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State Board · Class 12 · Chemistry

Lesson 8: Transition and Inner Transition Elements

7 concepts · 56 practice questions with answers · Free to read

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Electronic configuration of d-block elements

From daily life

Iron, copper, gold and silver — everyday metals are mostly transition elements.

Simple definition

A transition element has an incomplete d-subshell in the atom or an ion; general configuration (n–1)d¹⁻¹⁰ ns¹⁻².

ConfigurationFirst transition series (3d): configurationSc3d¹4s²Ti3d²4s²V3d³4s²Cr3d⁵4s¹Mn3d⁵4s²Fe3d⁶4s²Co3d⁷4s²Ni3d⁸4s²Cu3d¹⁰4s¹Zn3d¹⁰4s²Exceptions: Cr (3d⁵4s¹), Cu (3d¹⁰4s¹) — half/full d stabilityTransition element: incomplete d (in atom or ion); Zn, Cd, Hg are not4s electrons leave first: Fe²⁺ = 3d⁶; Mn²⁺ = 3d⁵

In the figure : The 3d series.

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Electronic configuration of d-block elements

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Rule 1: General configuration — Sc 3d¹4s² to Zn 3d¹⁰4s²; 4s fills first but leaves first during ionisation.

  • Easy : Fe: [Ar]3d⁶4s².
  • Medium : Fe²⁺: [Ar]3d⁶.
  • Exam-type : Mn²⁺: [Ar]3d⁵.

Rule 2: Exceptions — Cr: 3d⁵4s¹ and Cu: 3d¹⁰4s¹ — half/full d is more stable (exchange energy).

  • Easy : Cr ≠ 3d⁴4s².
  • Medium : Cu ≠ 3d⁹4s².
  • Exam-type : Pd: 4d¹⁰5s⁰.

Rule 3: Who is a transition element? — Zn, Cd, Hg have d¹⁰ even in ions ⇒ not transition; Sc counts because the atom has 3d¹.

  • Easy : Cu²⁺ 3d⁹ ⇒ transition.
  • Medium : Zn²⁺ 3d¹⁰ ⇒ not.
  • Exam-type : d-block ≠ transition.

Reasoning: why Cr is 3d⁵4s¹

  • Step 1: The 3d and 4s energies are close.
  • Step 2: Half-filled 3d⁵ has high exchange energy.
  • Step 3: One 4s electron moves to 3d, giving 3d⁵4s¹.

CET level (MCQ)

  • CET 1) Configuration of Cr? 1) [Ar]3d⁵4s¹ 2) [Ar]3d⁴4s² 3) [Ar]3d⁶ 4) [Ar]3d⁵4s² — Answer: 1).
  • CET 2) Not a transition element? 1) Zn 2) Fe 3) Cu 4) Ni — Answer: 1).

Real-life use: Coins, wires, catalysts, jewellery.

Remember

Cr 3d⁵4s¹; Cu 3d¹⁰4s¹; 4s leaves first; Zn is not transition.

Common mistake

Assuming 3d electrons leave first.

CET trap

4s leaves first ⇒ Fe²⁺ = 3d⁶.

Shortcut

Memory clue: "Cr and Cu are exceptions: half and full".

Spot the difference

ElementConfiguration
Cr3d⁵4s¹
Mn3d⁵4s²
Fe3d⁶4s²
Cu3d¹⁰4s¹

Practice questions

Easy
  1. 1Define transition element.
  2. 2Configuration of Cr.
  3. 3Configuration of Cu.
Medium
  1. 4Why is Zn not transition?
  2. 5Configuration of Fe²⁺.
  3. 6Configuration of Mn²⁺.
Challenging
  1. 7Name the 3d series.
  2. 8Why exceptions?
Show answers (try them yourself first)
  1. 1Incomplete d.
  2. 23d⁵4s¹.
  3. 33d¹⁰4s¹.
  4. 4Full d¹⁰.
  5. 53d⁶.
  6. 63d⁵.
  7. 7Sc–Zn.
  8. 8Half/full stability.

Oxidation states of the first transition series

From daily life

MnO₂ in a battery and green rust on copper — one metal, many states.

Simple definition

Many states because (n–1)d and ns electrons are close in energy; the maximum equals (3d + 4s) electrons (Mn = +7).

OxidationOxidation states (large dot = common)+1+2+3+4+5+6+7ScTiVCrMnFeCoNiCuZnMn: +2 to +7 (widest); maximum = number of (3d + 4s) electronsIn d-block states differ by 1 (by 2 in p-block)

In the figure : Oxidation states.

Rule 1: Range of states — From Sc +3 up to Mn +7, then falls; Mn: +2 to +7; steps of 1.

  • Easy : Ti: +4 common.
  • Medium : V: +5 (V₂O₅).
  • Exam-type : Cr: +3, +6.

Rule 2: Stable states — Mn²⁺ (3d⁵), Fe³⁺ (3d⁵), Cu²⁺ are common; high states oxidise (MnO₄⁻, Cr₂O₇²⁻), low states reduce.

  • Easy : Fe²⁺ ⇒ Fe³⁺ is easy.
  • Medium : Cu⁺ disproportionates in water.
  • Exam-type : Zn is only +2.

Rule 3: Compared with p-block — d-block states differ by 1, p-block by 2; higher states become more stable down a group.

  • Easy : Mn: 2, 3, 4, 6, 7.
  • Medium : Cl: 1, 3, 5, 7.
  • Exam-type : Os: +8.

Calculation: Cr in K₂Cr₂O₇ and Mn in KMnO₄

  • Step 1: K₂Cr₂O₇: 2(+1) + 2x + 7(–2) = 0.
  • Step 2: 2x = 12 ⇒ x = +6.
  • Step 3: KMnO₄: +1 + x – 8 = 0 ⇒ x = +7.

CET level (MCQ)

  • CET 1) Maximum oxidation state of Mn? 1) +7 2) +6 3) +5 4) +4 — Answer: 1).
  • CET 2) Cr in K₂Cr₂O₇? 1) +6 2) +3 3) +7 4) +2 — Answer: 1).

Real-life use: Batteries, catalysts, pigments.

Remember

Mn has the widest range (+2 to +7); maximum = (3d + 4s) electrons; Zn only +2.

Common mistake

Taking +8 as Fe's maximum.

CET trap

Fe: +2 and +3 are common (+6 is rare).

Shortcut

Memory clue: "Mn reaches seven".

Spot the difference

ElementCommon states
Cr+3, +6
Mn+2, +4, +7
Fe+2, +3
Cu+1, +2

Practice questions

Easy
  1. 1Why many states?
  2. 2States of Mn.
  3. 3Maximum state rule.
Medium
  1. 4Zn only +2.
  2. 5Cr in K₂Cr₂O₇.
  3. 6Mn in KMnO₄.
Challenging
  1. 7d vs p block.
  2. 8Behaviour of Cu⁺.
Show answers (try them yourself first)
  1. 1d and s close.
  2. 2+2 to +7.
  3. 3(3d + 4s) electrons.
  4. 4Full d¹⁰.
  5. 5+6.
  6. 6+7.
  7. 71 vs 2.
  8. 8Disproportionates.

Atomic radii, ionisation enthalpy and melting points

From daily life

A tungsten filament survives about 3400 °C — d-bonding at its best.

Simple definition

Trends: across 3d, radius falls slightly, IE rises slowly, m.p. rises to Cr.

TrendsFirst ionisation enthalpy (kJ/mol), Sc to Zn631Sc656Ti650V653Cr717Mn762Fe758Co736Ni745Cu906ZnIE rises slowly (similar Zeff); Zn highest (full 3d¹⁰4s²)Radii: fall Sc to Cr, near-constant Fe–Ni, rise at Cu–Znm.p. rises to Cr (more d–d bonding), dips at Mn, lowest at Zn (693 K)

In the figure : IE trend.

Rule 1: Atomic radii — Falls Sc to Cr, near-constant Fe–Ni, rises at Cu–Zn (3d repulsion); 4d and 5d radii are almost equal.

  • Easy : 4d ≈ 5d because of lanthanoid contraction.
  • Medium : Fe, Co, Ni have similar radii.
  • Exam-type : Zn is larger.

Rule 2: Ionisation enthalpy — First IE: 631 (Sc) to 906 kJ/mol (Zn), a slow rise as 3d adds little shielding; Zn is highest (full shell).

  • Easy : Mn: 717.
  • Medium : Zn: 906.
  • Exam-type : Cr has a high second IE.

Rule 3: Melting point and hardness — More unpaired d electrons ⇒ stronger metallic bonds ⇒ higher m.p.; rises to Cr, dips at Mn (stable 3d⁵), Zn lowest (693 K).

  • Easy : W has the highest m.p.
  • Medium : Hg is liquid.
  • Exam-type : Zn, Cd, Hg are low.

Reasoning: why Zn melts at a low temperature

  • Step 1: Zn is 3d¹⁰4s² — all d electrons paired.
  • Step 2: d electrons do not join metallic bonding.
  • Step 3: Weaker bonding ⇒ low m.p. (693 K).

CET level (MCQ)

  • CET 1) Highest IE? 1) Zn 2) Fe 3) Cu 4) Sc — Answer: 1).
  • CET 2) Liquid metal? 1) Hg 2) Zn 3) Cd 4) Cu — Answer: 1).

Real-life use: Filaments, cutting tools, thermometers.

Remember

IE rises slowly; Zn is highest; m.p. rises to Cr; Zn is lowest.

Common mistake

Assuming Mn melts higher than Cr.

CET trap

Mn dips (stable 3d⁵).

Shortcut

Memory clue: "more unpaired electrons ⇒ stronger bond".

Spot the difference

PropertyTrend
Radiusfalls, then rises
IErises slowly
m.p.↑ to Cr

Practice questions

Easy
  1. 1Radius trend.
  2. 2IE trend.
  3. 3Why highest IE for Zn?
Medium
  1. 4Why m.p. rises?
  2. 5m.p. of Zn.
  3. 6Why 4d ≈ 5d?
Challenging
  1. 7Highest m.p. metal.
  2. 8Dip at Mn.
Show answers (try them yourself first)
  1. 1Falls to Cr.
  2. 2Slow rise.
  3. 3Full shell.
  4. 4Unpaired d electrons.
  5. 5693 K.
  6. 6Lanthanoid contraction.
  7. 7W.
  8. 8Stable 3d⁵.

Magnetism, colour, catalysis, interstitial compounds and alloys

From daily life

Blue copper sulfate, fridge magnets and industrial catalysts — all d-block properties.

Simple definition

Magnetic moment μ = √[n(n+2)] BM (n = unpaired electrons); colour arises from d–d transitions.

MagnetismMagnetism and colourIonde⁻ nμ (BM)ColourCu²⁺3d⁹11.73blueNi²⁺3d⁸22.83greenCr³⁺3d³33.87greenFe²⁺3d⁶44.90greenMn²⁺3d⁵55.92pale pinkZn²⁺3d¹⁰00colourlessμ = √[n(n + 2)] BMColour: d–d transitions; d⁰ and d¹⁰ are colourless (Sc³⁺, Zn²⁺)Catalysts: Fe (Haber), V₂O₅ (contact), Ni (hydrogenation)Interstitial: TiC, Fe₃H; alloys: steel, brass (Cu + Zn)

In the figure : Magnetism and colour.

Rule 1: Magnetism — Unpaired electrons ⇒ paramagnetic; all paired ⇒ diamagnetic; μ: n = 1 ⇒ 1.73; 2 ⇒ 2.83; 3 ⇒ 3.87; 4 ⇒ 4.90; 5 ⇒ 5.92 BM.

  • Easy : Cu²⁺ (3d⁹) ⇒ 1.73.
  • Medium : Mn²⁺ (3d⁵) ⇒ 5.92.
  • Exam-type : Zn²⁺ ⇒ 0.

Rule 2: Colour — Incomplete d shell absorbs visible light for d–d jumps and the complementary colour is seen; d⁰ and d¹⁰ are colourless.

  • Easy : Cu²⁺ blue.
  • Medium : Ni²⁺ green.
  • Exam-type : Zn²⁺, Sc³⁺ colourless.

Rule 3: Catalysis, interstitial compounds, alloys — Catalysts: Fe (Haber), V₂O₅ (contact), Ni (hydrogenation); interstitial: TiC, Fe₃H (hard, conducting); alloys: brass (Cu + Zn), steel.

  • Easy : Variable states + surface ⇒ catalysis.
  • Medium : Interstitial ⇒ hard, dense.
  • Exam-type : Similar radii ⇒ alloys.

Calculation: μ of Fe³⁺ (3d⁵)

  • Step 1: 3d⁵ ⇒ n = 5.
  • Step 2: μ = √[5 × 7] = √35.
  • Step 3: μ = 5.92 BM.

CET level (MCQ)

  • CET 1) μ of Ni²⁺ (BM)? 1) 2.83 2) 1.73 3) 3.87 4) 0 — Answer: 1).
  • CET 2) Colourless ion? 1) Zn²⁺ 2) Cu²⁺ 3) Ni²⁺ 4) Fe²⁺ — Answer: 1).

Real-life use: Catalysts, magnets, pigments, alloys.

Remember

μ = √[n(n+2)]; d–d colour; d⁰/d¹⁰ colourless; Fe for Haber, V₂O₅ for contact.

Common mistake

Thinking Zn²⁺ is coloured.

CET trap

d¹⁰ ⇒ no d–d jump possible.

Shortcut

Memory clue: "n = 1…5 ⇒ 1.73, 2.83, 3.87, 4.90, 5.92".

Spot the difference

nμ (BM)
11.73
22.83
33.87
44.90
55.92

Practice questions

Easy
  1. 1Formula for μ.
  2. 2μ of Cu²⁺.
  3. 3μ of Mn²⁺.
Medium
  1. 4Why colour?
  2. 5Colourless ions.
  3. 6Three catalysts.
Challenging
  1. 7Interstitial compounds.
  2. 8Brass.
Show answers (try them yourself first)
  1. 1√[n(n+2)].
  2. 21.73.
  3. 35.92.
  4. 4d–d transition.
  5. 5Zn²⁺, Sc³⁺.
  6. 6Fe, V₂O₅, Ni.
  7. 7TiC, Fe₃H.
  8. 8Cu + Zn.

Potassium permanganate and potassium dichromate

From daily life

Purple KMnO₄ for rinsing wounds and Cr compounds in tanning — both transition compounds.

Simple definition

Strong oxidants: MnO₄⁻ (Mn +7) and Cr₂O₇²⁻ (Cr +6).

PermanganateKMnO₄ and K₂Cr₂O₇KMnO₄: deep purple; tetrahedral MnO₄⁻1) 3MnO₂ + 6KOH + KClO₃ → 3K₂MnO₄ (green) + KCl + 3H₂O2) CO₂ / acid: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂OAcid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OMnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂OSelf-indicator (pale pink end point)K₂Cr₂O₇: orange; Cr₂O₇²⁻ ⇌ 2CrO₄²⁻ (with pH)4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂Acid: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂OCr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂ONeeds an indicator (diphenylamine)Ex. 20 mL of 0.02 M KMnO₄ ⇒ 2 mmol Fe²⁺ (5 × 0.4)0.49 g K₂Cr₂O₇ (M = 294) ⇒ 10 mmol Fe²⁺ ⇒ 0.56 g Fe

In the figure : KMnO₄ and K₂Cr₂O₇.

Live figure

Potassium permanganate and potassium dichromate

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Live figure

Potassium permanganate and potassium dichromate

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Rule 1: KMnO₄ — MnO₂ + KOH + KClO₃ (fused) ⇒ K₂MnO₄ (green); CO₂ or acid: 3MnO₄²⁻ + 4H⁺ ⇒ 2MnO₄⁻ + MnO₂ + 2H₂O; purple, tetrahedral.

  • Easy : Acidic: Mn²⁺ (pale pink).
  • Medium : Neutral/alkaline: MnO₂.
  • Exam-type : Self-indicator.

Rule 2: K₂Cr₂O₇ — Chromite ore ⇒ Na₂CrO₄ ⇒ acid ⇒ Na₂Cr₂O₇ ⇒ KCl ⇒ K₂Cr₂O₇; pH: CrO₄²⁻ (yellow) ⇌ Cr₂O₇²⁻ (orange).

  • Easy : 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O.
  • Medium : Cr³⁺ is green.
  • Exam-type : External indicator needed.

Rule 3: Titration — MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O; Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

  • Easy : MnO₄⁻: 5 e⁻.
  • Medium : Cr₂O₇²⁻: 6 e⁻.
  • Exam-type : KMnO₄ is used in H₂SO₄, not HCl.

Calculation: 20 mL of 0.02 M KMnO₄ titrates 25 mL of Fe²⁺ solution

  • Step 1: n(MnO₄⁻) = 20 × 0.02 = 0.4 mmol.
  • Step 2: n(Fe²⁺) = 5 × 0.4 = 2 mmol.
  • Step 3: M = 2/25 = 0.08 M.

CET level (MCQ)

  • CET 1) Moles of Fe²⁺ oxidised by 1 mol MnO₄⁻ (acidic)? 1) 5 2) 6 3) 3 4) 1 — Answer: 1).
  • CET 2) Product when Cr₂O₇²⁻ is reduced in acid? 1) Cr³⁺ 2) Cr²⁺ 3) CrO₄²⁻ 4) Cr — Answer: 1).

Real-life use: Disinfection, titration, leather industry.

Remember

MnO₄⁻: 1 ⇒ 5 Fe²⁺; Cr₂O₇²⁻: 1 ⇒ 6 Fe²⁺; KMnO₄ is a self-indicator.

Common mistake

Titrating KMnO₄ in HCl.

CET trap

HCl is oxidised to Cl₂, so use H₂SO₄.

Shortcut

Memory clue: "permanganate 5, dichromate 6".

Spot the difference

Ione⁻Fe²⁺
MnO₄⁻55
Cr₂O₇²⁻66

Practice questions

Easy
  1. 1Preparation of KMnO₄.
  2. 2Preparation of K₂Cr₂O₇.
  3. 3Effect of pH.
Medium
  1. 4KMnO₄ reduction in acid.
  2. 5K₂Cr₂O₇ reduction in acid.
  3. 6Fe²⁺ M from 20 mL 0.02 M.
Challenging
  1. 7Fe from 0.49 g K₂Cr₂O₇.
  2. 8What is a self-indicator?
Show answers (try them yourself first)
  1. 1Fuse MnO₂ + KOH + KClO₃.
  2. 2Chromite + Na₂CO₃.
  3. 3Yellow ⇌ orange.
  4. 4Mn²⁺.
  5. 5Cr³⁺.
  6. 60.08 M.
  7. 70.56 g.
  8. 8Changes colour itself.

Metallurgy and extraction of iron

From daily life

Bridges, rails and utensils — iron (steel) is everywhere, extracted from its ore.

Simple definition

Reduction: Fe₂O₃ + 3CO → 2Fe + 3CO₂ in a blast furnace with coke and CaCO₃ giving slag.

FurnaceExtraction of iron: blast furnaceore + coke + CaCO₃ ↓≈ 500 Kore loses its moisture≈ 900 KFe₂O₃ + 3CO → 2Fe + 3CO₂≈ 1200 KCaCO₃ → CaO + CO₂CaO + SiO₂ → CaSiO₃ (slag)≈ 1500 KFe₂O₃ + 3C → 2Fe + 3CO≈ 2000 KC + ½O₂ → COΔH = –220 kJCO rises ↑airSlag floats on top; pig iron (≈ 4 % C) at the bottom160 kg Fe₂O₃ ⇒ 112 kg Fecast, wrought iron, steel

In the figure : Blast furnace.

Rule 1: Metallurgy steps — Ore ⇒ concentration ⇒ roasting/calcination ⇒ reduction ⇒ refining; iron ore: haematite Fe₂O₃.

  • Easy : Calcination: without air.
  • Medium : Roasting: in air.
  • Exam-type : Coke = reductant + fuel.

Rule 2: Reactions in the blast furnace — Combustion zone (≈ 2000 K): C + ½O₂ → CO (ΔH = –220 kJ); reduction zone (≈ 900 K): Fe₂O₃ + 3CO → 2Fe + 3CO₂; slag zone (≈ 1200 K): CaCO₃ → CaO + CO₂, CaO + SiO₂ → CaSiO₃.

  • Easy : Light slag floats on top.
  • Medium : Hot air blast from below.
  • Exam-type : Product: pig iron (≈ 4 % C).

Rule 3: Forms of iron — Pig iron ≈ 4 % C; cast iron (≈ 4 % C, hard, brittle); wrought iron (< 0.2 % C, very soft); steel (0.2–2 % C).

  • Easy : Steel: some C + Cr, Ni.
  • Medium : Stainless: Cr, Ni.
  • Exam-type : More C ⇒ brittle.

Calculation: Fe (Fe = 56) from 160 kg of Fe₂O₃ (M = 160)

  • Step 1: n = 160 kg / 160 = 1 kmol (1000 mol).
  • Step 2: 1 mol Fe₂O₃ ⇒ 2 mol Fe ⇒ 2 kmol.
  • Step 3: 2 × 56 = 112 kg Fe.

CET level (MCQ)

  • CET 1) Slag in the blast furnace? 1) CaSiO₃ 2) CaCO₃ 3) CaO 4) SiO₂ — Answer: 1).
  • CET 2) Main ore of iron? 1) haematite 2) bauxite 3) galena 4) cinnabar — Answer: 1).

Real-life use: Construction, machines, vehicles.

Remember

Fe₂O₃ + CO ⇒ Fe; coke; CaCO₃ ⇒ CaSiO₃; pig iron ≈ 4 % C.

Common mistake

Treating coke as only fuel.

CET trap

Coke gives the reductant CO.

Shortcut

Memory clue: "coke burns, CO forms, iron comes out".

Spot the difference

TypeC %
Pig iron≈ 4
Wrought iron≈ 0.2
Steel0.1–1.5

Practice questions

Easy
  1. 1Parts of the blast furnace.
  2. 2Three main reactions.
  3. 3What is slag?
Medium
  1. 4Role of coke.
  2. 5Calcination vs roasting.
  3. 6Forms of iron.
Challenging
  1. 7Fe from 160 kg Fe₂O₃.
  2. 8What is steel?
Show answers (try them yourself first)
  1. 1Ore, coke, limestone.
  2. 2C → CO₂ → CO → Fe.
  3. 3CaSiO₃.
  4. 4Reductant + fuel.
  5. 5Without air / in air.
  6. 6Pig, cast, wrought, steel.
  7. 7112 kg.
  8. 8Fe + a little C.

Lanthanoids, actinoids and postactinoids

From daily life

Lanthanoids feature in lighter flints and magnets, actinoids in nuclear power.

Simple definition

Lanthanoids: Ce–Lu, 4f filling; actinoids: Th–Lr, 5f filling, radioactive.

LanthanoidLanthanoid contractionZ: 57 → 71Ln³⁺ radius (pm)La³⁺ 103Lu³⁺ 86Poor 4f shielding ⇒ Zeff ↑ ⇒ radius ↓ (≈ 17 pm)Lanthanoids (Ce–Lu)Actinoids (Th–Lr)4f5f+3 (Ce +4, Eu +2)+3 to +7stable (except Pm)radioactiveMischmetal ≈ 95 % Ln + Fe: lighter flints, Mg alloysPost-actinoids: Z = 104 to 118 (Rf, Db, Sg …)

In the figure : Lanthanoid contraction.

Rule 1: Lanthanoids — Configuration [Xe]4fⁿ5d⁰⁻¹6s²; main state +3 (Ce +4, Eu +2); coloured ions; mischmetal ≈ 95 % Ln + Fe.

  • Easy : Ce: 4f¹5d¹6s².
  • Medium : Lighter flint: mischmetal.
  • Exam-type : Colours vary La³⁺ → Lu³⁺.

Rule 2: Lanthanoid contraction — 4f electrons shield poorly ⇒ Z_eff rises ⇒ Ln³⁺ radius falls from La³⁺ 103 to Lu³⁺ 86 pm (≈ 17 pm); results: 4d ≈ 5d, Zr ≈ Hf.

  • Easy : Zr and Hf are hard to separate.
  • Medium : Lanthanoids are hard to separate.
  • Exam-type : Basicity of Ln(OH)₃ falls.

Rule 3: Actinoids and postactinoids — 5f filling; states +3 to +7; all radioactive; U and Pu are nuclear fuels; postactinoids Z > 103 (Rf, Db, Sg).

  • Easy : Th–Lr: 14 elements.
  • Medium : U: reactors.
  • Exam-type : Z > 103 are synthetic.

Reasoning: why lanthanoid contraction occurs

  • Step 1: From La to Lu, 4f fills and Z increases.
  • Step 2: 4f orbitals shield the nucleus poorly.
  • Step 3: Z_eff rises ⇒ radius shrinks about 17 pm.

CET level (MCQ)

  • CET 1) Lanthanoids fill? 1) 4f 2) 5f 3) 3d 4) 4d — Answer: 1).
  • CET 2) Lanthanoid contraction makes which pair similar in size? 1) Zr and Hf 2) Ti and Zr 3) Fe and Co 4) Na and K — Answer: 1).

Real-life use: Lighter flints, magnets, nuclear energy.

Remember

Ln: 4f, +3; contraction ⇒ Zr ≈ Hf; Ac: 5f, radioactive.

Common mistake

Blaming 5f for lanthanoid contraction.

CET trap

It is the poor shielding of 4f.

Shortcut

Memory clue: "4f ⇒ Ln; 5f ⇒ An".

Spot the difference

SeriesFilling
Lanthanoid4f
Actinoid5f

Practice questions

Easy
  1. 1What are lanthanoids?
  2. 2Lanthanoid contraction.
  3. 3Consequences of contraction.
Medium
  1. 4Feature of actinoids.
  2. 5Mischmetal.
  3. 6Main state of Ln.
Challenging
  1. 7What are postactinoids?
  2. 8Ln vs An.
Show answers (try them yourself first)
  1. 1Ce–Lu, 4f.
  2. 2Radius falls La → Lu.
  3. 3Zr ≈ Hf.
  4. 4Radioactive.
  5. 595 % Ln + Fe.
  6. 6+3.
  7. 7Z > 103.
  8. 84f vs 5f.

Quick revision of the whole lesson

ConceptKey formula / fact
Magnetic momentμ = √[n(n+2)] BM
Cr config.3d⁵4s¹
Cu config.3d¹⁰4s¹
MnO₄⁻ acidicMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Cr₂O₇²⁻ acidicCr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Iron reductionFe₂O₃ + 3CO → 2Fe + 3CO₂
SlagCaO + SiO₂ → CaSiO₃
Lanthanoids4f, +3
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🎯 Most important points for the exam

  • 3d configuration.
  • Oxidation states.
  • Trends.
  • Magnetism, colour.
  • KMnO₄, K₂Cr₂O₇.
  • Iron.
  • Ln, An.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.