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

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

The colour of a transition metal complex comes from a d-d transition. The ligands split the d orbitals into two sets of different energy (crystal field splitting), and an electron absorbs visible light to jump from the lower set to the higher one. The colour seen is the complementary colour of the light absorbed.

Titanium (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); losing both 4s electrons and one 3d electron gives \( \ce{Ti^3+} \): \( \mathrm{3d^{1}} \). In \( \ce{[Ti(H2O)6]^3+} \) the six water ligands split the 3d orbitals into \( t_{2g} \) and \( e_g \) sets. The single electron absorbs green-yellow light to go from \( t_{2g} \) to \( e_g \), so the complex looks violet.

On heating, the water molecules are driven off. Without ligands there is no crystal field splitting, so there is no d-d transition, no visible light is absorbed, and the substance becomes colourless.

Why the other options are incorrect

• Option (2) d-d transition occur at higher wavelength. — Incorrect. A d-d transition at another wavelength would only change the colour; the substance would still be coloured.

• Option (3) \(3d^1\) electron of \(Ti^{3+}\) is lost. — Incorrect. Removing water does not oxidise the metal; titanium stays in the +3 state with its 3d¹ electron. The colour is lost because nothing splits the d orbitals.

• Option (4) d-d transition occur at lower wavelength. — Incorrect. As with Option (2), a shift in wavelength gives a different colour, not a colourless substance.

NCERT (2026-27 edition):
Colour in Coordination Compounds – …