The calculated spin only magnetic moment of Cr^2+ ion is :

1. 3.87 BM

2. 4.90 BM

3. 5.92 BM

4. 2.84 BM

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

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Answer: 2. 4.90 BM

Spin-only moment: $\mu=\sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

Cr (Z=24): $3d^54s^1$. $\mathrm{Cr^{2+}}$: $3d^4$, so $n=4$.

$$\mu=\sqrt{4\times6}=\sqrt{24}=4.90\ \mathrm{BM}$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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