Spin only magnetic moment of $F e^{2 +}$ is:

1.73 BM

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3.87 BM

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4.90 BM

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5.92 BM

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Solution

The correct option is C 4.90 BM
Spin only magnetic moment $(μ) = \sqrt{n (n + 2)} B M$

In $F e^{2 +},$ 4 unpaired electrons are present.

$μ = \sqrt{4 (4 + 2)} = \sqrt{24} B M = 4.90 B M$

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