Calculate the value of magnetic moment of $V^{+ 2}$ ?

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Solution

$V^{2 +}$ has outer electronic configuration of $3 d^{3}$ with 3 unpaired electrons.
The magnetic moment $μ_{e f f} = \sqrt{n (n + 2)}$ B.M.
$μ_{e f f} = \sqrt{3 (3 + 2)}$ B.M.
$μ_{e f f} = \sqrt{15}$ B.M.
$μ_{e f f} = 3.87$ B.M.

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