Question

Spin only magnetic moment of the compound $Hg\left[Co{\left(SCN\right)}_4\right]$ is :

• A. $\sqrt{3}$
• B. $\sqrt{1}5$
• C. $\sqrt{2}4$
• D. $\sqrt{8}$

Answer 1

The correct option is B $\sqrt{1}5$

Cobalt is present as $C{o}^{+2}$ which has [Ar] $4s^{0}3d^7$ configuration, which means it has 3 unpaired electrons. So the spin only magnetic moment of the compound is $\sqrt{3(3+2)}=\sqrt{15}$