The calculated spin only magnetic moment of Cr 2 + ion is

The calculated spin only magnetic moment of

{Cr}^{2 +}

ion is

$4.90 BM$
$5.92 BM$
$2.84 BM$
$3.87 BM$

Solution

${Cr}^{+ 2} = 3 d^{4}$

n = number of unpaired electrons.

$μ = \sqrt{n (n + 2)} B . M . = \sqrt{4 (6)} = \sqrt{24} B . M .$

$= 4.90 B . M .$

Asked in: MHT CET Full Test 9

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