The reaction of neutral \(MnO_4^-\) and \(I^-\) gives:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(MnO_2\) and \(IO_3^-\)
The product of permanganate depends on the medium. In acid it is reduced to \( \ce{Mn^{2+}} \) (5 electrons); in neutral or faintly alkaline solution it is reduced only to MnO2 (Mn +7 → +4, 3 electrons):
\( \ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-} \)
Iodide is oxidised to iodate (I −1 → +5, 6 electrons). Multiplying the permanganate half-reaction by 2 balances the 6 electrons:
\( \ce{I^- + 6OH^- -> IO3^- + 3H2O + 6e^-} \)
\( \ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \)
So the products are MnO2 and IO3−.
Why the other options are incorrect
Option (1) \(Mn^{2+}\) and \(I_2\) — Incorrect. These are the products in acidic solution (\( \ce{10I^{-}} \) + \( \ce{2MnO4^{-}} \) + \( \ce{16H^{+}} \) → \( \ce{2Mn^{2+}} \) + \( \ce{8H2O} \) + \( \ce{5I2} \)).
Option (2) \(MnO_2\) and \(I_2\) — Incorrect. \( \ce{MnO2} \) is right for the neutral medium, but there iodide is oxidised further, to iodate, not to iodine.
Option (4) \(Mn^{2+}\) and \(IO_3^-\) — Incorrect. \( \ce{Mn^{2+}} \) is formed only in acid solution, where iodide gives iodine.
NCERT (2026-27 edition):
Potassium Permanganate – …
