Q4. Name the lanthanoids that show the +4 state and those that show the +2 state, and relate this to their electronic configurations. (NCERT Exercise 4.32.)
+4: Ce, and, NCERT adds, Pr, Nd, Tb and Dy, but only in oxides MO2 (page 109); beyond the NCERT text, TbF4 is also known (Wikipedia, “Terbium(IV) fluoride”). Ce4+ reaches 4f0 and Tb4+ 4f7. +2: Eu, Yb and Sm. Eu2+ is 4f7 and Yb2+ 4f14. The empty, half-filled and full f sets are extra stable, but these ions still tend to return to +3, so Ce4+ and Tb4+ oxidise while Eu2+ and Yb2+ reduce (NCERT page 109). NCERT’s own Example 4.10 (page 113) names cerium as the lanthanoid best known for +4.
Q5. Write the electronic configurations of Pm3+, Ce4+ and Lu2+. (NCERT Exercise 4.1, parts (ii), (iv) and (vi).)
Pm3+: [Xe] 4f4. Ce4+: [Xe] (4f0). Lu2+: [Xe] 4f14 5d1 (NCERT Table 4.9, page 110).
Q6. Write the electronic configuration of the element with Z = 61, and say which block and series it belongs to. (Part of NCERT Exercise 4.34.)
Z = 61 is promethium, [Xe] 4f5 6s2 (NCERT Table 4.9, page 110): a lanthanoid, in the f-block.
Q7. Find the number of unpaired electrons and the spin-only magnetic moment of Tb3+ (Z = 65).
Tb3+ is 4f8. Seven f orbitals take one electron each and the eighth pairs up, so 6 are unpaired: μ = √(6 × 8) = √48 = 6.93 BM.
Q8. What is the lanthanoid contraction, and what are its consequences? (NCERT Exercise 4.7.)
The steady decrease in atomic and ionic radii from lanthanum to lutetium, caused by the poor shielding of one 4f electron by another as the nuclear charge rises (NCERT page 109). Its main consequence is that the 4d and 5d transition series have almost the same radii (Zr 160 pm, Hf 159 pm), so pairs such as zirconium and hafnium have very similar properties, occur together in nature and are hard to separate (NCERT pages 94 and 109). Beyond the NCERT text, the smooth fall in size also lets the lanthanoid ions be separated by chromatography (Wikipedia, “Lanthanide contraction”).