The calculated spin magnetic moment of gaseous ${\mathrm{Fe}}^{2+}$is 

The electronic configuration of ${ }_{26}{\mathrm{Fe}}^{2+}$is $3{\mathrm{d}}^6$. Hence, it contains 4 unpaired electrons. Thus, $\mu /{\mu }_{\mathrm{B}}=\sqrt{4\left(6\right)}=4.90$