Question

The spin only magnetic moment of $F{e}^{2+}$ ion (in BM) is approximately

• A. 4
• B. 5
• C. 7
• D. 6

Answer 1

The correct option is B 5

$Fe(Z=26):\left[Ar\right]3d^{6}4s^2$

$F{e}^{2+}:\left[Ar\right]3d^6$

Therefore, there are 4 unpaired electrons in $F{e}^{2+}$.
As spin only magnetic moment is $\mu =\sqrt{n(n+2)}B.M.$ where n is no. of unpaired electron.
$\mu =\sqrt{4(4+2)}=\sqrt{24}=4.89B.M.$