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Q.

The oxidation state of Sulphur in the anions $S_{2} O_{4}^{- 2}$ , ${SO}_{3}^{2 -} and S_{2} {O_{6}}^{2 -}$ follows the order:

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a

$S_{2} O_{6}^{2 -} < {SO}_{4}^{2 -} < {SO}_{3}^{2 -}$

b

$S_{2} O_{4}^{2 -} < S_{2} {O_{6}}^{2 -} < {SO}_{3}^{2 -}$

c

${SO}_{3}^{2 -} < S_{2} {O_{4}}^{2 -} < S_{2} {O_{3}}^{2 -}$

d

$S_{2} O_{4}^{2 -} < {SO}_{3}^{2 -} < S_{2} O_{6}^{2 -}$

answer is B.

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Detailed Solution

$\begin{aligned} S_{2} O_{6}^{2 -} = + 5, {SO}_{3}^{2 -} = + 4 \\ S_{2} O_{4}^{2 -} = + 3 \end{aligned}$

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The oxidation state of Sulphur in the anions S2O4-2 , SO32− and S2O62−follows the order: