Read the passage carefully and answer the Questions.

$KMnO_4$ is prepared by the fusion of $MnO_2$ with an alkali metal hydroxide and an oxidizing agent like $KNO_3$ to give a dark- green manganate ion which disproportionate to give permanganate as follows.

$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$

$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$

On heating $KMnO_4$ decomposes at 513 K to give $K_2MnO_4$. Permanganate ion is tetrahedral and diamagnetic. Acidified $KMnO_4$ acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.

Reaction of iodide ion ($I^-$) with $KMnO_4$ will.....

liberate $I_2$ in acidic solution

The correct answer is Option (1) → liberate $I_2$ in acidic solution **

Reaction of iodide ion with $KMnO_4$

Acidified $KMnO_4$ is a very strong oxidizing agent. Therefore it oxidizes iodide ions ($I^-$) to iodine ($I_2$).

1. Oxidation of iodide

Iodide loses electrons and forms iodine.

$2I^- \rightarrow I_2 + 2e^-$

This is oxidation because electrons are lost.

2. Reduction of permanganate in acidic medium

In acidic solution, permanganate is reduced from $Mn^{7+}$ to $Mn^{2+}$.

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Here permanganate gains electrons, so it is reduction.

3. Balancing electrons

• Iodide releases 2 electrons
• Permanganate requires 5 electrons

To balance electrons:

Multiply iodide reaction by 5

Multiply permanganate reaction by 2

Oxidation:

$10I^- \rightarrow 5I_2 + 10e^-$

Reduction:

$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$

4. Overall balanced reaction

$2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$

5. Final observation

• Iodide ions are oxidized to iodine
• Iodine ($I_2$) is liberated in acidic solution

Final answer

The correct answer is Option (1) $\rightarrow$ liberate $I_2$ in acidic solution.