Spin only magnetic moment in BM of [Fe(CO)4(C2O4)]+ is:

🧠 Strong-Field Ligands Around Fe(III) → Low-Spin

$[\mathrm{Fe(CO)_4(C_2O_4)}]^+$: charge balance $x + 4(0) + (-2) = +1 \Rightarrow x = +3$. So Fe(III), $\mathrm{d^5}$.

Ligands: 4 CO (neutral, strong-field, π-acceptor) + oxalate $\mathrm{C_2O_4^{2-}}$ (bidentate, mid-strong field). Total donor atoms: 4 + 2 = 6 → octahedral.

Strong-field environment forces $\mathrm{d^5}$ low-spin: $\mathrm{t_{2g}^5 e_g^0}$ → 1 unpaired.

$\mu = \sqrt{1 \times 3} = \sqrt{3} \approx 1.73$ BM.

🗺️ Why CO Always Goes Low-Spin

CO is one of the strongest π-acceptor ligands. Its $\Delta_o$ contribution is so large that even d⁵ Fe(III) — which usually goes high-spin with halides or water — pairs up to 1 unpaired.

⚡ The "CO + Fe(III) = Low-Spin d⁵" Anchor

Whenever you see CO ligands on Fe(III), expect low-spin: $\mu \approx 1.73$ BM. Same applies to CN⁻ on Fe(III) ($[\mathrm{Fe(CN)_6}]^{3-}$).

⚠️ Oxalate Is Mid-Strength, But CO Dominates

Oxalate alone might give high-spin Fe(III), but when CO is also present, CO's strong field dominates the average ligand-field strength → low-spin.

$\boxed{\text{Answer: (4) 1.73}}$