The difference in the 'spin-only' magnetic moment values of KMnO₄ and the manganese product formed during titration of KMnO₄ against oxalic acid in acidic medium is ________ BM. (nearest integer)

Solution

Chemistry ma’am · 4 steps

Find the oxidation state of Mn before and after the titration, then count unpaired d electrons in each.

1. The reaction. In acidic medium, permanganate oxidises oxalic acid to CO₂ and is itself reduced all the way to Mn²⁺: 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
2. Count unpaired electrons. Mn is [Ar] 3d⁵ 4s². In KMnO₄ it is +7, so it is 3d⁰ with no unpaired electrons. In Mn²⁺ it is 3d⁵, so all five d electrons are unpaired.

   μ = √n(n+2) BM

3. Put in numbers. For KMnO₄, n = 0 gives μ = 0. For Mn²⁺, n = 5 gives μ = √35 = 5.92 BM.
4. So. The difference is 5.92 - 0 = 5.92 BM, which rounds to 6.

The common slip: 4. Taking the product as MnO₂ (Mn⁴⁺, 3d³) gives √15 = 3.87, which rounds to 4. That product forms only in neutral or weakly alkaline medium.

The idea to keep. Permanganate's product depends on the medium: Mn²⁺ in acid, MnO₂ in neutral or weakly alkaline, MnO₄²⁻ in strongly alkaline. Fix the product first, then count unpaired electrons.

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