JEE Main 2022 · 25 June · evening · Single correct · +4 / −1
The metal ion (in gaseous state) with lowest spin-only magnetic moment value is :
- (a) V²⁺
- (b) Ni²⁺
- (c) Cr²⁺
- (d) Fe²⁺
Answer and solution
Answer (b) Ni²⁺
Solution
Chemistry ma’am · 3 steps
Do not count d electrons, count unpaired ones: fewer unpaired electrons means a smaller moment.
- Free gaseous ion. A gaseous ion has no ligands around it, so there is no crystal field splitting. The d electrons follow Hund's rule and stay unpaired as long as they can.
μ = √n(n+2) BM
Ion Configuration Unpaired electrons n μ (BM) V²⁺ [Ar] 3d³ 3 √15 = 3.87 Ni²⁺ [Ar] 3d⁸ 2 √8 = 2.83 Cr²⁺ [Ar] 3d⁴ 4 √24 = 4.90 Fe²⁺ [Ar] 3d⁶ 4 √24 = 4.90 - Why Ni is lowest. In 3d⁸, five electrons fill the five d orbitals singly and the next three pair up, leaving only 2 unpaired. That is the fewest of the four ions.
- So. Ni²⁺ has the lowest spin-only magnetic moment, 2.83 BM, so the answer is (b).
The idea to keep. The spin-only moment depends only on unpaired electrons. Past d⁵ the electrons start pairing, so the count of unpaired electrons rises up to d⁵ and then falls again.
Answer from the NTA final answer key. Solution by PrepWiser.