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 order of the ligands in their increasing crystal field splitting:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(Br^- \lt H_2O \lt NH_3 \lt CN^- \lt CO\)
Ligands can be arranged by the size of the crystal field splitting \( \Delta_o \) they produce. This experimentally found order is the spectrochemical series. Halide ions are weak field ligands at the start of the series, oxygen donors such as water come next, nitrogen donors such as ammonia are stronger, and ligands that can also accept electrons back from the metal (\( \ce{CN-} \), CO) give the largest splitting.
Picking out the five given ligands in the order of the series:
\( \displaystyle \ce{Br-} \lt \ce{H2O} \lt \ce{NH3} \lt \ce{CN-} \lt \ce{CO} \)
Shortcut: The weakest ligand, the halide, must come first and CO last. Only Option (2) starts with \( \ce{Br-} \) and ends with CO.
NCERT (2026-27 edition):
Crystal Field Theory – …
