Spin only magnetic moment of an octahedral complex of Fe$$^{2+}$$ in the presence of a strong field ligand in BM is:

Iron(II), i.e., $$\text{Fe}^{2+}$$, has the electronic configuration [Ar] $$3d^6$$, giving it 6 electrons in the $$d$$-subshell.

In an octahedral complex with a strong field ligand, the crystal field splitting energy $$\Delta_o$$ is large enough to force electron pairing. For a $$d^6$$ metal ion in a strong field octahedral environment, all six $$d$$ electrons pair up in the lower $$t_{2g}$$ orbitals:

$$t_{2g}^6 \, e_g^0$$

This results in zero unpaired electrons. The spin-only magnetic moment formula is:

$$\mu = \sqrt{n(n+2)} \text{ BM}$$

where $$n$$ is the number of unpaired electrons. With $$n = 0$$:

$$\mu = \sqrt{0(0+2)} = 0 \text{ BM}$$

Therefore, the spin-only magnetic moment of the octahedral $$\text{Fe}^{2+}$$ complex in the presence of a strong field ligand is $$0$$ BM, making it diamagnetic.