The spin only magnetic moment of $[Z C l_{4}]^{2 -}$ is $3.87 B M$ where $Z$ is:

M n

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N i

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C o

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C u

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Solution

The correct option is C $C o$
$μ = 3.87 = \sqrt{n (n + 2)} \Rightarrow n = 3$

$[Z C l_{4}]^{2 -}$ ; oxidation state of $Z$ is $+ 2$

No. of unpaired electrons is $3$ .

Hence, $Z^{2 +} = d^{7}$

So, element is $C o$ .

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