On heating, $[Ti(H_2O)_6] Cl_3$ becomes colorless because........
Answer & explanation
Correct answer: option 1
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.