What is the sum of the lowest oxidation states of sulphur in the following species:
$S_{4} O_{6}^{2 -}, S_{2} O_{3}^{2 -}, S_{2} O_{8}^{2 -}, S_{2} O_{7}^{2 -}$

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Solution

Minimum OS of sulphur in $S_{4} O_{6}^{2 -} = 0$
Minimum OS of sulphur in $S_{2} O_{3}^{2 -} = - 2$
Minimum OS of sulphur in $S_{2} O_{8}^{2 -} = + 6$
Minimum OS of sulphur in $S_{2} O_{7}^{2 -} = + 6$
Hence, sum of all the above oxidation states $= 0 - 2 + 6 + 6 = 10$

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