The correct order of [FeF6]3-, [CoF6]3-, [Ni(CN)4]2- and [Ni(CO)4] complexes based on the number of unpaired electrons…

🧠 Count Unpaired Electrons in Each Complex

🗺️ Evaluate

| Complex | Metal/d-config | Geometry/Spin | Unpaired | |---|---|---|---| | $[\mathrm{FeF_6}]^{3-}$ | Fe(III) d⁵, F⁻ weak → HS | Oct, $t_{2g}^3 e_g^2$ | 5 | | $[\mathrm{CoF_6}]^{3-}$ | Co(III) d⁶, F⁻ weak → HS | Oct, $t_{2g}^4 e_g^2$ | 4 | | $[\mathrm{Ni(CN)_4}]^{2-}$ | Ni(II) d⁸, CN⁻ strong | Sq pl, $d_{xy}^2 d_{xz}^2 d_{yz}^2 d_{z^2}^2$ | 0 | | $[\mathrm{Ni(CO)_4}]$ | Ni(0) d¹⁰ | Td (CO strong forces low-spin sp³) | 0 |

🗺️ Order

$5 > 4 > 0 = 0$ → $[\mathrm{FeF_6}]^{3-} > [\mathrm{CoF_6}]^{3-} > [\mathrm{Ni(CN)_4}]^{2-} = [\mathrm{Ni(CO)_4}]$.

⚡ Sq Pl Ni(II) Always Diamagnetic

For Ni(II) with strong-field ligands like CN⁻, sq pl forms with d⁸ all paired in 4 lowest orbitals → diamagnetic. Td Ni(II) (with weak field) has 2 unpaired ($e^4 t_2^4$).

⚠️ Ni(0) is d¹⁰

Ni(CO)₄ is Ni(0), d¹⁰, fully filled — diamagnetic regardless of geometry.

$\boxed{\text{Answer: (4) } [\mathrm{FeF_6}]^{3-} > [\mathrm{CoF_6}]^{3-} > [\mathrm{Ni(CN)_4}]^{2-} = [\mathrm{Ni(CO)_4}]}$