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?

$[Zn(H_2O)_6]^{2+}$

The correct answer is Option (3) → $[Zn(H_2O)_6]^{2+}$

The colour of transition metal complexes generally arises due to d–d electronic transitions.

For d–d transitions to occur:

• The metal ion must have partially filled d orbitals.
• If the d subshell is completely filled or completely empty, d–d transitions are not possible, so the complex is colourless.

[Ti(H₂O)₆]³⁺

Ti³⁺ configuration:

Ti = [Ar] 3d² 4s²

Ti³⁺ = 3d¹

Since there is one d electron, d–d transition can occur, so the complex is coloured.

CuSO₄·5H₂O

Cu²⁺ configuration:

Cu = [Ar] 3d¹⁰ 4s¹

Cu²⁺ = 3d⁹

Partially filled d orbitals allow d–d transitions, so the compound is blue coloured.

[Zn(H₂O)₆]²⁺

Zn²⁺ configuration:

Zn = [Ar] 3d¹⁰ 4s²

Zn²⁺ = 3d¹⁰

The d orbitals are completely filled, so no d–d transition occurs.

Thus, the complex is colourless.

[Ni(H₂O)₆]²⁺

Ni²⁺ configuration:

Ni = [Ar] 3d⁸ 4s²

Ni²⁺ = 3d⁸

Since the d orbitals are partially filled, d–d transitions occur, making the complex coloured.