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  4. The oxidation states of sulphur in the anions SO 3 2-, S 2 O 4 2- and S 2 O 6 2- follow the order -

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Q. The oxidation states of sulphur in the anions $SO _{3}{ }^{2-}, S _{2} O _{4}{ }^{2-}$ and $S _{2} O _{6}{ }^{2-}$ follow the order -

AIPMTAIPMT 2003Redox Reactions

Solution:

Oxidation state of S in $SO_3^{ 2 - } $
$ x + ( - 2 \times 3 ) = - 2$
$ x = + 6 - 2 = + 4$
Oxidation state of S in $S_2 O_4^{ 2 - } $
$ 2 x 4- (- 2 \times 4) = - 2 $
$ 2x = + 8 - 2 = + 6$
$ x = \frac{ + 6 }{ 2 } = + 3 $
Oxidation state of S in $ S_2 O_6^{ 2 - } $
$ 2 x 4 - (- 2 x\times 6) = - 2 $
$ 2x = 4 - 12 - 2 = 10$
$ x = \frac{10}{ 2} = - 5 $
Hence, increasing order of oxidation states of S is
$S_2O_4^{ 2 - } < SO_3^{ 2 - } < S_2O_6^{ 2 - } $