Q.Why salts of Sc3⊕, Ti4⊕, V5⊕ are colourless ?

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.)

OR (why transition elements form complexes readily): Transition metal ions readily form complexes because they (i) have a comparatively small ionic size and high nuclear charge, giving them a high charge/size ratio (charge density), which makes them strongly polarising and able to attract electron pairs from ligands, and (ii) have empty (or partially filled) d orbitals of suitable energy that can accept lone pairs of electrons donated by ligands to form coordinate (dative) bonds.

$Ti^{4+}$ 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.

$Ti$ is $[Ar]3d^24s^2$; $Ti^{4+}$ has lost all 4 valence electrons, leaving $[Ar]3d^0$. With no d electrons at all, there is no possible d–d transition, so $Ti^{4+}$ compounds are colourless.