JEE Chemistry 2023 Question with Solution

Given: KMnO$4$ oxidises I$^-$ in acidic and neutral/faintly alkaline medium.

Find: The oxidation products of iodide ion in the two media.

From the solution working, in acidic medium:

$$2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{I}_2 + 8\text{H}_2\text{O}$$

So, iodide is oxidised to $\text{I}_2$ in acidic medium.

In neutral/faintly alkaline solution:

$$2\text{MnO}_4^- + \text{I}^- + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^-$$

So, iodide is oxidised to $\text{IO}_3^-$ in neutral/faintly alkaline medium.

The solution states the products are $\text{I}_2$ and $\text{IO}_3^-$. Although the page also shows "The Correct Option is B", that label conflicts with the listed option texts. Matching the products to the options, the correct option is A.

In acidic medium, permanganate acts as a stronger oxidising agent and converts iodide to molecular iodine:

$$\text{I}^- \rightarrow \text{I}_2$$

In neutral or faintly alkaline medium, iodide is oxidised further to iodate:

$$\text{I}^- \rightarrow \text{IO}_3^-$$

Therefore, the required pair is $\text{I}_2$ and $\text{IO}_3^-$, which corresponds to option A.