Read the passage carefully and answer the Questions.
According to Valence Bond 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. While the VB theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it suffers from the many shortcomings. The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.
Which of the following represents the correct pair?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → inner orbital complex - low spin complex
In VB theory an octahedral complex uses one of two sets of d orbitals for hybridisation:
\( \mathrm{d^{2}sp^{3}} \) uses the inner (n−1)d orbitals. To free two of them, the metal electrons pair up, so the complex is an inner orbital, low spin (spin paired) complex, e.g. \( \ce{[Co(NH3)6]^3+} \).
\( \mathrm{sp^{3}d^{2}} \) uses the outer nd orbitals. The inner d electrons stay unpaired, so the complex is an outer orbital, high spin (spin free) complex, e.g. \( \ce{[CoF6]^3-} \).
So “inner orbital complex – low spin complex” is the correct pair.
Why the other options are incorrect
Option (1) \(d^2sp^3\)-spin free complex — Incorrect. \( \mathrm{d^{2}sp^{3}} \) needs paired inner electrons, so it gives a spin paired complex.
Option (3) outer orbital complex - low spin complex — Incorrect. Outer orbital complexes keep their inner electrons unpaired; they are high spin.
Option (4) \(d^2sp^3\)- outer orbital complex — Incorrect. \( \mathrm{d^{2}sp^{3}} \) uses the inner (n−1)d orbitals; the outer orbital type is \( \mathrm{sp^{3}d^{2}} \).
NCERT (2026-27 edition):
Valence Bond Theory – …
