Iodine oxidises $S_{2} O_{3}^{2 -}$ ion to ' $X$ ', change in oxidation state of sulphur

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$+ 4$

$+ 2$

$+ 0.5$

$+ 3$

answer is C.

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Detailed Solution

Iodine oxidises $S_{2} O_{3}^{2 -}$ ion to $S_{4} O_{6}^{2 -}$ ion. The oxidation state of sulphur changes from $+ 2$ to $+ 2.5$ . Thus there will be a change of $+ 0.5$ oxidation state overall.

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