The magnetic moment of a transition metal of $3 d$ -series is $6.92 B M$ . Its electronic configuration will be:

{(3 d)}^{4} {(4 s)}^{2}

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{(3 d)}^{5} {(4 s)}^{1}

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{(3 d)}^{10}

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{(3 d)}^{5} {(4 s)}^{0}

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Solution

The correct option is D ${(3 d)}^{5} {(4 s)}^{1}$
The magnetic moment, $μ_{s} = \sqrt{n (n + 2)} B M$ where n is the number of unpaired electrons.

Option B has 6 unpaired electrons and the corresponding magnetic moment is $6.92 μ_{B}$ .

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1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}

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The least magnetic moment will be shown by the transition metal ion with the outer electronic configuration:

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