A compound in which a metal ion M^x+ (atomic number=25) has a spin only magnetic moment of $\sqrt{24}\mathrm{BM}.$ The oxidation state of the metal ion or the value of ‘x’ is:

Spin only magnetic moment;

$\begin{array}{l}\mu =\sqrt{n(n+2)}=\sqrt{24}\\ n^2+2n-24=0\\ (n+6)(n-4)=0\\ n=-6\text{ or }4\end{array}$

Hence, the number of unpaired electrons is 4.
Electronic configuration of

$\mathrm{M}(\mathrm{Z}=25)=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{5}4{\mathrm{s}}^2$

Since, four unpaired electrons are present, therefore M present in +3 oxidation state.

$\therefore {\mathrm{M}}^{3+}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{4}4{\mathrm{s}}^0$

So, the value of x is 3.