$\mathrm{KMnO}_4$ oxidises $\mathrm{I}^{-}$ in acidic and neutral/faintly alkaline solutions, respectively, to :

<p>Potassium permanganate ($\mathrm{KMnO}_4$) is a strong oxidizing agent that can oxidize iodide ions ($\mathrm{I}^{-}$) to different products depending on the pH of the solution.</p> <p>In acidic solutions, $\mathrm{KMnO}_4$ oxidizes $\mathrm{I}^{-}$ to molecular iodine ($\mathrm{I}_2$) according to the following equation:</p> $$ 2\mathrm{MnO}_4^{-} + 16\mathrm{H}^{+} + 10\mathrm{I}^{-} \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O} + 5\mathrm{I}_2 $$ <p>In neutral or faintly alkaline solutions, the oxidation product is different because the $\mathrm{I}^{-}$ is oxidized to iodate ion ($\mathrm{IO}_3^{-}$). The reaction in a neutral or slightly alkaline solution is as follows:</p> $$ 2\mathrm{MnO}_4^{-} + 2\mathrm{H}_2\mathrm{O} + 10\mathrm{I}^{-} \rightarrow 2\mathrm{MnO}_2 + 4\mathrm{OH}^{-} + 5\mathrm{IO}_3^{-} $$ <p>So according to the reactions above, in acidic solutions, $\mathrm{I}^{-}$ is oxidized to $\mathrm{I}_2$ while in neutral or faintly alkaline solutions, $\mathrm{I}^{-}$ is oxidized to $\mathrm{IO}_3^{-}$. Therefore, the correct answer is:</p> <p>Option C: $\mathrm{I}_2 ~\&~ \mathrm{IO}_3^{-}$</p>