Spin only magnetic moment of Mn ^ 4 + ion is nearly

Electronic configuration of $Mn$ is \left(Z = 25\right) = \left[Ar\right]3d^{5}4s^{2}$</p><p>$∴$Electronic configuration of${Mn}^{4 +}$is$= \left[Ar\right]3d^{3}$</p><p>Thus,${Mn}^{4 +}$has$3$-unpaired electron and its spin only magnetic moment \left(\mu\right)$ is, $\mu = \sqrt{n\left(n + 2\right)}BM$

where, $n =$ number of unpaired electrons.

$\mu = \sqrt{3 ( 3 + 2 )} . BM ⇒ \mu = \sqrt{15} = 3 . 87BM \approx 4BM$