The geometry and magnetic property of the complex \([Ni(CO)_4]\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Tetrahedral, Diamagnetic
In \( \ce{[Ni(CO)4]} \) CO is a neutral ligand, so nickel is in the zero oxidation state: Ni (Z = 28) is [Ar] 3d84s2, with 2 unpaired 3d electrons in the free atom.
CO is a strong field ligand. It makes the two 4s electrons pair up in the 3d orbitals, giving 3d10 with 0 unpaired electrons. The empty 4s and three 4p orbitals then take the four electron pairs from four CO molecules:
3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ | 4s 4p: four pairs from CO ⇒ sp3
sp3 hybridisation gives a tetrahedral complex, and with no unpaired electron it is diamagnetic.
Why the other options are incorrect
Option (1) Square planar, Diamagnetic — Incorrect. Square planar geometry needs dsp2 hybridisation with one empty 3d orbital; in Ni(0) all five 3d orbitals are full, so no 3d orbital is free.
Option (2) Square planar, Paramagnetic — Incorrect. Both parts are wrong; the complex is neither square planar nor has any unpaired electron.
Option (4) Tetrahedral, Paramagnetic — Incorrect. The shape is right, but CO pairs all the electrons (3d10, 0 unpaired), so the complex is not paramagnetic.
NCERT (2026-27 edition):
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.”