An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in higher oxidation state is _____________ B.M. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Split +6 manganese into +7 and +4, then count the d electrons left on the +7 one.

1. Disproportionation. In manganate, MnO₄²⁻, manganese is +6. In acid it disproportionates: some Mn goes up to +7 in permanganate and some goes down to +4 in MnO₂.

   3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

2. Count d electrons. Mn has the configuration [Ar] 3d⁵ 4s², so it has 7 valence electrons. Mn⁷⁺ in MnO₄⁻ has lost all of them, so it is 3d⁰ with no unpaired electrons.
3. Apply the formula. With n = 0, the spin-only moment is μ = √n(n+2) = √0 = 0 B.M. This is why KMnO₄ is diamagnetic, even though it is deeply coloured (its colour comes from charge transfer, not d–d transitions).
4. So. The higher-oxidation-state product, MnO₄⁻, has a spin-only magnetic moment of 0 B.M.

The common slip: 4. Taking MnO₂ (Mn⁴⁺, 3d³) instead of permanganate gives √15 = 3.87, which rounds to 4, but +4 is the lower oxidation state.

The idea to keep. For spin-only moment, first fix the oxidation state, then count leftover d electrons. A d⁰ ion like Mn⁷⁺ or Cr⁶⁺ always gives zero, however intense its colour.

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