question_answer

The calculated spin only magnetic moment of \[C{{r}^{2+}}\] ion is                                  [NEET 2020]

A)  4.90 BM          

B)       5.92 BM

C)  2.84 BM          

D)       3.87 BM

Correct Answer: A

Solution :

[a] Electronic configuration of \[Cr[Ar]\text{ }3{{d}^{5}}4{{s}^{1}}\]

Electronic configuration of \[C{{r}^{2+}}[Ar]3{{d}^{4}}\]

Number of unpaired \[{{e}^{}}=4\]

Spin only magnetic moment \[=\sqrt{n(n+2)}\]

n = number of unpaired \[{{e}^{}}\]

Spin only magnetic moment \[=\sqrt{4(4+2)}\]

\[=\sqrt{24}\,BM\]

\[=4.9\,\,BM\]