Explanation

(c)
The electronic configuration of Mn is
_{25}Mn =[Ar] $3d^{5}\ 4s^{2}$
${Mn}^{4 +}$ =[Ar]$3d^{3}$
Thus, three unpaired electrons are present .
Spin only magnetic moment , $\mu = \sqrt{n(n + 2)}$
$n$=3
$\mu\ = \sqrt{3(3 + 2)}$
=$\ \sqrt{15}\ = 3.87$
$\approx \$4 BM