Thank you for reporting, we will resolve it shortly
Q.
The spin magnetic moment of cobalt in the compound $Hg[Co(SCN)_4]$ is
The d-and f-Block Elements
Solution:
In $Hg[Co(SCN)_4]$, the oxidation state of $Co$ is $+2$.
In $Co^{2+}; 1s^2 2s^2 2p^6 3s^2 3p^6 3d^7$, we have three unpaired
electrons so its spin magnetic moment will be
$\mu = \sqrt{3(3 + 2)} B.M. = \sqrt{ 3 \times 5} B.M.$
$= \sqrt{15}B.M$.