Consider the following reaction
MnO₂+KOH+O₂ → A+H₂ O .
Product 'A' in neutral or acidic medium disproportionate to give products 'B' and 'C' along with water. The sum of spin-only magnetic moment values of B and C is ________ BM. (nearest integer) (Given atomic number of Mn is 25)

Solution

Chemistry ma’am · 3 steps

Find the oxidation state of Mn in each product, count the unpaired d electrons, then add the moments.

1. Identify A. Fusing MnO₂ with KOH in air oxidises Mn from +4 to +6. The product is green potassium manganate, so A is K₂MnO₄: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O.
2. Disproportionation. Manganate is stable only in strong alkali. In neutral or acidic medium, Mn(+6) goes both up to +7 and down to +4. So B and C are permanganate and manganese dioxide.

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

   Product	Mn oxidation state	d electrons	μ = √n(n+2)
   MnO₄⁻	+7	3d⁰, n = 0	0 BM
   MnO₂	+4	3d³, n = 3	√15 = 3.87 BM

3. So. The sum is 0 + 3.87 = 3.87 BM, which rounds to 4.

The common slip: 3. Adding the numbers of unpaired electrons (0 + 3) instead of the magnetic moments gives 3. Each n has to go into √n(n+2) first.

The idea to keep. For spin-only moments, first fix the metal's oxidation state, then write its dⁿ count. Mn is 3d⁵ 4s², so Mn(+7) is d⁰, Mn(+6) is d¹ and Mn(+4) is d³.

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