Assign oxidation number to the underlined element in $H_{2} {S - -}_{2} O_{7}$

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Solution

Let X be the oxidation state of S in $H_{2} S_{2} O_{7}$ .
We know that oxidation no. for $H = + 1$ and $O = - 2$
$2 (+ 1) + 2 X + 7 (- 2) = 0$
$X = + 6$
Hence, the oxidation state of S in $H_{2} S_{2} O_{7}$ is +6.

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