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F Block Lanthanoids Actinoids

🎓 Class 12 Chemistry CBSE Theory Ch 4 – The d- and f-Block Elements ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: F Block Lanthanoids Actinoids

આ મૂલ્યાંકન આના પર આધારિત હશે: F Block Lanthanoids Actinoids

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F Block Lanthanoids Actinoids

The Hidden Block Beneath the Periodic Table

If you look at the bottom of the periodic table you will see two long strips, separated from the rest. These are the f-block or inner transition elements — fourteen lanthanoids (4f, Ce → Lu) and fourteen actinoids (5f, Th → Lr). They earn their place outside the main grid because their valence electrons enter the deeply buried 4f or 5f orbitals.

Roadmap: Lanthanoids — electronic configurations, oxidation states, lanthanoid contraction and its consequences, general characteristics. Actinoids — electronic configurations, ionic sizes, oxidation states, and a side-by-side comparison with lanthanoids.

8.17 The Lanthanoids (4f Series)

8.17.1 Electronic Configurations

All atoms of the lanthanoid series share an outer 6s² shell. The number of 4f electrons varies progressively from 0 (in La itself) up to 14 (in Lu). The general configuration is:

[Xe] 4f^x 5d^(0 or 1) 6s² (x runs 0–14)

The 5d¹ slot is occupied for a few special members (La, Ce, Gd, Lu) but most lanthanoids place all electrons in 4f and leave 5d empty. The most stable ions of all lanthanoids are Ln³⁺, which have the simple form 4f^n (n = 0 to 14):

ZElementSymbolAtom configLn³⁺ configAtomic r / pmLn³⁺ r / pm
57LanthanumLa5d¹6s²4f⁰187106
58CeriumCe4f¹5d¹6s²4f¹183103
59PraseodymiumPr4f³6s²4f²182101
60NeodymiumNd4f⁴6s²4f³18199
61PromethiumPm4f⁵6s²4f⁴18198
62SamariumSm4f⁶6s²4f⁵18096
63EuropiumEu4f⁷6s²4f⁶19995
64GadoliniumGd4f⁷5d¹6s²4f⁷18094
65TerbiumTb4f⁹6s²4f⁸17892
66DysprosiumDy4f¹⁰6s²4f⁹17791
67HolmiumHo4f¹¹6s²4f¹⁰17689
68ErbiumEr4f¹²6s²4f¹¹17588
69ThuliumTm4f¹³6s²4f¹²17487
70YtterbiumYb4f¹⁴6s²4f¹³17386
71LutetiumLu4f¹⁴5d¹6s²4f¹⁴

Eu has anomalously large atomic radius (199 pm) because its 4f⁷ shell is half-filled — extra-stable, reluctant to delocalise into bonding.

8.17.2 Atomic and Ionic Sizes — The Lanthanoid Contraction

Look at the trend of Ln³⁺ radius: from La³⁺ (106 pm) to Lu³⁺ (~85 pm) the size shrinks by about 21 pm — a steady decrease across the entire series. This is the famous lanthanoid contraction.

Cause. As we go from Ce to Lu, each successive electron enters the 4f sub-shell. The 4f orbitals have an unusual angular shape — they are rather diffuse and shield other 4f electrons very poorly from the nucleus. As nuclear charge increases by one each step, the 4f electrons feel almost the full extra pull, and so the entire 4f sub-shell (and the surrounding 5s5p shells) contracts.

85 90 95 100 105 Ln³⁺ radius / pm Atomic number Z (La=57 ... Lu=71) La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Steady decrease ≈ 21 pm from La³⁺ to Lu³⁺
Fig 8.9: Lanthanoid contraction — ionic radii of Ln³⁺ ions decrease almost linearly from La to Lu.

Consequences of the Lanthanoid Contraction

  1. 4d ≈ 5d sizes. The third transition series elements (Hf, Ta, W, Re...) are nearly the same size as the corresponding second series (Zr, Nb, Mo, Tc...). Famous case: Zr (160 pm) and Hf (159 pm) have almost identical radii — they occur together in nature (in zircon ore) and are notoriously difficult to separate.
  2. Increasing covalent character of M–O bonds from La to Lu — smaller, more polarising Ln³⁺.
  3. Increasing basicity of hydroxides decreases from La(OH)₃ (most basic) to Lu(OH)₃ (least basic).
  4. Identical chemical behaviour within the lanthanoid series, making mixed-rare-earth chemistry routine but mutual separation extremely difficult.

8.17.3 Oxidation States

Almost universally, lanthanoids show +3 as their characteristic oxidation state. A few elements show +2 or +4 in addition, driven by the stability of empty (f⁰), half-filled (f⁷) or fully filled (f¹⁴) configurations:

  • +4 states: Ce⁴⁺ (4f⁰, noble-gas configuration — the most important; used as the analytical oxidant Ce(SO₄)₂). Pr⁴⁺, Nd⁴⁺, Tb⁴⁺ (4f⁷ — half-filled, oxidising), Dy⁴⁺ — but only in the solid state (oxides MO₂).
  • +2 states: Eu²⁺ (4f⁷ — half-filled), Yb²⁺ (4f¹⁴ — full), Sm²⁺ (4f⁶ — close to half-filled). Eu²⁺ is a strong reducing agent, reverting to Eu³⁺.

Cerium(IV)/Cerium(III) has E° = +1.74 V — high enough to oxidise water, but the reaction is so slow that Ce(IV) is a usable analytical reagent.

8.17.4 General Characteristics & Chemical Reactions

All lanthanoids are silvery-white, soft metals that tarnish rapidly in air. Hardness rises across the series (Sm is steel-hard). Melting points lie between 1000 K and 1200 K. The chemistry is broadly comparable to the alkaline-earth metals at the start of the series and to aluminium by the end:

2 Ln + 3 H₂O → 2 Ln(OH)₃ + 3 H₂↑ (slow, faster on heating) 2 Ln + 6 H⁺ → 2 Ln³⁺ + 3 H₂↑ (with dilute acids) 2 Ln + 3 X₂ → 2 LnX₃ (X = halogen) 4 Ln + 3 O₂ → 2 Ln₂O₃ (burn in O₂) 2 Ln + N₂ → 2 LnN (heat with N₂) Ln + S → Ln₂S₃ (heat with S) Ln + C → LnC₂ / Ln₃C / Ln₂C₃ (carbides) Ln + H₂ → LnH₂ / LnH₃ (combine with H₂ on warming)
Ln Ln₂O₃ burns in O₂ LnN heated with N₂ LnH₂ / LnH₃ + H₂, warm LnX₃ + halogens Ln₂S₃ heat with S Ln(OH)₃ + H₂ + H₂O / dilute acid
Fig 8.10: Summary of the chemical reactions of a generic lanthanoid metal Ln.

Colours and magnetism. Most Ln³⁺ ions (except 4f⁰: La³⁺, Ce⁴⁺ and 4f¹⁴: Yb²⁺, Lu³⁺) are coloured because of f-f transitions; the absorption bands are very narrow (the 4f orbitals are well-shielded so ligand-field effects are small). All ions other than f⁰ and f¹⁴ are paramagnetic.

8.17.5 Uses of the Lanthanoids

  • Mischmetall (≈95% Ln + 5% Fe + traces of S, C, Ca, Al) — used in lighter flints, bullets, magnesium-based alloys.
  • Ln-oxide catalysts in petroleum cracking.
  • Individual Ln-oxides as phosphors in colour TV / monitor screens (Eu³⁺ red, Tb³⁺ green).
  • Nd-Fe-B alloy for the world's strongest permanent magnets.

8.18 The Actinoids (5f Series)

8.18.1 Members and Radioactivity

The actinoids run from Ac (Z = 89) to Lr (Z = 103). All are radioactive. Earlier members (Th, U) have long half-lives and occur in nature; later members (Es, Fm, Md, No, Lr) have half-lives ranging from days to a few minutes — they exist only as nanogram quantities synthesised in nuclear reactors or particle accelerators. This makes their chemistry both difficult to study and dangerous to handle.

ZElementSymbolAtom configM³⁺ configM³⁺ r / pm
89ActiniumAc6d¹7s²5f⁰111
90ThoriumTh6d²7s²5f¹
91ProtactiniumPa5f²6d¹7s²5f²
92UraniumU5f³6d¹7s²5f³103
93NeptuniumNp5f⁴6d¹7s²5f⁴101
94PlutoniumPu5f⁶7s²5f⁵100
95AmericiumAm5f⁷7s²5f⁶99
96CuriumCm5f⁷6d¹7s²5f⁷99
97BerkeliumBk5f⁹7s²5f⁸98
98CaliforniumCf5f¹⁰7s²5f⁹98
99EinsteiniumEs5f¹¹7s²5f¹⁰
100FermiumFm5f¹²7s²5f¹¹
101MendeleviumMd5f¹³7s²5f¹²
102NobeliumNo5f¹⁴7s²5f¹³
103LawrenciumLr5f¹⁴6d¹7s²5f¹⁴

8.18.2 Electronic Configurations & Ionic Sizes

All actinoids carry 7s² with variable occupation of 5f and 6d. The 14 added electrons formally enter the 5f sub-shell, but irregularities (e.g. Am [Rn]5f⁷7s², Cm [Rn]5f⁷6d¹7s²) reflect the extra stability of f⁰, f⁷ and f¹⁴ occupancies.

An actinoid contraction exists analogous to the lanthanoid contraction, but is greater per element — because 5f electrons shield the nuclear charge even more poorly than 4f electrons. The 5f orbitals are not as deeply buried as the 4f, so 5f electrons can participate in bonding to a far greater extent.

8.18.3 Oxidation States

Although +3 is common to all actinoids, the early members exhibit a remarkable range. The maximum oxidation state climbs with Z up to Np (+7), then falls again:

ElementAcThPaUNpPuAmCmBkCfEs+
Oxidation states33,43,4,53,4,5,63,4,5,6,73,4,5,6,73,4,5,63,43,43,43

Why so many states? The 5f, 6d and 7s sub-shells of the actinoids lie close in energy, so different numbers of electrons can be lost without much extra cost. By contrast, the lanthanoids have 4f buried much more deeply below 5d/6s, so only the outer electrons participate — limiting them to the +3 state with rare exceptions.

8.18.4 General Characteristics & Comparison with Lanthanoids

FeatureLanthanoids (4f)Actinoids (5f)
Common oxidation stateAlmost exclusively +3+3 common; many also +4, +5, +6, +7
Burying of f-orbitals4f deeply buried — barely participate in bonding5f more exposed — participate in bonding more
Contraction per elementSmaller (lanthanoid contraction)Larger (actinoid contraction; 5f shields worse)
Colour intensityPale; very narrow f-f bandsStronger; broader bands due to 5f-ligand mixing
Magnetic behaviourRoughly follows √[J(J+1)] — orbital + spinComplex; lower magnetic moment than f-equivalent Ln
RadioactivityOnly Pm radioactiveAll radioactive
Ionisation enthalpies (early)HigherLower (5f electrons less tightly held)
ReactivitySoft, silvery, react with H₂O slowlyHighly reactive when finely divided

Interactive: Lanthanoid / Actinoid Configuration & Magnetism Tool L3 Apply

Pick any f-block element. The tool returns its symbol, ground-state configuration, common oxidation state, and the spin-only magnetic moment of its M³⁺ ion (treating only the unpaired f-electrons, ignoring orbital contribution).

Pick an f-block element to begin.
Activity 8.4 — Why Are Zr and Hf So Hard to Separate?L4 Analyse

Aim: Connect the lanthanoid contraction to a practical industrial-chemistry problem — the separation of zirconium from hafnium.

Procedure:

  1. List the atomic radii of Zr (Group 4, Period 5) and Hf (Group 4, Period 6).
  2. Without lanthanoid contraction, what radius would you predict for Hf simply from "going down a group"?
  3. Explain (i) why ion-exchange chromatography or solvent extraction are needed to separate them, (ii) why their oxides ZrO₂ and HfO₂ behave almost identically in chemical reactions.

Predict: How does the lanthanoid contraction make Hf chemistry "look like" Zr chemistry?

Zr radius = 160 pm, Hf radius = 159 pm. Going from Period 5 to Period 6 should add a whole shell, predicting Hf ~175 pm. Instead, the 14 lanthanoids slipped in between, each contracting the size by ~1.5 pm; the cumulative shrinkage almost exactly cancels the period-down increase.

Result: Zr⁴⁺ and Hf⁴⁺ have nearly identical charge density, so they form chemically indistinguishable salts (ionic potential, hydrolysis behaviour, complexes). Standard chemical methods cannot tell them apart. Industrial separation requires liquid–liquid extraction with TBP/HNO₃, or repeated ion-exchange chromatography on a cation-exchange column — tedious and expensive (one reason hafnium is rarer in commerce than zirconium).

Worked Example 8.11 L3 Apply

Q. Use Hund's rule to derive the electronic configuration of Ce³⁺ ion and calculate its spin-only magnetic moment.

Ce (Z = 58) atom = [Xe] 4f¹ 5d¹ 6s². Removing 3 electrons (6s², 5d¹ first): Ce³⁺ = [Xe] 4f¹.

One unpaired f-electron → n = 1 → μ = √(1·3) = 1.73 BM. (Note: experimental moments include orbital angular momentum and are often higher.)

Worked Example 8.12 L4 Analyse

Q (In-text 4.10). The actinoid contraction is greater per element than the lanthanoid contraction. Why?

The 5f orbitals shield the outer electrons from the rising nuclear charge even less effectively than the 4f orbitals. As Z rises step by step across the actinoids, the outer 7s/6d electrons feel a sharper increase in effective nuclear charge, so they are pulled in more strongly per element than in the lanthanoid series.

Worked Example 8.13 L2 Understand

Q. Name a lanthanoid that exhibits the +4 oxidation state and one that exhibits the +2 state.

+4: Cerium — Ce⁴⁺ has 4f⁰ (noble-gas-like). +2: Europium — Eu²⁺ has 4f⁷ (half-filled). Both states owe their existence to the special stability of f⁰, f⁷ or f¹⁴ configurations.

Competency-Based Questions L3 L4

Stimulus: A geochemist analyses a "rare-earth" mineral and finds that nearly all the rare-earth metals appear in the +3 oxidation state, while a separate sample of pitchblende (a uranium ore) shows U in +4 and +6 states. She also notes that Hf concentrations are persistently 1–3% of Zr concentrations across her samples.

Q1. (MCQ) Which of the following ions has the smallest radius?

  • (a) La³⁺
  • (b) Eu³⁺
  • (c) Tb³⁺
  • (d) Lu³⁺
(d) Lu³⁺ ≈ 85 pm — the smallest because of cumulative lanthanoid contraction.

Q2. (MCQ) Which of the following atomic numbers correspond to an inner transition element?

  • (a) 29
  • (b) 59
  • (c) 74
  • (d) 104
(b) 59 (Pr — a lanthanoid). Z = 29 (Cu) and 74 (W) are d-block; Z = 104 (Rf) is also d-block (transactinide). 59 is a 4f-block element.

Q3. (SA) Why do Zr and Hf show almost identical chemical behaviour?

Lanthanoid contraction reduces the expected size increase from Period 5 to Period 6. Hf (159 pm) ends up almost the same size as Zr (160 pm). Same Group 4 → same valence (4+); same size → same ionic potential and lattice/complex chemistry → essentially identical reactions.

Q4. (LA) List three differences between the chemistry of the actinoids and that of the lanthanoids.

(i) Oxidation states — actinoids show a wide range (+3 to +7); lanthanoids almost exclusively +3. (ii) Bonding by f-electrons — 5f orbitals are larger and less buried, so they participate in bonding (covalent character, organometallic compounds); 4f electrons stay localised. (iii) Radioactivity — every actinoid is radioactive; only Pm is radioactive among the lanthanoids. Many actinoids exist only in nanogram quantities.

Q5. (HOT) Why is Eu²⁺ unusually stable compared with the +2 ions of its neighbours?

Eu has the configuration [Xe] 4f⁷ 6s². Losing only the two 6s electrons gives Eu²⁺ = 4f⁷, a half-filled f-shell with maximum exchange energy. The third ionisation step would have to break this stability, costing extra energy — so Eu²⁺ persists where most other Ln²⁺ disappear quickly.

Assertion–Reason Questions L4 L5

Choose: A) Both A and R true and R explains A · B) Both true but R does not explain A · C) A true, R false · D) A false, R true.

Assertion (A): The +3 oxidation state is the most common across the entire lanthanoid series.

Reason (R): The 4f electrons are deeply buried below 5d and 6s, so only the two 6s and the one 5d electron participate in bonding.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Removal of the three outer (6s² + 5d¹ or one 4f) electrons gives the +3 state in nearly every lanthanoid.

Assertion (A): Actinoid contraction is greater than lanthanoid contraction per element.

Reason (R): The 5f electrons provide poorer shielding than the 4f electrons.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Poorer 5f shielding means the outer electrons feel a stronger pull as Z rises across the actinoid series.

Assertion (A): Hf and Zr cannot be separated by simple chemical methods.

Reason (R): Their atomic and ionic radii are nearly identical due to the lanthanoid contraction that occurs between them in the periodic table.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Identical sizes give identical charge density and chemistry — separation requires ion exchange or solvent extraction.

Frequently Asked Questions - F Block Lanthanoids Actinoids

What is the main concept covered in F Block Lanthanoids Actinoids?
In NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements), "F Block Lanthanoids Actinoids" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is F Block Lanthanoids Actinoids useful in real-life or applied chemistry?
Real-life applications of "F Block Lanthanoids Actinoids" from NCERT Class 12 Chemistry Chapter 4 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for F Block Lanthanoids Actinoids?
Key reactions in "F Block Lanthanoids Actinoids" (NCERT Class 12 Chemistry Chapter 4 The d- and f-Block Elements) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 4?
NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements) is structured so each part builds chemical understanding sequentially. "F Block Lanthanoids Actinoids" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from F Block Lanthanoids Actinoids?
CBSE board questions from "F Block Lanthanoids Actinoids" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "F Block Lanthanoids Actinoids" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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Chemistry Class 12 Part I – NCERT (2025-26)
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