let the oxidation number of sulphur be $x$. Considering $+1$ for Na and $-2$ for $O$ atoms, we have the equation-


There is a double bond between two sulphur atoms.
Oxidation number of $S-1=0$
Let the oxidation number of $S-2$ be $x$
$2(+1)+3\times (-2)+x+1\left(0\right)=0$
$x–4=0$
$x=+4$
$S-1$ sulphur oxidation state $=0$
$S-2$ sulphur oxidation state $=+4$

B)
The structure of $N{a}_2{S}_4{O}_6$ is given below:

In the given structure, two central sulphur atoms have zero oxidation state because electron pair forming the $S-S$ bond remain in the centre.
Let oxidation state of other $S$ atoms be $x$. The oxidation number of $Na$ and $O$ atoms are $+1$ and $-2$ respectively. We have, $2(+1)+6(-2)+2x+2\left(0\right)=0$
$x-5=0$
$x=+5$ (for both $S-1$ and $S-4$)

C)
Let the oxidation number of sulphur be $x$.
Considering $+1$ for Na and $-2$ for $O$ atoms, we form the equation-
$2(+1)+x+(-2)\times 3=0$
$2+x–6=0$
$x=+4$

D)
Let the oxidation number of sulphur be $x$.
Considering $+1$ for $Na$ and $-2$ for $O$ atoms, we form the equation-
$2+x+(-2)\times 4=0$
$2+x–8=0$
$x–6=0$
$x=+6$
Hence oxidation number of sulphur in $N{a}_{2}S{O}_4$ is $+6$.