In alkaline medium, MnO₄⁻ oxidises I⁻ to

Solution

Chemistry ma’am · 4 steps

In base, permanganate becomes MnO₂ and iodide is pushed right up to iodate.

1. What permanganate becomes. In neutral or alkaline medium, MnO₄⁻ is reduced to MnO₂, so Mn goes from +7 to +4. Each Mn gains 3 electrons. In acidic medium it would go all the way to Mn²⁺ instead.
2. What iodide becomes. Under these conditions I⁻ (-1) is oxidised to iodate, IO₃⁻ (+5). Each iodide loses 6 electrons, so two MnO₄⁻ are needed for every I⁻.

   2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

3. Check the balance. Electrons: 2 × 3 = 6 gained and 6 lost. Charge: -3 on both sides. Oxygen: 9 on both sides. Hydrogen: 2 on both sides.
4. So. In alkaline medium MnO₄⁻ oxidises I⁻ to IO₃⁻, which is option (b).

The idea to keep. Permanganate with iodide depends on the medium. Acidic gives Mn²⁺ and I₂. Neutral or alkaline gives MnO₂ and IO₃⁻.

Answer from the NTA final answer key. Solution by PrepWiser.