Q.Why is $Sc^{3+}$ colourless while $Ti^{3+}$ coloured? (Atomic number Sc = 21, Ti = 22)

Colour in transition-metal compounds arises from d-d transitions; complex-forming tendency arises from small size, high charge, and available empty d orbitals.

Why coloured: Most transition metal ions have partially filled d orbitals ($d^1$ to $d^9$). In the free ion these five d orbitals are degenerate, but in a compound/complex, surrounding ligands split them into sets of different energies (crystal field splitting). An electron can absorb a photon of visible light and get promoted from a lower-energy d orbital to a higher-energy one (a d-d transition). Since only certain wavelengths of visible light are absorbed for this transition, the compound transmits/reflects the complementary colour, making it appear coloured. (Ions with either empty, $d^0$, or completely filled, $d^{10}$, d orbitals — like $Sc^{3+}$ or $Zn^{2+}$ — are usually colourless because no d-d transition is possible.)

…

$Ti^{4+}$ is $d^0$ — no d electrons means no d–d transition, hence colourless.

Colour in transition-metal complexes arises because partially filled d orbitals allow an electron to be promoted from a lower-energy to a higher-energy d orbital (d–d transition) on absorbing visible light, and the complementary colour is seen.

…