Question

Calculate oxidation number of $S$ in $N{a}_2{S}_2{O}_3$ & $N{a}_2{S}_4{O}_6$ respectively.

• A. -2, -2.5
• B. +2.5, +2
• C. +2, +2.5
• D. +2, +1

Answer 1

The correct option is C +2, +2.5

charge on oxygen atom is =-2
charge on sodium atom is =+1
(i) Let oxidation number of S-atom in $N{a}_2{S}_2{O}_3$ is $x$.
$\therefore (+1)\times 2+\left(x\right)\times 2+(-2)\times 3=0$
$\Rightarrow x=+2$
(ii) Let oxidation number of S-atom in $N{a}_2{S}_4{O}_6$ is $x$
$\therefore (+1)\times 2+\left(x\right)\times 4+(-2)\times 6=0\Rightarrow x=+2.5$

It is important to note here that $N{a}_2{S}_2{O}_3$ have two S-atoms and there are four S-atoms in $N{a}_2{S}_4{O}_6.$ However none of the sulphur atoms in both the compounds have +2 or + 2.5 oxidation number, it is the average of oxidation number, which reside on each sulphur atom.