Question

This spin magnetic moment of cobalt in $Hg\left[Co\right(SCN{)}_4]$ is :

• A. $\sqrt{3}$
• B. $\sqrt{8}$
• C. $\sqrt{15}$
• D. $\sqrt{24}$

Answer 1

The correct option is C $\sqrt{15}$

In $Hg\left[CO\right(SCN{)}_4]$, $Co$ exists as $C{o}^{2+}$

$SC{N}^{-}$ ligand provides four electron pair to show $s{p}^3$-hybridization in $\left[Co\right(SCN{)}_4{]}^{2-}$ and thus three unpaired electrons exist on $C{o}^{2+}$

$\because$ Magnetic moment = $\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}$

So, the correct option is $C$