How many of the following metal ions have similar value of spin only magnetic moment in gaseous state? ______________
(Given : Atomic number V, 23; Cr, 24; Fe, 26; Ni, 28)
V³⁺, Cr³⁺, Fe²⁺, Ni³⁺

Solution

Chemistry ma’am · 2 steps

Write each ion's 3d configuration, count the unpaired electrons, and look for a matching count.

1. Remove 4s first. When a transition metal forms a cation, the 4s electrons leave before the 3d ones. In the gaseous state there are no ligands, so Hund's rule fills the 3d orbitals singly before any pairing.

   μ = √n(n+2) BM

   Ion	Configuration	Unpaired n	μ (BM)
   V³⁺	[Ar] 3d²	2	√8 = 2.83
   Cr³⁺	[Ar] 3d³	3	√15 = 3.87
   Fe²⁺	[Ar] 3d⁶	4	√24 = 4.90
   Ni³⁺	[Ar] 3d⁷	3	√15 = 3.87

2. So. Only Cr³⁺ and Ni³⁺ share the same moment, 3.87 BM, so the answer is 2.

The common slip: 0. Treating Ni³⁺ as low-spin d⁷ with one unpaired electron leaves no matching pair and gives 0. A gaseous ion has no ligand field, so it stays high spin.

The idea to keep. Spin-only moment depends only on the number of unpaired electrons. Ions with dⁿ and d¹⁰⁻ⁿ configurations, like d³ and d⁷, have the same unpaired count in the free ion.

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