Question

Why is \( \mathrm{[Ti(H_2O)_6]^{3+}} \) coloured ? [Atomic number : Ti = 22]

Hint:

Ti3+ is 3d1, so the lone d electron absorbs visible light in a d-d transition from t2g to eg and the complex shows a colour.

Answer 1

Step 1: Find the d-electron count of the metal ion:
Titanium (Z = 22) is \( \mathrm{[Ar]\,3d^2\,4s^2} \).
Ti3+ loses three electrons, so its configuration is \( \mathrm{3d^1} \). It has one electron in the d subshell.

Step 2: Apply crystal field splitting:
Six water ligands form an octahedral field. The five d orbitals split into a lower set \( t_{2g} \) and a higher set \( e_g \).
The single d electron sits in the \( t_{2g} \) level, so the ground state is \( t_{2g}^1 e_g^0 \).

Step 3: Explain the absorption of light:
The gap between \( t_{2g} \) and \( e_g \) is small. Light from the visible region has just the right energy to lift the electron from \( t_{2g} \) to \( e_g \).
This jump is called a d-d transition. The ion absorbs part of the white light, mostly blue-green.

Step 4: Link absorption to colour:
The light that is not absorbed passes through. We see this complementary colour, which is purple (violet).

Final Answer:
\( \mathrm{[Ti(H_2O)_6]^{3+}} \) has one unpaired d electron. It absorbs visible light for the d-d transition, so it appears coloured.
\[ \boxed{\text{d-d transition of the } 3d^1 \text{ electron}} \]