On heating, $[Ti(H_2O)_6] Cl_3$ becomes colorless because........

Crystal field splitting is lost.

The correct answer is Option (1) → Crystal field splitting is lost.

Statement: On heating, $[Ti(H_{2}O)_{6}]Cl_{3}$ becomes colorless because crystal field splitting is lost.

Reasoning:

The violet colour of the $Ti^{3+}$ complex is due to $d-d$ transitions between split $d$-orbitals in the octahedral crystal field created by water ligands. On heating, coordinated water molecules are removed, the ligand field collapses, crystal field splitting disappears, and hence no $d-d$ transition occurs — the compound becomes colorless.

Option-wise analysis

Option 1: Crystal field splitting is lost.

Statement: This is correct.

Reasoning: Without ligands, the $d$-orbitals are no longer split, so $d-d$ transitions (which give colour) cannot occur.

Option 2: $d-d$ transition occur at higher wavelength.

Statement: Incorrect.

Reasoning: A shift in wavelength would change the colour, not make the compound colorless. Colour disappears only when $d-d$ transitions stop, not merely shift.

Option 3: $3d^{1}$ electron of $Ti^{3+}$ is lost.

Statement: Incorrect.

Reasoning: Heating removes water ligands, not the electron from $Ti^{3+}$. The oxidation state of titanium does not change here.

Option 4: $d-d$ transition occur at lower wavelength.

Statement: Incorrect.

Reasoning: Like Option 2, a wavelength shift would still produce colour. The observed colourlessness requires absence of crystal field splitting and hence no $d-d$ transition.