Solid potassium permanganate is heated to 513 K. Which of the following sets of products is formed?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \ce{K2MnO4} \), \( \ce{MnO2} \) and \( \ce{O2} \)
On heating, potassium permanganate decomposes at 513 K. It is an internal redox reaction: part of the manganese is reduced and part of the oxygen is oxidised to oxygen gas.
\( \displaystyle \ce{2KMnO4} \xrightarrow{\text{513 K}} \ce{K2MnO4 + MnO2 + O2} \)
One Mn goes from +7 to +6 in green potassium manganate, K2MnO4 (gains 1 electron).
The other Mn goes from +7 to +4 in MnO2 (gains 3 electrons).
Two oxide ions (−2) become one O2 molecule (0), giving up 4 electrons; so 1 + 3 = 4 electrons gained balance the 4 electrons lost.
Why the other options are incorrect
Option (2) \( \ce{K2O} \), \( \ce{MnO2} \) and \( \ce{O2} \) — Incorrect. Only half of the manganese is reduced to MnO2; the potassium stays combined as potassium manganate, K2MnO4, not as K2O.
Option (3) \( \ce{K2MnO4} \), \( \ce{Mn2O3} \) and \( \ce{O2} \) — Incorrect. The reduced manganese ends up as MnO2 (+4), not Mn2O3 (+3).
Option (4) \( \ce{K2O} \) and \( \ce{Mn2O7} \) — Incorrect.
This would leave Mn at +7 with no change at all, and no oxygen would be given off.
Mn2O7 is the covalent oxide of Mn(VII), not a product of heating KMnO4.
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of KClO4. The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K. 2KMnO4 → K2MnO4 + MnO2 + O2”