Question

Answer the following question.
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 an acidic solution to give a 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).

Answer 1

Potassium permanganate is prepared by fusion of MnO₂ with an alkali metal hydroxide and an oxidizing agent like KNO₃. This produces the dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

\[\ce{2MnO2 + 4KOH + O2 -> \underset{\text{(A) Dark Green}}{2K2MnO4}+ 2H2O}\] 

\[\ce{3MnO^2-_4 + 4H^+ ->\underset{\text{(B) purple}}{2MnO^-_4} + MnO2 + 2H2O}\]

In acid solutions:
Iodine is librated from potassium iodine: 

\[\ce{10I^- +\underset{\text{(B)}}{2MnO^-_4}+ 16H^+ -> 2Mn^2+ + 8H2O +\underset{\text{(D)}}{5l2}}\]

In neutral or faintly alkaline solution:
A notable reaction is the oxidation of iodine to iodate: 

\[\ce{\underset{\text{(B)}}{2MnO^-_4}+ H2O + I^- -> 2MnO2 + 2OH^- + \underset{\text{(C)}}{IO^-_3}}\].