f-Block Elements — Lanthanoids and Actinoids
D and F Block Elements: f-Block Elements — Lanthanoids and Actinoids
What you'll learn
- What lanthanoids and actinoids are, and where they are placed in the periodic table
- The f-orbital filling pattern for both series
- What lanthanoid contraction is, its cause, and its far-reaching effects
- How to compare lanthanoids and actinoids (oxidation states, radioactivity, magnetism)
- Why actinoids show more variable oxidation states than lanthanoids
- Uses of lanthanoids and actinoids in modern technology
Level 1 Foundations
What are f-Block Elements?
f-Block elements are those in which the last electron enters an f-orbital:
| Series | Elements | f-orbital filling | Period |
|---|---|---|---|
| Lanthanoids | Ce (58) to Lu (71) — 14 elements | 4f (1 to 14) | 6th period |
| Actinoids | Th (90) to Lr (103) — 14 elements | 5f (1 to 14) | 7th period |
La (Z=57) and Ac (Z=89) are often included in discussions but have empty f-orbitals.
Lanthanoids — 4f Series
Lanthanoids: La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu
Memory phrase: "La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu"
General electronic configuration: [Xe] 4f^(1–14) 5d^(0–1) 6s²
- Most lanthanoids have 4f electrons filling with 5d⁰ or 5d¹
- Notable exceptions: La ([Xe]5d¹6s²), Ce ([Xe]4f¹5d¹6s²), Gd ([Xe]4f⁷5d¹6s² — half-filled 4f)
Common oxidation state: +3 in all lanthanoids (most stable)
- Ce also shows +4 (CeO₂ — oxidising agent); Eu and Sm show +2
Actinoids — 5f Series
Actinoids: Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr
General electronic configuration: [Rn] 5f^(0–14) 6d^(0–2) 7s²
- All actinoids are radioactive — many are artificially produced transuranium elements
- Th, Pa, U, and Np occur naturally in significant amounts; elements beyond Np (Z=93) are synthetic
Common oxidation states: +3, +4, +5, +6 (actinoids show a wider range than lanthanoids)
- U: +3, +4, +5, +6 (UO₂²⁺ — uranyl ion is the most stable)
- Np, Pu also show +7
Lanthanoid Contraction
Definition: The steady decrease in ionic radii of lanthanoid ions (La³⁺ to Lu³⁺) from 106 pm to 86 pm as the atomic number increases.
Cause:
- As 4f electrons are added across the series (Ce to Lu), nuclear charge increases by +1 each step
- 4f orbitals have poor shielding efficiency — they have a complex radial distribution and shield other electrons (and each other) poorly from the increasing nuclear charge
- The effective nuclear charge experienced by outer electrons increases more than expected → electrons are drawn inward → ionic radius decreases
Magnitude: ~1–2 pm decrease per element across the lanthanoid series (total contraction ~20 pm)
Effects of Lanthanoid Contraction
-
Similar radii of 5d elements (periods 5 and 6): The lanthanoid contraction in the 4f series before the 5d series (period 6) compresses the 5d period 6 atoms so that they have almost identical radii to their period 5 counterparts (4d elements).
4d element Radius 5d element Radius Zr (40) 160 pm Hf (72) 159 pm Nb (41) 146 pm Ta (73) 146 pm Mo (42) 139 pm W (74) 139 pm This similarity in size makes pairs like Zr/Hf and Nb/Ta chemically almost identical — very difficult to separate.
-
Basicity decreases across lanthanoid series:
- Smaller ionic size → greater polarising power of Ln³⁺ → weaker base
- La(OH)₃ is the strongest base; Lu(OH)₃ is the weakest base
-
Separation of lanthanoids is difficult — their similar sizes give nearly identical properties.
Actinoids vs Lanthanoids — Comparison Table
| Property | Lanthanoids | Actinoids |
|---|---|---|
| f-orbital | 4f (filling) | 5f (filling) |
| Radioactivity | Non-radioactive (except Pm) | All radioactive |
| Common OS | +3 | +3, +4, +5, +6 (wider range) |
| Magnetic behaviour | Paramagnetic (unpaired 4f) | Strongly paramagnetic |
| Orbital involvement in bonding | 4f rarely involved in bonding | 5f can participate in bonding |
| Complexes | Form fewer complex types | Form more diverse complexes |
| Origin | Mostly natural | Many synthetic (transuranic) |
| Shielding of f-orbitals | Poor (4f) | Even poorer (5f — closer in energy to d) |
Why actinoids show more oxidation states:
- 5f, 6d, and 7s orbitals have similar energies in actinoids — all can participate in bonding
- In lanthanoids, the 4f orbitals are deeply buried and not available for bonding; only 5d and 6s contribute
- The energy difference between 5f and 6d is smaller than between 4f and 5d
Uses of Lanthanoids and Actinoids
Lanthanoids:
- Rare-earth magnets (Nd₂Fe₁₄B): Strongest permanent magnets — used in EV motors, headphones, hard drives
- Phosphors: Eu³⁺ (red), Tb³⁺ (green) in LED screens and fluorescent lamps
- Misch metal (Ce, La, Nd alloy): Used in lighter flints, pyrophoric alloys
- CeO₂: Polishing agent for glass; in catalytic converters
- Lanthanoid-doped glass: Used in lasers (Nd:YAG laser for eye surgery)
Actinoids:
- U-235: Nuclear fission fuel in nuclear power plants (fissile material)
- Pu-239: Used in nuclear weapons and as reactor fuel
- Th-232: Fertile material in thorium reactors (India's 3-stage nuclear program)
- Am-241: Used in smoke detectors (α-emitter)
Level 2 JEE Depth
Why 5f Orbitals Are More Available for Bonding Than 4f
In lanthanoids, 4f orbitals are contracted and buried close to the nucleus (deeply core-like):
- They don't participate significantly in chemical bonding
- Crystal field effects on 4f are small
- Magnetic properties are governed by isolated 4f electrons
In actinoids, 5f orbitals are more diffuse and have higher radial extent:
- Energy gap between 5f, 6d, and 7s is small → orbital mixing → more oxidation states
- Actinoid complexes show stronger crystal field effects than lanthanoid complexes
- 5f participation in covalent bonding is possible (especially in UF₆, uranyl ion)
Lanthanoid Contraction — Quantitative JEE Perspective
The total ionic radius decrease from La³⁺ (106.1 pm) to Lu³⁺ (86.1 pm) = 20 pm over 14 elements.
This is due to incomplete shielding — each additional 4f electron adds charge shielding of less than +1 unit (shielding constant for 4f-4f ~ 0.35 per electron, much less than inner-core shielding).
For JEE: The question "Why are Zr and Hf almost identical in size?" has the one-line answer: lanthanoid contraction.
Oxidation State Stability Comparison
Lanthanoids:
- +3 is universally stable (loss of 6s² and one 4f/5d electron)
- Ce(+4): stable because Ce⁴⁺ achieves [Xe] configuration (empty 4f) — extra stability
- Eu(+2): Eu²⁺ has [Xe]4f⁷ — half-filled 4f, extra stable
- Yb(+2): Yb²⁺ has [Xe]4f¹⁴ — completely filled 4f, extra stable
Actinoids:
- Pa(+5), U(+6), Np(+7) — accessible because 5f, 6d, 7s are all close in energy
- Higher OS stabilised by strongly oxidising conditions and electronegative ligands (F, O)
- Stability of +3 increases going from Pa → Lr (similar to lanthanoids towards the end of series)
Colour of f-Block Ions
Both lanthanoid and actinoid ions are often coloured:
- Colour arises from f-f transitions (within 4f or 5f energy levels) and also charge-transfer transitions
- f-f transitions are Laporte-forbidden but weakly observed due to vibronic coupling
- Actinoid colours are more intense than lanthanoid colours due to greater f-orbital involvement
Worked Examples
Example 1: Lanthanoid Contraction Explains Zr/Hf Separation Difficulty
Problem: Why are Zr and Hf (period 5 and period 6 of group 4) almost
identical in chemical properties and very difficult to separate?
Answer:
Before the 5d series in period 6, the 14 lanthanoid elements
(Ce to Lu, filling 4f¹ to 4f¹⁴) cause the lanthanoid contraction.
Each lanthanoid adds a 4f electron with poor shielding → increasing
effective nuclear charge → decreasing atomic/ionic radius.
Total contraction ≈ 20 pm across 14 lanthanoids.
This cancels the expected increase in size from period 5 → period 6.
Result:
- Zr (period 5, group 4): atomic radius = 160 pm
- Hf (period 6, group 4): atomic radius = 159 pm ← nearly identical
Same size → same charge-to-radius ratio → nearly identical chemistry
→ Zr and Hf always occur together in nature and are extremely difficult
to separate (requires ion exchange chromatography or solvent extraction).
Example 2: Why Actinoids Show More Oxidation States Than Lanthanoids
Problem: Explain why U shows +4, +5, +6 oxidation states while the
corresponding lanthanoid Nd shows mainly +3.
Comparison:
Nd (Z=60): [Xe] 4f⁴ 6s²
Configuration of 4f is DEEPLY buried in the core
Energy of 4f >> energy of 5d/6s for bonding purposes
→ Only 5d and 6s electrons available → max OS usually +3
U (Z=92): [Rn] 5f³ 6d¹ 7s²
5f, 6d, and 7s have similar energies in early actinoids
All these electrons can potentially be involved in bonding
→ U can lose 3, 4, 5, or 6 electrons (OS +3 to +6)
→ UF₆ (U in +6) and UO₂²⁺ (uranyl ion, U in +6) are stable
Conclusion: 5f orbital energy similarity to 6d/7s in actinoids
enables wider range of oxidation states.
Common Mistakes
| Mistake | Why it's wrong | Correct approach |
|---|---|---|
| Thinking lanthanoid contraction increases size of 5d elements | It decreases the size — the contraction compresses 5d period 6 elements | Lanthanoid contraction makes 5d period 6 elements SMALLER than expected, matching period 5 sizes |
| Saying all lanthanoids are radioactive | Only Promethium (Pm, Z=61) is radioactive; all others are stable | All actinoids are radioactive; lanthanoids are mostly stable |
| Stating lanthanoids show the same wide OS as actinoids | 4f is too buried for bonding; lanthanoids mainly show +3 | Only Ce(+4), Eu(+2), Sm(+2), Yb(+2) are notable exceptions |
| Confusing f-f colour with d-d colour | In d-block, colour is from d-d transitions; in f-block, colour is from f-f transitions (and charge transfer) — both are possible in complexes | Always specify which transition gives colour when answering about specific elements |
Quick Check
- How many elements are in the lanthanoid series? Name the first and last.
- What is the cause of lanthanoid contraction?
- Give one consequence of lanthanoid contraction on 5d transition metals.
- Why are all actinoids radioactive but most lanthanoids are not?
- (Stretch) Ce shows a +4 oxidation state and Eu shows a +2 oxidation state. Using the concept of extra stability of empty, half-filled, and completely filled f-subshells, explain why each shows this extra oxidation state.
NCERT Link & Exam Connections
- NCERT Class 12, Chapter 8, Sections 8.5 (Lanthanoids) and 8.6 (Actinoids)
- Table 8.6 and 8.7 (NCERT) — lanthanoid and actinoid properties
- JEE Main: Direct MCQs on lanthanoid contraction definition, effects, actinoid vs lanthanoid OS comparison
- JEE Advanced: Explanation-type questions on why 5d properties resemble 4d, why actinoids show variable OS
- NEET: Definition of lanthanoid contraction, uses (nuclear reactors, magnets, phosphors)
Study strategy: Draw a flowchart: Lanthanoid contraction (cause: poor 4f shielding) → effects (smaller Ln³⁺ radii, similar 4d/5d radii, harder to separate pairs like Zr/Hf, decrease in basicity). The comparison table between lanthanoids and actinoids is high-yield — memorise it as a table.
Practice in Drishti
Work through the f-Block Elements question sets in Chemistry D-Block chapter. Start with Easy (definitions, uses), then Medium (lanthanoid contraction effects, OS comparison), then Hard (JEE-level electronic configuration and OS reasoning).
Ask Drishti AI
Ask: "Can you show me a step-by-step explanation of why Zr and Hf have nearly the same atomic radius, using lanthanoid contraction?" for a visual diagram with numbers.
Track Your Progress
Complete all 5 Quick Check questions. If Q5 is unclear, revisit the "Oxidation State Stability" section under Level 2. Track your f-block mastery score in Drishti before moving to revision.
Next Steps
- Revise: Full D-Block chapter summary (transition metals + properties + f-block)
- Practice: Mixed JEE-level MCQs across all three d-block notes (Hard difficulty)
- Next chapter: Coordination Compounds (ligands, IUPAC naming, isomerism, crystal field theory)
Key Takeaways (TL;DR)
- What you'll learn
- Level 1 Foundations
- Level 2 JEE Depth
- Worked Examples
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