Part (i): Why transition metal compounds are coloured.
Step 1: The key feature is the presence of unpaired electrons in the \(d\) subshell of the metal ion. Under the electric field of surrounding ligands the degenerate \(d\) orbitals are no longer equal in energy; they separate into a lower and a higher group with an energy gap \(\Delta\).
Step 2: This gap \(\Delta\) corresponds to the energy of visible light. A \(d\) electron absorbs the matching wavelength and jumps to the higher level (a \(d\text{-}d\) excitation), and the eye sees the leftover complementary colour, giving the characteristic colours of ions such as \(Cu^{2+}\) (blue) and \(Fe^{3+}\) (yellow).
Step 3: If there is no unpaired \(d\) electron to excite, as in \(d^0\) (\(Sc^{3+}\)) or \(d^{10}\) (\(Zn^{2+}\)) ions, no such absorption occurs and the compound is white or colourless.
Part (ii): Lanthanoid contraction.
Step 1: Moving along the 4f series each new electron goes into a deep-lying 4f orbital. Because 4f electrons screen the nucleus very poorly, every added proton is felt more strongly by the outer shells.
Step 2: The rising effective nuclear charge draws the outer electrons inward, so both atomic and \(M^{3+}\) ionic radii shrink gradually from lanthanum to lutetium; this gradual shrinkage is the lanthanoid contraction.
Step 3: Its effects include the near-identical sizes of Zr/Hf and Nb/Ta in the 4d and 5d series (hard to separate), the steady fall in basicity of the lanthanoid hydroxides, and the overall close similarity of the lanthanoids to one another.