The magnetic moment of Fe 2 + is B . M .

We know

$\mathrm{\mu } = \sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}$

where, $\mathrm{n}$ is the number of unpaired electrons

The electronic configuration of ${\mathrm{Fe}}^{2+} = \left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^0$

According to the Aufbau's principal the unpaired electrons present in ${\mathrm{Fe}}^{2+}$ is $4$

Thus,

$$\begin{gathered} \mathrm{\mu } = \sqrt{4\times \left(6\right)} \mathrm{B}.\mathrm{M}. \\ \mathrm{\mu }= 4.9 \mathrm{B}.\mathrm{M}. \end{gathered}$$