Coordination Compounds Ques 31
31. The calculated spin only magnetic moment of $Cr^{2+}$ ion is $\mu_{\text{eff}} = \sqrt{n(n+2)}~\text{BM}$
[2020]
(a) $4.90 BM$
(b) $5.92 BM$
(c) $2.84 BM$
(d) $3.87 BM$
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Answer:
Correct Answer: 31.(d)
Solution:
(a) Electronic configuration of $Cr^{2+}$ is $[Ar] 3d^5$
| — | — | — | — | — |
|---|---|---|---|---|
| 1 | 1 | 1 | 1 |
n=4
$ \mu_n = \sqrt{n(n+2)} $
$ \therefore \mu_n = \sqrt{4(4+2)} = \sqrt{24} BM = 4.9 \text{ BM} $
BETA
