Calculate the oxidation no. of sulphur in $H_{2} {S - -}_{4} O_{6}$ .

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Solution

Let X be the oxidation state of S in $H_{2} S_{4} O_{6}$ .
$2 (+ 1) + 4 X + 6 (2 -) = 0$
$X = + \frac{5}{2}$
Hence, the oxidation state of S in $H_{2} S_{4} O_{6}$ is +5/2.
But oxidation number cannot be fractional.
Terminal S atoms have +5 oxidation number and middle S atoms have 0 oxidation number.

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