KMnO₄ oxidises I⁻ in acidic and neutral/faintly alkaline solutions, respectively, to :

Solution

Chemistry ma’am · 4 steps

Same permanganate, same iodide, but the medium decides how far iodide gets pushed.

1. Acidic medium. Permanganate is reduced all the way to Mn²⁺ (+7 to +2), a 5-electron change. Each I⁻ gives up one electron and pairs up as I₂: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O.
2. Neutral or faintly alkaline. Permanganate stops at MnO₂ (+7 to +4), a 3-electron change. Iodine is not released here. Iodide goes all the way from -1 to +5 and ends up as iodate, IO₃⁻.

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

3. Check the electrons. Two Mn atoms each gain 3 electrons, so 6 are gained in total. One I atom loses 6 electrons going from -1 to +5. The two sides match.
4. So. The products are I₂ in acidic solution and IO₃⁻ in neutral or faintly alkaline solution. That is option (c).

The idea to keep. For KMnO₄ and iodide, remember the pairs: acid gives Mn²⁺ and I₂. Neutral or faintly alkaline gives MnO₂ and IO₃⁻.

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