$μ = \sqrt{15}$ is true for the pair:

C o^{+ 2}, C r^{+ 3}

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F e^{+ 2}, C r^{+ 3}

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F e^{+ 3}, C r^{+ 2}

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M n^{+ 2}, F e^{+ 2}

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Solution

The correct option is C $C o^{+ 2}, C r^{+ 3}$
The electronic configuration of $C o^{2 +}$ is; $[A r] 4 s^{0} 3 d^{7}$
Hence there are three unpaired electrons in $C o^{2 +}$
Now the elctronic configuration of $C r^{3 +}$ is; $[A r] 3 d^{3}$
Hence there are 3 unpaired electrons in $C r^{3 +}$
We know magnetic moment $μ = \sqrt{(n (n + 2)}$ $
where n is number of unpaired electrons.
So when n=3,
$μ = \sqrt{(3 (3 + 2)} = \sqrt{15}$
Hence option A is correct answer.

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