Spin only magnetic moment of an octahedral complex of ${\mathrm{Fe}}^{2+}$in the presence of a strong field ligand is B.M.

An octahedral complex's spin-only magnetic moment ${\mathrm{Fe}}^{2+}$. in the presence of a strong field ligand pairing occurs due to strong field ligand it have zero unpaired electron so magnetic moment is zero. Electrical arrangement of ${{}_{26}\mathrm{Fe}}^{2+}$ is $\left[\mathrm{Ar}\right]3{ \mathrm{d}}^{6}4{ \mathrm{s}}^0$.
The magnetic moment is now. $\left(\mu \right)=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ here n= number of unpaired electrons.
$\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{0(0+2)}=0\mathrm{BM}$