Question

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

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

Answer 1

Answer: (B)

$\sqrt{15}$

$Hg\left[Co\right(SCN{)}_4]$,

$Co$ is present as $C{o}^{2+}$.

The atomic number of cobalt is $27$.

$Co=\left[Ar\right]3d^{7}4s^2$

$C{o}^{2+}=\left[Ar\right]3d^7$

Unpaired electrons, $n=3$

Magnetic moment $\mu =\sqrt{n(n+2)}$

$\mu =\sqrt{3(3+2)}=\sqrt{15}$ BM.