The disproportionation of MnO₄²⁻ in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment (μ) value of B in BM is __________. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Manganate splits into permanganate (+7) and MnO₂ (+4). B is MnO₂, so count its d electrons.

1. Disproportionation. In MnO₄²⁻, Mn is in the +6 state. In acid it is unstable and goes both up and down: up to MnO₄⁻ (+7) and down to MnO₂ (+4).

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

2. Identify B. Two Mn atoms rise by 1 each and one Mn falls by 2, so the electrons balance. A is MnO₄⁻ (+7) and B, with the lower oxidation state, is MnO₂ (+4).
3. Count unpaired electrons. Mn is [Ar] 3d⁵ 4s². Removing 4 electrons gives Mn⁴⁺ as 3d³, which has n = 3 unpaired electrons.
4. So. μ = √n(n+2) = √(3 × 5) = √15 ≈ 3.87 BM, which rounds to 4.

The common slip: 6. Taking B as Mn²⁺ (3d⁵, 5 unpaired) gives √35 ≈ 5.92, so 6. Manganate in acid disproportionates to MnO₂, not Mn²⁺.

The idea to keep. In a disproportionation, find both products first and pick the right one. Then the spin-only moment needs only the d-electron count: μ = √n(n+2) BM.

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