Select the option with correct property:

🧠 Two Tetrahedral Ni Complexes, Opposite Magnetic States

Both $[\mathrm{Ni(CO)_4}]$ and $[\mathrm{NiCl_4}]^{2-}$ are tetrahedral, but their oxidation states differ:

• $[\mathrm{Ni(CO)_4}]$: Ni(0), $\mathrm{d^{10}}$ → all electrons paired → diamagnetic ($\mu = 0$).
• $[\mathrm{NiCl_4}]^{2-}$: Ni(II), $\mathrm{d^8}$, tetrahedral. Tetrahedral $\mathrm{d^8}$: $\mathrm{e^4 t_2^4}$ → 2 unpaired → paramagnetic ($\mu = 2.83$ BM).

Match: Ni(CO)₄ diamagnetic, NiCl₄²⁻ paramagnetic → option (4).

🗺️ Why CO Drives Ni to Zero

CO is a neutral ligand and a strong π-acceptor. To stabilise the Ni–C π-back-bond, Ni donates its 4s and 3d electrons into CO π* orbitals. The formal oxidation state stays 0 (CO is neutral, complex is neutral) — and Ni ends up $\mathrm{d^{10}}$, fully paired.

⚡ The "Carbonyl = M(0) = $\mathrm{d^n_{max}}$" Pattern

Binary carbonyls — $[\mathrm{Ni(CO)_4}], [\mathrm{Fe(CO)_5}], [\mathrm{Cr(CO)_6}]$ — always have the metal in 0 oxidation state, with maximum d-electron count for that metal. They obey the 18-electron rule and are nearly always diamagnetic.

⚠️ Don't Lump Ni(CO)₄ and NiCl₄²⁻

Same metal, same coordination number (4), same geometry (tetrahedral) — opposite magnetic behaviour. The difference is the oxidation state, not the geometry.

$\boxed{\text{Answer: (4) Ni(CO)₄ diamagnetic, NiCl₄²⁻ paramagnetic}}$