According to the valence bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np, nd orbitals for hybridization to yield a set of equivalent orbitals of definite geometry. These hybridized orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. It is usually possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of the valence bond theory. Consider the formation of \([Co(NH_3)_5Cl]Cl_2\) and answer the following question:
The hybridization of cobalt in the above coordination entity is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(d^2 sp^3\)
Cobalt is +3 in the complex: \( \ce{Co^{3+}} \) is [Ar]3d6. The six ligands (five \( \ce{NH3} \) and one \( \ce{Cl^{-}} \)) need six empty orbitals. The six 3d electrons pair up in three 3d orbitals, leaving two inner 3d orbitals empty, and these combine with one 4s and three 4p orbitals:
3d: [↑↓] [↑↓] [↑↓] [ ×× ] [ ×× ] 4s: [ ×× ] 4p: [ ×× ] [ ×× ] [ ×× ]
(×× = electron pair donated by a ligand.) This is d2sp3 hybridisation: an inner orbital, low spin, octahedral complex, as for \( \ce{[Co(NH3)6]^{3+}} \).
Why the other options are incorrect
Option (1) sp3d2 — Incorrect. Uses the outer 4d orbitals and leaves four unpaired electrons; this is the high spin outer orbital complex \( \ce{[CoF6]^{3-}} \), which is paramagnetic.
Option (3) sp3d — Incorrect. Gives only five hybrid orbitals (trigonal bipyramidal), but six ligands are attached.
Option (4) dsp3 — Incorrect. Also gives only five hybrid orbitals, not the six needed here.
NCERT (2026-27 edition):
Valence Bond Theory – …
