Which of the following is a correct statement about [NiCl4]2− and [Ni(CO)4]?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [NiCl4]2− is tetrahedral and paramagnetic whereas [Ni(CO)4] is tetrahedral and diamagnetic.
By valence bond theory, the shape of a four-coordinate complex depends on whether the ligands force the d electrons to pair:
sp3 hybridisation → tetrahedral: used when no inner d orbital is freed.
dsp2 hybridisation → square planar: used when a strong field ligand pairs the electrons and frees one 3d orbital.
The magnetic behaviour follows from the unpaired electrons left: any unpaired electron makes the complex paramagnetic; none makes it diamagnetic.
[NiCl4]2−: x + 4(−1) = −2, so nickel is +2. Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8, three pairs and two single electrons in the five 3d orbitals. \( \ce{Cl^{-}} \) is a weak field ligand and cannot pair the electrons, so two electrons stay unpaired. The four \( \ce{Cl^{-}} \) ions use the 4s and 4p orbitals (sp3), so the complex is tetrahedral and paramagnetic.
[Ni(CO)4]: CO is neutral, so nickel is in the zero oxidation state and has the configuration of the Ni atom, [Ar]3d84s2. The strong field CO ligands push the 4s electrons into 3d, giving 3d10 with no unpaired electron. The empty 4s and 4p orbitals give sp3 hybridisation, so the complex is tetrahedral and diamagnetic.
Why the other options are incorrect
Option (1) [NiCl4]2− is tetrahedral and paramagnetic whereas [Ni(CO)4] is also tetrahedral and paramagnetic. — Incorrect. \( \ce{[Ni(CO)4]} \) has a 3d10 configuration with no unpaired electron, so it is diamagnetic.
Option (3) [NiCl4]2− is square planar and diamagnetic whereas [Ni(CO)4] is tetrahedral and diamagnetic. — Incorrect. Square planar and diamagnetic describes [Ni(CN)4]2−, where the strong field \( \ce{CN^{-}} \) pairs the electrons; weak field \( \ce{Cl^{-}} \) cannot do this.
Option (4) [NiCl4]2− is square planar and diamagnetic whereas [Ni(CO)4] is also square planar and diamagnetic. — Incorrect. Neither complex is square planar; \( \ce{[Ni(CO)4]} \) has no empty 3d orbital for dsp2 hybridisation.
NCERT (2026-27 edition):
Valence Bond Theory – …
