When MnO₂ is fused with KOH in the presence of KNO₃ as an oxidizing agent it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B) An alkaline solution of compound (B) oxidizes KI to compound (C) whereas an acidified solution of compound (B) oxidizes KI to (D) Identify (A), (B), (C) and (D).

MnO₂ reacts with KOH to potassium permanganate in presence of KNO₃ which acts as an oxidizing agent. The reaction is given as follows:

K₂MnO₄ is dark green in color implies that this is compound (A).

Potassium permanganate disproportionates in acidic solution to give an oxidized and reduced form of Mn ion

Compound (B) is KMnO₄

Compound (B) is oxidizes KI solution which implies that it should itself undergo reduction, and as permanganate ion has +7 oxidation state of manganese it can no longer undergo oxidation and acts as a strong oxidation agent that oxidizes KI. Under alkaline medium:

Oxidized compound KIO₃ is (C)

Under acidic conditions:

Compound (D) is I₂, and orange brown color is obtained.

NOTE: the balancing of above two reactions is done with the help of oxidation number balancing in acidic and basic medium as mentioned in the question.