Werner was the first to describe the bonding features in coordination compounds. But his theory could not answer basic questions like :
(i) Why only certain elements possess the remarkable property of forming coordination compounds ?
(ii) Why the bonds in coordination compounds have directional properties ?
(iii) Why coordination compounds have characteristic magnetic and optical properties ?
Many approaches have been put forth to explain the nature of bonding in coordination compounds viz. Valence Bond Theory (VBT), Crystal Field Theory (CFT), Ligand Field Theory (LFT) and Molecular Orbital Theory (MOT). We shall focus our attention on elementary treatment of the application of VBT and CFT to coordination compounds.
According to this theory, the metal atom or ion under the influence of ligands can use its (n−1)d, ns np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding.
Predict the shape of [Ni(CO)4] on the basis of VBT :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Tetrahedral
In valence bond theory the metal uses a set of hybrid orbitals to accept the ligand electron pairs, and the hybridisation decides the shape: 4 pairs → sp3 (tetrahedral) or dsp2 (square planar); 6 pairs → d2sp3 (inner orbital octahedral) or sp3d2 (outer orbital octahedral).
In \( \ce{[Ni(CO)4]} \), CO is neutral, so nickel is in the 0 oxidation state. Ni (Z = 28) has 28 − 18 = 10 electrons beyond the argon core: [Ar] 3d84s2. CO is a strong-field ligand, so the two 4s electrons pair up into 3d, giving 3d10 and leaving 4s and 4p empty:
Ni(0) ground state: 3d [↑↓] [↑↓] [↑↓] [↑] [↑] 4s [↑↓]
Ni in \( \ce{[Ni(CO)4]} \): 3d [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] 4s 4p: four sp3 orbitals filled by 4 CO pairs
The one 4s and three 4p orbitals form four sp3 hybrids, so the complex is tetrahedral and diamagnetic (no unpaired electron).
Why the other options are incorrect
Option (1) Square Planar — Incorrect. dsp2 needs an empty 3d orbital; in Ni(0) all five 3d orbitals are full (3d10). Square planar is the shape of \( \ce{[Ni(CN)4]^{2-}} \) (\( \ce{Ni^{2+}} \), 3d8).
Option (3) Inner octahedral — Incorrect. An octahedral complex needs six ligands, and d2sp3 needs two empty 3d orbitals; \( \ce{[Ni(CO)4]} \) has only four ligands.
Option (4) Outer octahedral — Incorrect. sp3d2 also needs six ligand pairs; with four CO ligands no octahedral arrangement is possible.
NCERT (2026-27 edition):
Valence Bond Theory (Table 5.2: Number of Orbitals and Types of Hybridisations)
| Coordination number | Type of hybridisation | Distribution of hybrid orbitals in space |
|---|---|---|
| 4 | sp3 | Tetrahedral |
| 4 | dsp2 | Square planar |
| 5 | sp3d | Trigonal bipyramidal |
| 6 | sp3d2 | Octahedral |
| 6 | d2sp3 | Octahedral |