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.
Which of the following complex is not expected to be coloured?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \([Zn(H_2O)_6]^{2+}\)
According to crystal field theory, the colour of a complex comes from a d–d transition: an electron in the lower \( t_{2g} \) set absorbs visible light and jumps to the higher \( e_g \) set. The complex shows the colour complementary to the light absorbed. Such a transition needs at least one d electron and a vacancy in the \( e_g \) set, so ions with \( d^0 \) or \( d^{10} \) configurations are colourless.
For zinc, Z = 30, the atom is \( \mathrm{[Ar]\,3d^{10}\,4s^{2}} \). Losing the two 4s electrons gives \( \ce{Zn^2+} \): \( \mathrm{3d^{10}} \).
Both \( t_{2g} \) and \( e_g \) are full, so no d–d transition is possible and \( [\ce{Zn(H2O)6}]^{2+} \) is colourless.
Option (1) \([Ti(H_2O)_6]^{3+}\) — Coloured. Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); losing three electrons gives \( \ce{Ti^3+} \): \( \mathrm{3d^{1}} \). The single electron can jump from \( t_{2g} \) to \( e_g \), and the complex is purple.
Option (2) \(CuSO_4.5H_2O\) — Coloured. Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); losing the 4s electron and one 3d electron gives \( \ce{Cu^2+} \): \( \mathrm{3d^{9}} \). With one vacancy in \( e_g \), d–d transition occurs and the salt is blue.
Option (3) \( [Zn(H_2O)_6]^{2+} \) — Not coloured (3d10). Zn2+ is 3d10, so no d–d transition is possible.
Option (4) \([Ni(H_2O)_6]^{2+}\) — Coloured. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \). The \( e_g \) set is half filled, and the complex is green.
NCERT (2026-27 edition):
Formation of Coloured Ions – …
