The spin only magnetic moment of [Mn(H₂ O)₆]²⁺ complexes is _________ B.M. (Nearest integer)
(Given : Atomic no. of Mn is 25)

Solution

Chemistry ma’am · 3 steps

Get the d-electron count of Mn²⁺ first, then let water's weak field decide high spin.

1. Configuration of Mn²⁺. Mn (Z = 25) is [Ar] 3d⁵ 4s². To make Mn²⁺, the two 4s electrons leave first, so Mn²⁺ is [Ar] 3d⁵.
2. Water is weak field. H₂O is a weak-field ligand, so the octahedral splitting is smaller than the pairing energy. The complex is high spin: the five d electrons occupy all five d orbitals singly, giving n = 5 unpaired electrons.

   μ = √n(n+2) B.M. = √(5 × 7) = √35 ≈ 5.92 B.M.

3. So. The spin-only magnetic moment is 5.92 B.M., which rounds to 6.

The common slip: 2. Treating the complex as low spin pairs four of the five electrons, leaving 1 unpaired and √3 = 1.73, which rounds to 2.

The idea to keep. For spin-only moments, count unpaired d electrons and use √n(n+2). Remove 4s electrons before 3d, and with weak-field ligands like H₂O assume high spin.

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