Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ____________ B.M. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

In acid, permanganate always ends up as Mn²⁺, so count the unpaired d electrons of Mn²⁺.

1. Find the product. In acidic medium MnO₄⁻ (Mn in +7) takes 5 electrons and is reduced to Mn²⁺. Oxalic acid is oxidised to CO₂.

   2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

2. Count unpaired electrons. Mn is [Ar] 3d⁵ 4s². Removing the two 4s electrons gives Mn²⁺ as 3d⁵. With one electron in each of the five d orbitals, n = 5 unpaired electrons.
3. Spin-only formula. μ = √n(n+2) = √(5 × 7) = √35 ≈ 5.92 B.M.
4. So. The magnetic moment of Mn²⁺ is 5.92 B.M., which rounds to 6.

The common slip: 5. Reporting the number of unpaired electrons, 5, as the answer misses the last step. Put n = 5 into √n(n+2) to get 5.92, which rounds to 6.

The idea to keep. First fix the product from the medium: acidic gives Mn²⁺, neutral or weakly basic gives MnO₂, strongly basic gives MnO₄²⁻. Then use μ = √n(n+2).

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