15.1 Lanthanides and Actinides: Contraction, Magnetism, Spectra, Separation
★★★ 2021–2025: 27 parts, 261 marks (Civil Services 11 parts / 115 m · IFoS 9 / 71 m · Geo-Scientist 7 / 75 m). Magnetic moments and separation methods come up nearly every year in all three exams.
Syllabus lines. Civil Services and IFoS: “Lanthanides and actinides; separation, oxidation states, magnetic and spectral properties; lanthanide contraction”. Geo-Scientist: “f-block elements: electronic configuration, ionization enthalpies, oxidation states, variation in atomic and ionic (3+) radii, magnetic and spectral properties of lanthanides, separation of lanthanides (by ion-exchange method)”. Ligand-field ideas used here are in 13.2 Crystal field theory.
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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes
15.1.1 Configurations and oxidation states ★★
- Atoms: [Xe]4fn6s² for most; La [Xe]5d¹6s², Ce [Xe]4f¹5d¹6s², Gd [Xe]4f⁷5d¹6s², Lu [Xe]4f¹⁴5d¹6s² (a 5d electron where it keeps 4f empty, half-full or full). Pr is [Xe]4f³6s², Eu [Xe]4f⁷6s² CSE 2023 Q7(b) 10m.SN
- Ions: 6s and 5d go first, so Ln³⁺ = [Xe]4fn, n = 0 (La³⁺) to 14 (Lu³⁺). +3 dominates because the sum IE₁ + IE₂ + IE₃ is balanced by large lattice or hydration energies; the fourth IE is too high for most.SN
- Other states appear when they reach 4f⁰, 4f⁷ or 4f¹⁴: Ce⁴⁺ (4f⁰), Tb⁴⁺ (4f⁷), Pr⁴⁺ (in solids only); Eu²⁺ (4f⁷), Yb²⁺ (4f¹⁴), Sm²⁺ (4f⁶) IFoS 2022 Q7(c)(i) 8m.SN
- Why Eu²⁺ is stable: 4f⁷ is half-filled with extra exchange energy, so IE₃ of Eu is unusually high. Of Ce, Tb and Sm, Sm most readily gives +2 (Sm²⁺ is 4f⁶, one short of half-filled; SmI₂ is a common reagent); Ce and Tb prefer +4 IFoS 2024 Q7(c) 10m.SN
- Half-filled or filled shells are not the whole story: Sm²⁺ and Tm²⁺ exist without such shells, because the balance of ionisation, lattice and hydration energies decides.
15.1.2 The lanthanide contraction ★★★
- What: a steady fall in atomic and Ln³⁺ size across the series, 103 pm (La³⁺) → 86 pm (Lu³⁺), about 1 pm per element CSE 2024 Q5(e)(i) 5m.SN
- Why: each step adds one proton and one 4f electron. 4f orbitals are diffuse and shield one another poorly, so Zeff felt by the 5s²5p⁶ and valence electrons rises and the whole atom shrinks.SN
- Relativity adds a smaller part: in a heavy atom the s and p1/2 electrons that penetrate near the nucleus move fast, their mass rises and their orbitals contract; this contraction (and the indirect expansion of 4f and 5d) adds to the effect of poor shielding, but poor shielding remains the main cause IFoS 2024 Q5(d) 8m.
15.1.3 Consequences of the contraction ★★★
- 4d and 5d metals are the same size: Zr 160 / Hf 159 pm, Nb 146 / Ta 146, Mo 139 / W 139. The normal rise down a group is cancelled, so Zr and Hf have almost the same chemistry and occur together; Hf has twice the density (13.3 against 6.5 g cm⁻³). Hf goes below Zr because its outer configuration 5d²6s² matches 4d²5s² and it keeps +4 CSE 2024 Q5(e)(ii) 5m.SN
- Basicity falls from La(OH)₃ to Lu(OH)₃: smaller Lu³⁺ polarises OH⁻ more, M–O becomes more covalent. Order: Ce(OH)₃ > Nd(OH)₃ > Gd(OH)₃ > Lu(OH)₃; only the last (like Yb(OH)₃) dissolves in hot concentrated NaOH as [Lu(OH)₆]³⁻ IFoS 2023 Q8(c)(i) 7m Geo 2023 Q10(b) 10m.SN
- Hydrolysis rises: [Lu(H₂O)n]³⁺ is the stronger acid (pKa ≈ 7.6 against ≈ 8.5 for La³⁺), so aqueous LuCl₃ has the lower pH CSE 2022 Q7(b) 10m.SN
- Oxides: La₂O₃ is ionic and strongly basic (takes up CO₂ and water like CaO); Lu₂O₃ is more covalent and only weakly basic, so acidic character of the oxides rises La → Lu. Oxo salts (carbonates, sulphates, nitrates) become less stable to heat La → Lu, as the smaller cation polarises the anion more Geo 2025 Q8(b) 15m.
- Complexes get more stable La → Lu (log K for [Ln(EDTA)]⁻ rises from about 15.5 to 19.8), because charge density rises. Yet ΔH° of EDTA complexation is ≈ 0 and nearly constant: stronger Ln–ligand bonds are paid for by stronger Ln–water bonds that must break; the drive is entropic, from release of bound water (ΔG° = ΔH° − TΔS°) Geo 2024 Q6(c)(i) 5m.SN
- Y³⁺ (90 pm) is as large as Ho³⁺, so Y occurs and is separated with the heavy lanthanides; small regular size differences make separation possible (15.1.6).
15.1.4 Magnetic properties ★★★ every year
- 3d ions: the ligand field (≈ 10⁴ cm⁻¹) is far larger than spin–orbit coupling (≈ 10² cm⁻¹) and quenches the orbital moment: μ ≈ μs = √[n(n + 2)].
- 4f ions: 4f is buried under 5s²5p⁶; the ligand field is only ≈ 10² cm⁻¹ while spin–orbit coupling is ≈ 10³ cm⁻¹. L and S stay coupled to J, and the moment depends on J CSE 2025 Q5(h) 5m CSE 2021 Q8(b) 20m.SN
- Hund’s rules for 4fn: S as large as possible; L = sum of ml (fill 3, 2, 1, 0, −1, −2, −3); J = L − S for n < 7, J = L + S for n > 7.SN
- So f¹–f⁶ have low moments, f⁸–f¹³ high moments; only Gd³⁺ (L = 0) gives the spin-only value; La³⁺ and Lu³⁺ (and Ce⁴⁺, Yb²⁺) are diamagnetic.SN
| Ion | 4fn | Ground term | gJ | μJ / BM | found | Colour |
|---|---|---|---|---|---|---|
| La³⁺ | 4f0 | 1S0 | – | 0.00 | 0 | colourless |
| Ce³⁺ | 4f1 | 2F5/2 | 6/7 | 2.54 | 2.4 | colourless |
| Pr³⁺ | 4f2 | 3H4 | 4/5 | 3.58 | 3.5 | green |
| Nd³⁺ | 4f3 | 4I9/2 | 8/11 | 3.62 | 3.5 | lilac |
| Pm³⁺ | 4f4 | 5I4 | 3/5 | 2.68 | – | pink |
| Sm³⁺ | 4f5 | 6H5/2 | 2/7 | 0.85 | 1.5 | yellow |
| Eu³⁺ | 4f6 | 7F0 | – | 0.00 | 3.4 | pale pink |
| Gd³⁺ | 4f7 | 8S7/2 | 2 | 7.94 | 7.9 | colourless |
| Tb³⁺ | 4f8 | 7F6 | 3/2 | 9.72 | 9.7 | pale pink |
| Dy³⁺ | 4f9 | 6H15/2 | 4/3 | 10.65 | 10.5 | yellow |
| Ho³⁺ | 4f10 | 5I8 | 5/4 | 10.61 | 10.5 | yellow |
| Er³⁺ | 4f11 | 4I15/2 | 6/5 | 9.58 | 9.5 | lilac |
| Tm³⁺ | 4f12 | 3H6 | 7/6 | 7.56 | 7.3 | pale green |
| Yb³⁺ | 4f13 | 2F7/2 | 8/7 | 4.54 | 4.5 | colourless |
| Lu³⁺ | 4f14 | 1S0 | – | 0.00 | 0 | colourless |
Solved PYQs CSE 2025 Q8(d) 10m Geo 2025 Q10(c) 10m IFoS 2021 Q8(c) 10m
- Pr³⁺, 4f²: S = 1, L = 3 + 2 = 5, J = L − S = 4 (³H₄); g = 4/5; μ = 4/5√20 = 3.58 BM (found ≈ 3.5).
- Tb³⁺, 4f⁸: S = 3, L = 3, J = 6 (⁷F₆); g = 3/2; μ = 3/2√42 = 9.72 BM (found ≈ 9.7).
- La³⁺ 4f⁰ and Lu³⁺ 4f¹⁴: 0 BM; Gd³⁺ 4f⁷: S = 7/2, L = 0, g = 2, μ = √63 = 7.94 BM.
- Sm³⁺, 4f⁵: S = 5/2, L = 5, J = 5/2 (⁶H5/2); g = 2/7; μ = 0.85 BM. Eu³⁺, 4f⁶: S = 3, L = 3, J = 0: μ = 0. Found ≈ 1.5 and 3.4 BM: the next J levels lie only a few hundred cm⁻¹ higher and are partly occupied at room temperature.SN
With the uncoupled formula (15.1.3) Pr³⁺, Eu³⁺, Gd³⁺ give 6.16, 7.75 and 7.94 BM CSE 2023 Q7(b); quote both formulas when the question names μS+L. Among Ce³⁺, Eu³⁺, Tb³⁺, Yb³⁺ the calculated moment is zero only for Eu³⁺ (J = 0) IFoS 2023 Q8(c)(ii) 8m.
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