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Chem notesInorganicd/f-Block

Independent topic page

d- and f-Block Elements

Derive transition behaviour from d-electron count and master Cr/Mn chemistry through oxidation-state maps.

Portion InorganicFocus Reliable oxidation-state, colour, magnetism, and oxoanion questionsTarget 45–75sPortion share 23.7%Trend Stable
Transition-metal configurations, oxidation states, E° trends, catalysis and alloys; chromate/dichromate and permanganate; lanthanoid contraction and f-block oxidation states. E° comparisons and catalysis stems still start from OS and bonding — trends without half-cell pairing are incomplete.

How Advanced asks this

  • Find d count, unpaired electrons, colour/magnetic character.
  • Compare oxidation-state stability using configuration and E° data.
  • Balance dichromate/permanganate oxidation in acid, neutral, or base.
  • Explain catalysis, alloying, interstitial compounds, and atomisation trends.
  • Apply lanthanoid contraction and common Ln/An oxidation states.

Traps that leak marks

  • Counting 4s electrons in a cation.
  • Assuming every d-block ion is coloured.
  • Using acidic permanganate product in neutral medium.
  • Reading E° as an intrinsic reaction direction without pairing half-cells.
  • Confusing lanthanoid contraction with actinoid contraction.
  • Counting 4s electrons in a cation configuration
  • Assuming every d-block ion is coloured
  • Using acidic permanganate products in the wrong medium

01. A transition element has an incomplete d subshell in its atom or common cation; Zn-group exceptions matter.

02. Variable oxidation states arise because ns and (n−1)d energies are close; colour and magnetism follow d count and ligand environment.

03. Cr(VI) and Mn(VII) oxoanions are medium-sensitive redox systems, best handled by half-reactions.

04. Poor 4f shielding causes lanthanoid contraction, controlling radii, separation difficulty, and 4d/5d similarity.

05. Write configurations of atoms and ions carefully: ns electrons leave before (n−1)d when forming common cations.

06. Colour, magnetism and catalytic behaviour track incomplete d shells and accessible oxidation states — d⁰/d¹⁰ are usually colourless and diamagnetic for simple ions.

07. Lanthanoid contraction is the size story that joins Zr/Hf similarity and f-block oxidation-state patterns to periodic trends.

Official · 2026

Syllabus for this topic

Official syllabus points for d- and f-Block Elements only — not the full Inorganic portion map. Portion hub: Inorganic · Inorganic analysis.

Syllabus on this page is topic-scoped: d- and f-Block Elements. The parent portion (Inorganic) also covers other topics — use the portion hub for the full map.

Must cover (this topic)

  • d-block: OS stability, E°, interstitial, alloys, catalysis; Cr/Mn oxoanions
  • f-block: lanthanoid/actinoid contraction; OS; general characteristics

Gaps / watch-outs

No topic-specific exclusions flagged — finish the unit points below for d/f-Block.

Source: https://jeeadv.ac.in/documents/jee-advanced-2026-syllabus.pdf

Unit 01

d- and f-Block Elements

high

Exam focusd/f-block trends, oxidation states, alloys, and lanthanoid/actinoid notes.

Syllabus points

  • d-block: OS stability, E°, interstitial, alloys, catalysis; Cr/Mn oxoanions
  • f-block: lanthanoid/actinoid contraction; OS; general characteristics

Full Chemistry syllabus →

Cross-topic stems

Mixed joins for this topic

Advanced rarely asks a chapter in isolation. These are tagged multi-idea Chemistry stems that touch d- and f-Block Elements — practise the knowledge bridge, not only the single-chapter drill.

No tagged mixed stem matched this topic’s keywords yet. Still revise the joins listed under Trends, and scan the mixed joins board for Physical / Inorganic / Organic bridges.

All mixed joins →

Why methods matter

Common methods that change your score

Oxidation state and d-electron count predict colour, magnetism and redox behavior together.

01

Remove ns first

Gives correct transition-ion configurations.

Do thisOn ionisation, remove 4s electrons before 3d.

Why it works

Gives correct transition-ion configurations.

See it once

Fe²⁺ is [Ar]3d⁶, not 3d⁴4s².

02

d-count audit

d⁰/d¹⁰ ions are usually colourless; unpaired count drives μ.

Do thisConvert every oxidation state to dⁿ before judging colour/magnetism.

Why it works

d⁰/d¹⁰ ions are usually colourless; unpaired count drives μ.

See it once

Ti⁴⁺ is d⁰ and diamagnetic.

03

Potential direction test

Prevents memorised redox-strength reversals.

Do thisWrite both half-reactions as reductions, then use E°cell.

Why it works

Prevents memorised redox-strength reversals.

See it once

A positive E°cell identifies the spontaneous direction.

04

Medium locks manganese product

Mn product changes with medium.

Do thisCircle acidic/neutral/strongly alkaline before balancing MnO₄⁻.

Why it works

Mn product changes with medium.

See it once

Acid gives Mn²⁺; neutral gives MnO₂.

Processed examples

Sample questions · full working

Read the impact line first. Then notice which family/model was chosen. The solution is written the way your rough sheet should look in the hall.

01

bankFamily · Colour and magnetism

Sample question

Find the spin-only moment of high-spin Fe³⁺.

This stem rewards early recognition of the colour and magnetism model.

Model chosenDetermine d count and unpaired electrons.

Processed solution

  1. Fe³⁺ is 3d⁵.
  2. High-spin d⁵ has five unpaired electrons.
  3. μ=√[5(5+2)] BM.
  4. Thus μ=√35≈5.92 BM.

Answer5.92 BM

Takeaway: Oxidation state must precede magnetic calculation.

02

standardFamily · Permanganate or dichromate titration

Sample question

How many Fe²⁺ ions are oxidized per MnO₄⁻ in acidic medium?

This stem rewards early recognition of the permanganate or dichromate titration model.

Model chosenCompare half-reaction electron counts.

Processed solution

  1. MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O.
  2. Fe²⁺→Fe³⁺+e⁻.
  3. Five iron half-reactions supply five electrons.
  4. The ratio Fe²⁺:MnO₄⁻ is 5:1.

Answer5

Takeaway: The medium fixes permanganate’s electron demand.

03

twistFamily · Oxidation-state stability

Sample question

Using E°(Cu²⁺/Cu⁺)=0.15 V and E°(Cu⁺/Cu)=0.52 V, determine whether Cu⁺ disproportionates under standard conditions.

This Advanced-level variation couples hidden constraints with the oxidation-state stability model.

Model chosenCombine reduction of one Cu⁺ with oxidation of another.

Processed solution

  1. Cathodic half-reaction is Cu⁺+e⁻→Cu with E°=0.52 V.
  2. Anodic process is Cu⁺→Cu²⁺+e⁻, opposite to E°=0.15 V.
  3. E°cell=0.52−0.15=0.37 V.
  4. The positive potential makes 2Cu⁺→Cu²⁺+Cu spontaneous.
  5. Hence Cu⁺ disproportionates.

AnswerYes; E°=+0.37 V

Takeaway: Compare adjacent couples to test intermediate oxidation-state stability.

Reusable models

Common question families

Advanced stems cluster. Name the family in 10–15s, write the fast model, then run the same path. Do not invent a new plan each time.

01

Oxidation-state stability

Looks likeMost stable state or disproportionation of transition ions

Fast modelCombine electronic configuration with medium and ligand stabilization.

Same solve path

  1. Write d-electron counts.
  2. Inspect half/full-shell effects.
  3. Include oxidizing/reducing medium.
  4. Predict stable states.

Rough skeleton

  • d counts
  • medium
  • stable state

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

02

Colour and magnetism

Looks likeObserved colour or spin-only moment of an ion

Fast modelFind d count first; d⁰/d¹⁰ are usually colourless.

Same solve path

  1. Determine oxidation state.
  2. Write dⁿ configuration.
  3. Count unpaired electrons.
  4. Infer colour possibility and μ.

Rough skeleton

  • O.S.
  • dⁿ
  • unpaired/colour

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

03

Permanganate or dichromate titration

Looks likeRedox equivalent in acidic/basic medium

Fast modelWrite the correct medium-dependent reduction product.

Same solve path

  1. Select Mn/Cr product for the medium.
  2. Write half-reactions.
  3. Equalize electrons.
  4. Apply stoichiometry.

Rough skeleton

  • medium
  • half-reactions
  • equivalents

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

04

Lanthanoid contraction

Looks likeSize, basicity or separation comparison across 4f series

Fast modelUse poor 4f shielding as the common cause.

Same solve path

  1. Identify properties controlled by radius.
  2. Apply decreasing Ln³⁺ radius.
  3. Include similar Zr/Hf sizes if relevant.
  4. State the trend.

Rough skeleton

  • 4f shielding
  • radius trend
  • consequence

Why faster

Recognition fixes the model before arithmetic begins.

Refuse this

Do not start blind expansion or casework before identifying the governing model.

Hall timing

45–75s process

  1. 0–12s
    Classify the stem and underline the species, condition, and demanded comparison.

    Decide whether the controlling idea is trend, structure, equilibrium, redox, or a named preparation/test.

  2. 12–25s
    Write the governing configuration, equation, or exception before inspecting options.

    Fix oxidation state, charge balance, medium, and periodic position; do not answer from a vague trend.

  3. 25–60s
    Apply the shortest valid model and eliminate statements one clause at a time.

    For multi-correct questions, independently verify every species and every condition.

  4. 60–75s
    Audit charge, atoms, trend direction, colour/state, and syllabus exceptions.

    Reject conclusions that silently change medium, coordination number, temperature, or oxidation state.

After the model is locked

Quick formula bank

d count

dⁿ = group number − oxidation state (first-row guide)

Magnetism and electronic configuration.

Spin-only moment

μ = √[n(n+2)] BM

n is number of unpaired electrons.

Dichromate reduction

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Acidic dichromate oxidation.

Permanganate acidic

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Strongly acidic medium.

Permanganate neutral

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Neutral/weakly alkaline medium.

Chromate equilibrium

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O

Yellow chromate/orange dichromate interconversion.

Mental bank

  • 4s electrons leave before 3d.
  • Zn, Cd, Hg are not transition elements by the strict d-shell definition.
  • d⁰/d¹⁰ generally colourless and diamagnetic.
  • CrO₄²⁻ yellow; Cr₂O₇²⁻ orange; MnO₄⁻ purple.
  • Transition metals catalyse via variable OS/adsorption.
  • Lanthanoids mainly +3; Ce(IV), Eu(II), Yb(II) are notable.
  • Lanthanoid contraction makes Zr/Hf sizes similar.
  • Zn, Cd, Hg not transition by strict d-shell definition
  • CrO₄²⁻ yellow; Cr₂O₇²⁻ orange; MnO₄⁻ purple
  • Spin-only μ = √[n(n+2)] BM
  • Lanthanoids mainly +3; Ce(IV), Eu(II), Yb(II) notable

Make it automatic

Practice · checklist

Daily

  • Revise one d/f-block configuration and redox map from memory.
  • Solve 6 mixed single/multiple-correct statements.
  • Balance or justify two reactions without notes.

Weekly

  • Attempt one 25-minute d/f-block configuration and redox set.
  • Rebuild the exception/colour/reagent sheet.
  • Redo every error after hiding the answer.

100% checklist

  • Write neutral and ionic configurations correctly.
  • Calculate spin-only moments.
  • Balance Cr(VI)/Mn(VII) redox in each medium.
  • Interpret E° comparisons.
  • Explain catalysis/alloys/interstitial compounds.
  • Apply lanthanoid contraction and f-block OS trends.
  • Neutral and ionic configurations correct
  • Medium-correct Cr(VI)/Mn(VII) redox
  • Lanthanoid contraction applied where sizes converge
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