Spin only magnetic moment of an octahedral complex of Fe²⁺ in the presence of a strong field ligand in BM is:

🧠 Fe²⁺ + Strong Field = Low-Spin d⁶ = Diamagnetic

Fe²⁺ has $\mathrm{d^6}$. With a strong-field ligand (CN⁻, NO₂⁻, en, NH₃-on-Fe(II)-borderline), the field is large enough to force pairing in the $\mathrm{t_{2g}}$ set:

$\mathrm{t_{2g}^6 e_g^0}$ → 0 unpaired.

$\mu = 0$ BM.

🗺️ Why d⁶ Goes Fully Paired

In low-spin d⁶: 6 electrons fit perfectly into the three $\mathrm{t_{2g}}$ orbitals (2 each) → no electrons in $\mathrm{e_g}$ → no unpaired. This is the "perfect closure" — d⁶ low-spin is the most stable d-count for octahedral coordination chemistry.

⚡ The "d⁶ Low-Spin Trio"

Fe(II), Co(III), Ru(II), Os(II), Ir(III) — all $\mathrm{d^6}$ low-spin candidates. Most famous example: $[\mathrm{Fe(CN)_6}]^{4-}$ (potassium ferrocyanide) — diamagnetic, kinetically inert.

⚠️ Don't Confuse With High-Spin Fe(II)

$[\mathrm{Fe(H_2O)_6}]^{2+}$ is high-spin d⁶ → 4 unpaired → $\mu = 4.90$ BM. Same metal, opposite extreme. The question specifies strong-field ligand, locking the answer at 0.

$\boxed{\text{Answer: (3) 0}}$