The magnetic moment of a transition metal compound has been calculated to be $3.87 \mathrm{BM}$. The metal ion is

The magnetic moment, ${\mathrm{\mu }}_{\mathrm{s}}=\sqrt{\mathrm{n}(\mathrm{n}+2)}\mathrm{BM}$ where $\mathrm{n}$ is the number of unpaired electrons. 

$$\begin{gathered} {\mathrm{Cr}}^{+2} : \left[\mathrm{Ar}\right], 3{\mathrm{d}}^4, 4{\mathrm{s}}^{°} \mathrm{n}=4, \mathrm{\mu }=\sqrt{4(4+2)}=\sqrt{24}=4.89 \mathrm{BM} \\ {\mathrm{Mn}}^{+2} : \left[\mathrm{Ar}\right], 3{\mathrm{d}}^5, 4{\mathrm{s}}^{°} \mathrm{n}=5, \mathrm{\mu }=\sqrt{5(5+2)}=\sqrt{35}=5.91 \mathrm{BM} \\ {\mathrm{V}}^{+2} : \left[\mathrm{Ar}\right], 3{\mathrm{d}}^3, 4{\mathrm{s}}^{°} \mathrm{n}=3, \mathrm{\mu }=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM} \\ {\mathrm{Ti}}^{+2} : \left[\mathrm{Ar}\right], 3{\mathrm{d}}^2, 4{\mathrm{s}}^{°} \mathrm{n}=2, \mathrm{\mu }=\sqrt{2(2+2)}=\sqrt{8}=2.82 \mathrm{BM} \end{gathered}$$