Explanation

(d)
Spin only magnetic moment.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mu$=$\sqrt{n(n + 2)} = \sqrt{24}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n$^{2}+2$n -$24=0
($n$+6)($n -$4)=0
$\therefore$ $n$ =4
[$\because n$ =$-$6not possible.]
Here, $n$ is the number of unpaired electrons.
The electronic configuration of the metal ion $M^{x +}$ is
Z(25) =$1s^{2}\ ,2s^{2}\ ,2p^{6}\ ,3s^{2}\ ,3p^{6},4s^{2}\ ,3d^{5}$
Since, four unpaired electrons are present, the oxidation state must be +3.
$\therefore\ \ Z^{3 +}$(25)= $1s^{2}\ ,2s^{2}\ ,2p^{6}\ ,3s^{2}\ ,3p^{6},\ 3d^{4}$