Read the passage carefully and answer the questions:
The degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy, \( t_{2g} \) set and two orbitals of higher energy, \( e_g \) set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by Δo. Thus energy of the two \( e_g \) orbitals will increase by (3/5) Δo and that of three \( t_{2g} \) will decrease by (2/5) Δo. The crystal field splitting Δo depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields, in which case the splitting will be large, whereas others produce weak fields and consequently result in small splitting of d orbitals. Relative magnitude of crystal field splitting energy Δo and pairing energy, P (energy required for electron pairing in a single orbital) determine the formation of low spin (Δo > P) or high spin (Δo < P) complex. In tetrahedral coordination entity formation, the d-orbital splitting is inverted and is smaller as compared to the octahedral field splitting. The crystal field theory attributes the colour of the complex to d-d transition of the electron. The colour of the coordination compounds depends on the crystal field splitting. In the absence of ligand, crystal field splitting does not occur and hence the substance is colourless.
Choose the correct statement from the following:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\Delta_o \gt P\); strong field ligand; low spin complex
In an octahedral complex the first three d electrons go singly into the \( t_{2g} \) orbitals. For the fourth electron there are two choices, and the result depends on the crystal field splitting \( \Delta_o \) compared with the pairing energy P:
If \( \Delta_o \gt P \), it costs less energy to pair up in a \( t_{2g} \) orbital than to climb to \( e_g \). The electrons pair, the number of unpaired electrons is the minimum, and the complex is low spin. Ligands that cause such a large splitting are strong field ligands (\( \ce{CN-} \), CO, en, \( \ce{NH3} \)).
If \( \Delta_o \lt P \), the electron goes up into \( e_g \) unpaired, giving a high spin complex. Ligands with such a small splitting are weak field ligands (halide ions, \( \ce{H2O} \)).
So the consistent set is \( \Delta_o \gt P \), strong field ligand, low spin complex.
Why the other options are incorrect
Option (2) \(\Delta_o \gt P\); weak field ligand; low spin complex — Incorrect. A weak field ligand produces a small splitting, so it cannot make \( \Delta_o \) larger than P.
Option (3) \(\Delta_o \lt P\); weak field ligand; low spin complex — Incorrect. When \( \Delta_o \lt P \) the electrons stay unpaired, so the complex is high spin, not low spin.
Option (4) \(\Delta_o \lt P\); strong field ligand; high spin complex — Incorrect. High spin with \( \Delta_o \lt P \) is right, but that situation arises with weak field ligands, not strong field ligands.
NCERT (2026-27 edition):
Crystal Field Theory – …
