Part (i): Colour of transition metal compounds.
Step 1: Reason. Transition metal ions have partially filled (n-1)d orbitals. In the presence of ligands or in the crystal, the five d orbitals split into two sets of slightly different energy (\(t_{2g}\) and \(e_g\)).
Step 2: d-d transition. When visible light falls on the compound, an unpaired d electron absorbs a particular wavelength (energy) and is promoted from the lower to the higher d-orbital set. This is called a d-d transition.
Step 3: Observed colour. The compound transmits/reflects the remaining light, so it appears in the complementary colour of the light absorbed. Ions with completely empty (\(d^0\), e.g. \(Sc^{3+}\), \(Ti^{4+}\)) or completely filled (\(d^{10}\), e.g. \(Zn^{2+}\), \(Cu^+\)) d orbitals have no d-d transition and are therefore colourless.
Part (ii): Lanthanoid contraction.
Step 1: Statement. Across the lanthanoid series (from \(La\) to \(Lu\)), as the atomic number increases the atomic and ionic (\(M^{3+}\)) radii show a steady decrease. This regular decrease is called the lanthanoid contraction.
Step 2: Cause. The added electrons enter the inner 4f subshell. The 4f electrons have a poor shielding effect, so the effective nuclear charge felt by the outer electrons increases with atomic number, pulling the electron cloud closer and reducing the size.
Step 3: Consequences.
(a) Elements of the second (4d) and third (5d) transition series have almost the same size (e.g. Zr and Hf), so their properties are very similar and they are difficult to separate.
(b) There is a slow decrease in the basic strength of the hydroxides from \(La(OH)_3\) to \(Lu(OH)_3\).
(c) It causes similarity in properties of the lanthanoids, making their separation difficult.