A green solution of potassium manganate, K2MnO4, is made acidic. Whic…

The correct answer is Option (4) → \( \ce{MnO4^-} \) and \( \ce{MnO2} \)

Manganese(VI) in the manganate ion is unstable in acid solution. It undergoes disproportionation: part of it is oxidised to a higher state and part is reduced to a lower state.

• Manganate, MnO₄²⁻: Mn is +6.

• Permanganate, MnO₄⁻: Mn is +7 (oxidised, one electron lost per Mn).

• Manganese dioxide, MnO₂: Mn is +4 (reduced, two electrons gained per Mn).

• Electrons balance when two Mn are oxidised for every one reduced, so 3 manganate ions give 2 permanganate ions and 1 MnO₂.

    \( \displaystyle \ce{3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O} \)

This step is used in the preparation of potassium permanganate from the dark green K₂MnO₄.

Why the other options are incorrect

Option (1) \( \ce{MnO4^-} \) and \( \ce{Mn^2+} \) — Incorrect. Mn²⁺ is formed when acidified permanganate oxidises a reducing agent such as Fe²⁺ or I⁻. In acid, manganate gives MnO₂ (Mn +4) as its lower product.

Option (2) \( \ce{MnO4^-} \) only — Incorrect.

• If every Mn went from +6 to +7, the electrons would have to be taken by an oxidising agent.

• Permanganate alone is obtained when manganate is oxidised, as in the commercial electrolytic oxidation; simple acidification gives a lower product as well.

Option (3) \( \ce{MnO2} \) and \( \ce{Mn^2+} \) — Incorrect. Both have oxidation states below +6. Manganese would then only be reduced, with nothing oxidised, which is not possible without a reducing agent.

NCERT (2026-27 edition):
Example 4.9 – “When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution.

3Mn^VIO₄^2– + 4 H⁺ → 2 Mn^VIIO₄^– + Mn^IVO₂ + 2H₂O”

4.4.1 Oxides and Oxoanions of Metals – “Potassium permanganate is prepared by fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. This produces the dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.”