Magnetic moment of the compound $H g [C o (S C N)_{4}]$ is:

\sqrt{3}

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\sqrt{15}

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\sqrt{24}

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\sqrt{8}

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Solution

The correct option is C $\sqrt{15}$
Spin only magnetic moment of the compound $H g [C o (S C N)_{4}]$ is $\sqrt{15}$ .
$H g [C o (S C N)_{4}]$ $n = 3$ $μ_{s} = \sqrt{n (n + 2)} = \sqrt{3 \times 5} = \sqrt{15}$

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