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.
$\mu =\sqrt{4(4+2)}=\sqrt{24}=4.89\text{B.M.}$