The spin only magnetic moment of $[C o F_{6}]^{3 -}$ is:

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Solution

The complex $[C o F_{6}]^{3 -}$ is an octahedral complex.
Here the oxidation state of $C o$ is $+ 3$
Thus the elctronic configuration of $C o^{3 +} = [A r] 3 d^{6}$
So, according to crystal field splitting theory, the electronic configuration in d orbital will be $t_{2 g}^{4} e_{g}^{2}$
So number of unpaired electrons is = 4
Again spin only magnetic moment is given by, $\sqrt{n (n + 2)}$ ,
Where n =number of unpaired electron
So, the spin only magnetic moment of $[C o F_{6}]^{3 +} = \sqrt{4 (4 + 2)} = 4.89$

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