The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is ____________.

Solution

Chemistry ma’am · 4 steps

In basic medium, permanganate stops at MnO₂, not Mn²⁺, so manganese lands at +4.

1. Who does what. KMnO₄ is the oxidising agent, with Mn at +7. Here H₂O₂ acts as the reducing agent: its oxygen goes from -1 to 0 and leaves as O₂ gas.
2. Product in basic medium. In basic medium, MnO₄⁻ takes 3 electrons and becomes a brown precipitate of MnO₂. Each H₂O₂ gives 2 electrons, so 2 permanganate ions need 3 peroxide molecules.

   2MnO₄⁻ + 3H₂O₂ → 2MnO₂ + 3O₂ + 2OH⁻ + 2H₂O

3. Oxidation state. In MnO₂, x + 2(-2) = 0, so x = +4. That matches the 3-electron drop from +7.
4. So. Manganese in the product is in the +4 oxidation state, so the answer is 4.

The common slip: 2. Using the acidic-medium product Mn²⁺ gives 2. In basic medium, permanganate is reduced only as far as MnO₂.

The idea to keep. The medium decides where permanganate stops: acidic gives Mn²⁺ (+2), neutral or weakly basic gives MnO₂ (+4), strongly basic gives MnO₄²⁻ (+6).

Answer from the NTA final answer key. Solution by PrepWiser.