NEET 2026 (Phase 1) · ChemistryValence bondPrevious Year Question
The calculated spin only magnetic moment of $\ce{Cr^2+}$ ion is:
- A.$5.92$ BM
- B.$2.84$ BM
- C.$3.87$ BM
- D.$4.90$ BM✓
Correct Answer
(D) $4.90$ BM
Solution & Explanation
\textbf{Answer:} (D) $4.90$ BM \textbf{Solution:} The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. $\ce{Cr^2+}$ has the configuration $3d^4$, giving $n = 4$ unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$ BM. Hence the answer is option D.
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