Q. The spin magnetic moment of cobalt in the compound $Hg\left[\right.Co\left(\right.SCN\left(\left.\right)_{4}\left]\right.$ is

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$\sqrt{3}$

$\sqrt{8}$

$\sqrt{15}$

$\sqrt{24}$

Solution:

In the given complex, $Co$ is in $+2$ oxidation state
$_{27}Co=\left[\right.Ar\left]\right.^{18}3d^{7}4s^{2}$
$Co^{2 +}=\left[\right.Ar\left]\right.^{18}3d^{7}$
i.e., it has three unpaired electrons.
Here $Co$ is present as $Co^{2 +}$ ion which has $3$ unpaired electrons. So the spin magnetic moment will be $\sqrt{3 \left(\right. 3 + 2 \left.\right)}$ , i.e., $\sqrt{15}B.M$ .
So spin magnetic moment $= \sqrt{\text{n} \left(\text{n} + 2\right)} = \sqrt{3 \left(3 + 2\right)} = \sqrt{15} \text{B} . \text{M} .$

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