Assertion (A): The green manganate ion, MnO42−, is stable in acidic solution.
Reason (R): An oxidation state that becomes less stable relative to one lower and one higher oxidation state undergoes disproportionation.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) is false but (R) is true.
Assertion (A): The green manganate ion, MnO42−, is stable in acidic solution. — False. Manganese is +6 in manganate. In acidic (or neutral) solution Mn(VI) becomes unstable relative to Mn(VII) and Mn(IV), so manganate disproportionates into purple permanganate and manganese dioxide:
\( \displaystyle \ce{3MnO4^2- + 4H+ -> 2MnO4- + MnO2 + 2H2O} \)
Two Mn atoms go up from +6 to +7, and one comes down from +6 to +4. This step is used to make permanganate from the green K2MnO4.
Reason (R): An oxidation state that becomes less stable relative to one lower and one higher oxidation state undergoes disproportionation. — True. This is the meaning of disproportionation: the same element is oxidised and reduced at the same time.
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.”
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.”