Q. Spin only magnetic moment of the compound $Hg\lbrack Co(SCN)_{4}\rbrack$is
A) $\sqrt{3}$
B) $\sqrt{15}$
C) $\sqrt{24}$
D) $\sqrt{8}$
Explanation
(b)
In $Hg\left\lbrack Co(SCN)_{4} \right\rbrack,\ Co$ is present as $Co^{2 +}$. The configuration of $Co^{2 +}$ is given as following
$\lbrack Ar\rbrack 3d^{7}\ 4s^{0}$
[image:9_image225.emf]
$\therefore$ Magnetic moment ($\mu$)=$\sqrt{n(n + 2)}$
$\sqrt{3(3 + 2)} = \sqrt{15}$ BM
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.