What is the type of hybridisation of nickel in [Ni(CO)4]?

The correct answer is Option (2) → sp³

• Oxidation state of Ni: CO is a neutral ligand, so x + 4(0) = 0 and x = 0.

• Configuration: Ni (Z = 28) is [Ar]3d⁸4s².

• In the complex: \( \ce{[Ni(CO)4]} \) is diamagnetic, so all ten valence electrons are paired in the 3d orbitals (3d¹⁰), and no 3d orbital is free.

• Hybridisation: the four CO electron pairs go into one 4s and three 4p orbitals: sp³, four orbitals pointing to the corners of a tetrahedron.

Why the other options are incorrect

Option (1) dsp² — Incorrect. dsp² needs a vacant 3d orbital and gives a square planar complex, as in \( \ce{[Ni(CN)4]^2-} \) (Ni²⁺, 3d⁸); in Ni(0) with 3d¹⁰ no 3d orbital is empty.

Option (3) sp³d — Incorrect. It uses five orbitals and goes with coordination number 5 (trigonal bipyramidal); here there are only four ligands.

Option (4) d²sp³ — Incorrect. It uses six orbitals and goes with coordination number 6 (octahedral).

NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. … Similarly, [Ni(CO)₄] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.”