Spin only magnetic moment of ${\mathrm{Mn}}^{4+}$ ion is nearly

Electronic configuration of $\mathrm{Mn}$ is $\left(\mathrm{Z}=25\right)=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^2$

$\therefore$ Electronic configuration of ${\mathrm{Mn}}^{4+}$ is $=\left[\mathrm{Ar}\right]3{\mathrm{d}}^3$

Thus, ${\mathrm{Mn}}^{4+}$ has $3$-unpaired electron and its spin only magnetic moment $\left(\mathrm{\mu }\right)$ is, $\mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}$ $\mathrm{BM}$

where, $\mathrm{n}=$ number of unpaired electrons.

$\mathrm{\mu }=\sqrt{3(3+2)}. \mathrm{BM}\Rightarrow \mathrm{\mu }=\sqrt{15}=3.87\mathrm{BM}\approx 4\mathrm{BM}$