Identify the oxidizing behavior of $\mathrm{KMnO_4}$

The oxidation product of $\mathrm{I^-}$ depends on the medium because $\mathrm{KMnO_4}$ behaves differently in acidic and neutral/faintly alkaline solutions.

In acidic medium

In acidic solution, permanganate is a very strong oxidizing agent and is reduced as: $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

Iodide is oxidized to iodine: $\mathrm{2I^- \rightarrow I_2 + 2e^-}$

Thus, in acidic medium, $\mathrm{I^-}$ is oxidized to $\mathrm{I_2}$.

In neutral or faintly alkaline medium

In neutral/faintly alkaline medium, permanganate is reduced to $\mathrm{MnO_2}$: $\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$

Under these conditions, iodide is oxidized further to iodate: $\mathrm{I^- \rightarrow IO_3^-}$

So, in neutral/faintly alkaline medium, the oxidation product is $\mathrm{IO_3^-}$.

Match with options

Therefore, respectively in acidic and neutral/faintly alkaline solutions, the products are: $\mathrm{I_2 \; \& \; IO_3^-}$

This corresponds to Option C.

Check all options

• A: $\mathrm{IO_3^- \& IO_3^-}$ ❌ acidic medium does not give iodate here.
• B: $\mathrm{I_2 \& I_2}$ ❌ neutral/faintly alkaline medium gives iodate, not iodine.
• C: $\mathrm{I_2 \& IO_3^-}$ ✅ correct.
• D: $\mathrm{IO_3^- \& I_2}$ ❌ reversed order.

Comparison with stored answer

Stored correct answer = C.

My derived answer = C.

Hence, they agree.