Question

Spin only magnetic moment of the compound $Hg[Co(SCN)_4]$ is

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

Answer 1

Answer: (B)

$\sqrt {15}$

Answer 2

In $Hg[Co(SCN)_4], CO^{2+}$ has $3d^7$ configuration. $SCN^-$ produces weak ligand field, no pairing of electrons in d-orbitals occurs against Hund's rule, hence :$\mu =\sqrt {3(3+2)}BM=\sqrt {15} BM$