When neutral or faintly alkaline $\mathrm{KMnO}{}_4$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Solution & Explanation

\textbf{Answer:} (c) $\mathrm{IO}{}_3{}^{-}$ \textbf{Solution:} In neutral or faintly alkaline medium, $\mathrm{KMnO}{}_4$ acts as a moderate oxidising agent and oxidises iodide all the way to iodate: $2\mathrm{MnO}{}_4{}^{-}+\mathrm{H}{}_2\mathrm{O}+\mathrm{I}{}^{-}\to 2\mathrm{MnO}{}_2+2\mathrm{OH}{}^{-}+\mathrm{IO}{}_3{}^{-}$ Here $\mathrm{Mn}$ is reduced from +7 to +4 ($\mathrm{MnO}{}_2$), and iodide $\mathrm{I}{}^{-}$ (-1) is oxidised to iodate $\mathrm{IO}{}_3{}^{-}$ (+5). Hence 'X' is $\mathrm{IO}{}_3{}^{-}$. (In acidic medium iodide would instead be oxidised only to $\mathrm{I}{}_2$, but neutral/faintly alkaline conditions give $\mathrm{IO}{}_3{}^{-}$.)