The oxidation state of sulphur in $S_{2} O_{6}^{2 -}$ is :

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Solution

Let $x$ be the oxidation state of $S$ in $S_{2} O_{6}^{2 -}$ .
Since the overall charge on the complex is $- 2$ , the sum of oxidation states of all elements in it should be equal to $- 2$ .
$2 x + 6 (- 2) = - 2$
$x = + 5$

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