f-Block

f-Block

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

The f-block holds the two inner-transition series: lanthanoids (4f, Z = 58–71) and actinoids (5f, Z = 90–103). General configuration: [Xe]4f¹⁻¹⁴5d⁰⁻¹6s² for lanthanoids and [Rn]5f¹⁻¹⁴6d⁰⁻¹7s² for actinoids. The defining anomaly is the lanthanoid contraction — a steady ≈ 1.85 pm decrease in Ln³⁺ radius from La³⁺ to Lu³⁺, caused by the poor shielding of 4f electrons. Must-remember oxidation states: +3 for all lanthanoids, +4 for Ce (and to a lesser extent Pr, Tb), and +2 for Eu, Yb, Sm. f-orbitals are core-like, so bonding is largely ionic and f–f transitions give sharp, line-like colours. JEE Main almost every year asks the cause or a consequence of lanthanoid contraction (Zr–Hf similarity is the classic) and the stable oxidation states of Ce⁴⁺, Eu²⁺, Yb²⁺.

---

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Electronic Configuration Anomaly

4f orbitals fill after 6s and before 5d. The pattern [Xe]4fⁿ6s² is followed, but exceptions exist: La (4f⁰), Ce (4f¹5d¹6s²), Gd (4f⁷5d¹6s²), Lu (4f¹⁴5d¹6s²) — each gains extra stability from empty, half-filled, or fully filled f-subshells. For actinoids, 5f, 6d, and 7s energies are close, so 5f–6d occupancy is irregular (Th, Pa, U, Np each have 6d¹).

Lanthanoid Contraction

Across the 4f series, nuclear charge rises by 1 per element, but each added 4f electron shields the next one very poorly (shielding constant s ≈ 1 for f-electrons). Net Z_eff therefore climbs steadily, pulling the electron cloud inward. The ionic radius falls from ~103 pm (La³⁺) to ~86 pm (Lu³⁺), and the atomic radius contracts similarly.

Consequences Table

Consequence	Reason	Example
Zr ≈ Hf radii	Ln contraction makes 5d elements match 4d	Zr⁴⁺ 72 pm, Hf⁴⁺ 71 pm
Hard separation of Ln³⁺	Radii differ by < 1.5 pm/element	Ion-exchange needed for Nd/Pr
Decreasing basicity Ln(OH)₃	Smaller Ln³⁺ → stronger M–OH bond	La(OH)₃ most basic, Lu(OH)₃ least
Tighter packing in metals	Smaller radii raise density	Lu is the densest lanthanoid

Oxidation States and Redox Potentials

The +3 state is universal because the 6s² pair and one 4f/5d electron are lost easily. Deviations:

• Ce⁴⁺ (4f⁰) is a strong oxidant, E°(Ce⁴⁺/Ce³⁺) ≈ +1.61 V.
• Eu²⁺, Yb²⁺, Sm²⁺ (half/full f-shell stability) are reducing, E°(Eu³⁺/Eu²⁺) ≈ –0.35 V. Actinoids show wider oxidation states (+3 to +7) because 5f orbitals participate in bonding more than 4f.

Magnetic and Colour Properties

Ln³⁺ moments obey spin–orbit coupling, μ_eff = √[4J(J+1)] B.M. — J is obtained by L–S coupling, not the spin-only formula used for d-block. f–f transitions are Laporte-forbidden but visible because f-orbitals are shielded, producing sharp, line-like spectra unlike the broad bands of transition metals.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Conceptual Subtleties

f-orbital penetration is small: 4f has no radial node inside the 6s/5p/5s cloud, so 4f electrons sit inside the xenon core and behave like inner-core electrons. This is why lanthanoid chemistry is dominated by ionic bonding and +3, whereas 5f orbitals in early actinoids (Th–Pu) extend further and allow covalency (e.g., uranyl UO₂²⁺). After Am, 5f collapses inward and the later actinoids behave lanthanoid-like — the basis of the “actinide concept” abandoned after Cm.

Worked Comparison: Zr vs Hf

Zr ([Kr]4d²5s²) and Hf ([Xe]4f¹⁴5d²6s²) sit in Group 4. Without lanthanoid contraction, Hf would be much larger than Zr because of the added 18 protons and 32 electrons across two periods. Instead, the intervening lanthanoid contraction pulls Hf’s effective radius down, giving Zr⁴⁺ = 72 pm and Hf⁴⁺ = 71 pm. Consequence: nearly identical chemistry, very hard mutual separation, and a characteristic JEE trap question — “Why are Zr and Hf difficult to separate?” — answered by similar radii, not by lanthanoid properties of Zr/Hf themselves.

Common Mistakes to Avoid

• Writing Eu as +3 only — Eu²⁺ is the more stable reduced form in aqueous solution.
• Claiming Gd has 4f⁸ — Gd is 4f⁷5d¹6s², half-filled f-shell.
• Confusing actinoid contraction with lanthanoid contraction: both occur, but actinoid contraction is less uniform because 5f electrons shield more variably.
• Using the spin-only formula μ = √[n(n+2)] for Ln³⁺ — invalid; use the J-based formula.

JEE-Main Strategy

This topic carries ~3% weight, typically 1 question. High-yield items: (i) configuration of Gd and Lu, (ii) cause/effect of lanthanoid contraction, (iii) stable oxidation state of Ce/Eu/Yb, (iv) reason Zr ≈ Hf, (v) why f-block elements are coloured with sharp lines. Memorise the three +2 lanthanoids (Eu, Yb, Sm) and the single +4 lanthanoid (Ce) — these alone answer ~70% of MCQs on oxidation states.

Practice Prompts

1. “Arrange La³⁺, Gd³⁺, Lu³⁺ in order of ionic radius and justify using lanthanoid contraction.” — Answer: Lu³⁺ < Gd³⁺ < La³⁺; monotonic Z_eff increase.
2. “E°(Ce⁴⁺/Ce³⁺) is more positive than E°(Eu³⁺/Eu²⁺). What does this say about their redox behaviour?” — Answer: Ce⁴⁺ is a stronger oxidant than Eu³⁺, while Eu²⁺ is a strong reductant; both arise from the drive toward the stable 4f⁰ (Ce⁴⁺) and 4f⁷ (Eu²⁺) configurations.

---

Content adapted based on your selected roadmap duration. Switch tiers using the selector above.