Question

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

• A. 3.87 BM
• B. 4.90 BM
• C. 5.92 BM
• D. 2.84 BM

Answer 1

Answer: (B)

4.90 BM

Answer 2

Electronic configuration of $Cr=[Ar]3d^{5}\,4s^{1}$
Electronic configuration of $Cr^{+2}=[Ar]3d^{4}$
Number of unpaired electrons $= 4$
Spin only magnetic moment $=\sqrt{n\left(n+2\right)}BM$ as $n=4$
$\therefore\mu=\sqrt{4\left(4+2\right)}$
$=\sqrt{24}$
$=4.90\,BM$