ExamBro
ExamBro
CUET · CHEMISTRY · PYQ PAPER 2023

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 comple: to yield three orbitals of lower energy, \(t_{2 g}\) 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 \(\Delta_0\). Thus energy of the two \(e_g\) orbitals will increase by \((3 / 5) \Delta_0\) and that of three \(t_{2 g}\) will decrease by \((2 / 5) \Delta_0\). The crystal field splitting \(\Delta_0\) 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 \(\Delta_0\) and pairing energy, \(P\) (energy required for electron paring in a single orbital) determine the formation of low spin ( \(\Delta_0>P\) ) or high spin ( \(\Delta_0<P\) ) 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?

  1. A \(\left[ Ti \left( H _2 O \right)_6\right]^{3+}\)
  2. B \(CuSO _4 \cdot 5 H _2 O\)
  3. C \(\left[ Zn \left( H _2 O \right)_6\right]^{2+}\)
  4. D \(\left[ Ni \left( H _2 O \right)_6\right]^{2+}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[ Zn \left( H _2 O \right)_6\right]^{2+}\)

Step-by-step Solution

Detailed explanation

\(Zn^{2+}\): \([Ar] 3d^{10}\) No d-d transitions are possible for a \(d^{10}\) configuration. Therefore, \(\left[ Zn \left( H _2 O \right)_6\right]^{2+}\) is not expected to be coloured.
Same subject
Explore more questions on app