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Answer: 3. tetrahedral geometry and diamagnetic

In \([\mathrm{Ni(CO)_4}]\) Ni is in the zero oxidation state with configuration \(3d^8 4s^2\). CO is a strong-field ligand, so the 4s electrons shift into 3d, giving \(3d^{10}\).

With no d vacancy, Ni uses one 4s and three 4p orbitals: \(sp^3\) hybridisation, so the shape is tetrahedral. With no unpaired electrons, it is diamagnetic.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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