In the neutral or faintly alkaline medium, $\mathrm{KMnO}{}_4$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from

Solution & Explanation

\textbf{Answer:} (a) $+7$ to $+4$. \textbf{Solution:} In neutral or faintly alkaline medium, $\mathrm{KMnO}{}_4$ oxidises iodide ($\mathrm{I}{}^{-}$) to iodate ($\mathrm{IO}{}_3{}^{-}$): $2\mathrm{MnO}{}_4{}^{-}+\mathrm{H}{}_2\mathrm{O}+\mathrm{I}{}^{-}\to 2\mathrm{MnO}{}_2+2\mathrm{OH}{}^{-}+\mathrm{IO}{}_3{}^{-}$ Manganese starts in $\mathrm{MnO}{}_4{}^{-}$ as Mn($+7$) and is reduced to $\mathrm{MnO}{}_2$, where Mn is in the $+4$ state. So the oxidation state of manganese changes from $+7$ to $+4$. Hence option (a) is correct.