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 type of hybridisation state in [CoF6]3−
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → sp3d2
For a six-coordinate (octahedral) complex, VBT allows two hybridisations: d2sp3 (inner orbital, uses the 3d orbitals, needs electron pairing) with strong-field ligands, and sp3d2 (outer orbital, uses the 4d orbitals, no pairing) with weak-field ligands.
In \( \ce{[CoF6]^{3-}} \): x + 6(−1) = −3, so cobalt is +3. Co (Z = 27) has 27 − 18 = 9 electrons beyond the argon core: [Ar] 3d74s2; removing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): [Ar] 3d6.
\( \ce{F^{-}} \) is a weak-field ligand, so the 3d electrons stay unpaired by Hund’s rule and no 3d orbital is free:
\( \ce{Co^{3+}} \) in \( \ce{[CoF6]^{3-}} \): 3d [↑↓] [↑] [↑] [↑] [↑] 4s, 4p, 4d: six sp3d2 orbitals filled by 6 \( \ce{F^{-}} \) pairs
The 4s, three 4p and two 4d orbitals hybridise: sp3d2, an outer orbital (high spin) octahedral complex with 4 unpaired electrons (paramagnetic).
Why the other options are incorrect
Option (1) sp3 — Incorrect. sp3 gives only four hybrid orbitals (tetrahedral), but six \( \ce{F^{-}} \) ligands need six.
Option (2) dsp2 — Incorrect. This also gives four orbitals (square planar), as in \( \ce{[Ni(CN)4]^{2-}} \).
Option (4) d2sp3 — Incorrect. This inner orbital hybridisation needs two empty 3d orbitals, which requires pairing by a strong-field ligand, as in diamagnetic \( \ce{[Co(NH3)6]^{3+}} \); \( \ce{F^{-}} \) does not cause pairing.
NCERT (2026-27 edition):
Valence Bond Theory – “… The paramagnetic octahedral complex, [CoF6]3− uses outer orbital (4d) in hybridisation (sp3d2). It is thus called outer orbital or high spin or spin free complex.”