By the end of this chapter you'll be able to…

  • 1Write electronic configurations of d- and f-block elements including Cr and Cu exceptions
  • 2Explain variable oxidation states, colour, and magnetic and catalytic properties
  • 3Describe the preparation and oxidising action of K2Cr2O7 and KMnO4
  • 4Explain lanthanoid contraction and its consequences
  • 5Compare lanthanoids and actinoids
💡
Why this chapter matters
Transition metals are chemistry's workhorses -- they catalyse reactions, form coloured compounds, and show variable oxidation states. Understanding the d- and f-block, lanthanoid contraction, and key oxidants like K2Cr2O7 and KMnO4 is essential for inorganic chemistry.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

d- and f-Block Elements

'Transition metals are the WORKHORSES of chemistry — they catalyse reactions, form coloured compounds, and display multiple oxidation states.'

1. Chapter Overview

The d-block elements (Groups 3 to 12) are known as TRANSITION METALS — they have incompletely filled d-orbitals in the ground state or common oxidation states. The f-block elements (LANTHANOIDS and ACTINOIDS) have incompletely filled f-orbitals. Topics include: ELECTRONIC CONFIGURATION of d- and f-block elements, PERIODIC TRENDS (atomic radii, ionisation enthalpy, oxidation states), MAGNETIC PROPERTIES, FORMATION OF COLOURED IONS, CATALYTIC PROPERTIES, the preparation and properties of POTASSIUM DICHROMATE (K₂Cr₂O₇) and POTASSIUM PERMANGANATE (KMnO₄), and the LANTHANOID CONTRACTION.


2. Electronic Configuration

d-Block

  • General: (n−1)d¹⁻¹⁰ ns¹⁻² (except for anomalies in Cr, Cu, etc.).
  • Anomalies: Cr (3d⁵4s¹ — HALF-FILLED d subshell), Cu (3d¹⁰4s¹ — FULLY FILLED d subshell). 'Half-filled and fully-filled d subshells provide EXTRA STABILITY.'

f-Block

  • Lanthanoids (4f): Ce to Lu (4f¹⁻¹⁴5d⁰⁻¹6s²).
  • Actinoids (5f): Th to Lr (5f⁰⁻¹⁴6d⁰⁻²7s²). All are RADIOACTIVE.

3. General Properties of Transition Metals

Physical Properties

  • HIGH melting and boiling points (due to strong metallic bonding via d-electrons).
  • HARD and DENSE.
  • GOOD conductors of heat and electricity.

Variable Oxidation States

  • 'Transition metals show MULTIPLE oxidation states because the (n−1)d and ns orbitals have SIMILAR energies — electrons can be removed from both.'
  • Examples: Mn (+2 to +7), Cr (+2 to +6), Fe (+2, +3).
  • Stability trends: Higher oxidation states are more stable with O, F (e.g., Mn₂O₇ exists, MnF₇ exists); lower oxidation states are more stable in aqueous solution.

Formation of Coloured Ions

  • 'The colour of transition metal ions is due to d−d TRANSITIONS — electrons absorb visible light and jump between d-orbitals.'
  • The colour depends on: (1) the metal, (2) its oxidation state, (3) the ligands attached.
  • Ti³⁺ (violet), Cu²⁺ (blue), Ni²⁺ (green), Fe³⁺ (yellow/brown).

4. Magnetic Properties

TypeBehaviourExamples
ParamagneticATTRACTED by magnetic field (unpaired electrons)Cu²⁺ (1 unpaired), Fe²⁺ (4 unpaired)
DiamagneticWEAKLY REPELLED (no unpaired electrons)Zn²⁺, Sc³⁺
FerromagneticSTRONGLY ATTRACTED (permanent magnetic moment)Fe, Co, Ni

5. Catalytic Properties

  • Transition metals are EXCELLENT catalysts because they:
    1. Have VARIABLE oxidation states — can accept and donate electrons easily.
    2. Provide SURFACE for adsorption of reactants.
  • Examples: Fe (Haber process — NH₃ synthesis). Ni (hydrogenation of oils). Pt (catalytic converters). V₂O₅ (Contact process — H₂SO₄ manufacture).

6. Important Compounds

Potassium Dichromate (K₂Cr₂O₇)

  • Preparation: Chromite ore (FeCr₂O₄) → Na₂CrO₄ → Na₂Cr₂O₇ → K₂Cr₂O₇.
  • Properties: ORANGE crystalline solid. STRONG OXIDISING AGENT in acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
  • Colour change: Orange (Cr₂O₇²⁻) → Green (Cr³⁺).
  • Tests: Oxidises Fe²⁺ to Fe³⁺, SO₂ to SO₄²⁻, I⁻ to I₂.

Potassium Permanganate (KMnO₄)

  • Preparation: Pyrolusite (MnO₂) → K₂MnO₄ → KMnO₄ (electrolytic oxidation).
  • Properties: DEEP PURPLE crystalline solid. STRONG OXIDISING AGENT.
  • Oxidation in different media:
MediumHalf-ReactionColour Change
AcidicMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OPurple → Colourless/Pale pink
NeutralMnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻Purple → Brown (MnO₂)
AlkalineMnO₄⁻ + e⁻ → MnO₄²⁻Purple → Green (manganate)

7. Lanthanoid Contraction

  • 'The STEADY DECREASE in atomic and ionic radii across the lanthanoid series (from La to Lu).'
  • Cause: The 4f electrons have POOR SHIELDING EFFECT — as nuclear charge increases, the outer electrons are pulled INWARDS.
  • Consequences:
    1. Similar chemical properties across the series.
    2. Difficult SEPARATION of lanthanoids.
    3. Post-lanthanoid elements (Zr, Hf; Nb, Ta) have ALMOST IDENTICAL radii — extremely similar properties.

Comparison: Lanthanoids vs Actinoids

PropertyLanthanoids (4f)Actinoids (5f)
Electronic configuration4f¹⁻¹⁴5d⁰⁻¹6s²5f⁰⁻¹⁴6d⁰⁻²7s²
RadioactivityMostly non-radioactiveALL radioactive
Oxidation statesMainly +3 (some +2, +4)+3 to +7 (more VARIABLE)
ColourMany have characteristic coloursColoured
SeparationVery DIFFICULT (similar properties)Slightly easier (more variation)

8. Common Mistakes

  1. Cr and Cu electronic configuration exceptions: Cr is NOT 3d⁴4s². It is 3d⁵4s¹ (half-filled stability). Cu is 3d¹⁰4s¹ (fully filled d).
  2. Oxidation states: Not ALL transition metals show the same range. Sc only shows +3. Zn only shows +2.
  3. Colour in transition metals: Colour is due to d-d transitions — requires at LEAST ONE unpaired d-electron. Sc³⁺ and Zn²⁺ are COLOURLESS (d⁰ and d¹⁰).
  4. KMnO₄ titrations: In ACIDIC medium, MnO₄⁻ is REDUCED to Mn²⁺ (pale pink, almost colourless). In NEUTRAL, it forms MnO₂ (brown precipitate).

9. CBSE Exam Focus

  1. Electronic configuration of d-block elements — general and exceptions (Cr, Cu)
  2. General properties — oxidation states, colour, magnetic, catalytic
  3. K₂Cr₂O₇ — preparation, properties as oxidising agent
  4. KMnO₄ — preparation, oxidising action in acidic/neutral/alkaline media
  5. Lanthanoid contraction — cause and consequences
  6. Comparison of lanthanoids and actinoids

10. Self-Test

Q1: Write the electronic configuration of Cr (Z=24) and Cu (Z=29). A1: Cr: [Ar]3d⁵4s¹ (NOT 3d⁴4s²). Cu: [Ar]3d¹⁰4s¹ (NOT 3d⁹4s²).

Q2: Why are transition metal ions coloured? A2: Due to d-d transitions involving unpaired d-electrons. Electrons in lower d-orbitals absorb visible light and get excited to higher d-orbitals. The COMPLEMENTARY colour is transmitted/reflected, giving the ion its colour.

Q3: Balance the following in acidic medium: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺. A3: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺.

Q4: Why is Zn not considered a transition metal? A4: Zn has electronic configuration [Ar]3d¹⁰4s². Its d-orbitals are FULLY FILLED (d¹⁰) in the ground state and common oxidation state (+2). It does NOT show variable oxidation states or form coloured ions. 'A transition metal MUST have incompletely filled d-orbitals.'

Q5: What is lanthanoid contraction? Write two consequences. A5: The gradual DECREASE in atomic/ionic radii from La to Lu due to poor shielding by 4f electrons. Consequences: (1) Lanthanoids have VERY SIMILAR chemical properties — difficult to separate. (2) Post-lanthanoid pairs (Zr/Hf, Nb/Ta) have ALMOST IDENTICAL atomic radii.


11. Conclusion

Transition metals are the MOST VERSATILE elements in the periodic table:

  • ELECTRONS: 'The d-orbitals give them variable oxidation states, colour, and magnetic properties.'
  • CATALYSIS: 'They can accept and donate electrons with ease — the hallmark of a good catalyst.'
  • COMPOUNDS: 'K₂Cr₂O₇ and KMnO₄ are POWERFUL oxidising agents — workhorses of analytical chemistry.'
  • f-BLOCK: 'Lanthanoids and actinoids — the inner transition series — with unique properties of their own.'

'd- and f-block elements are the colour, magnetism, and catalytic POWER of the periodic table.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

d-block general configuration
(n-1)d^1-10 ns^1-2
Cr = 3d5 4s1 and Cu = 3d10 4s1 (extra stability).
Dichromate oxidation (acidic)
Cr2O7 2- + 14H+ + 6e- -> 2Cr3+ + 7H2O
Orange to green colour change.
Permanganate oxidation (acidic)
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Purple to colourless; n = 5 in acidic medium.
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Writing Cr as 3d4 4s2 and Cu as 3d9 4s2
Cr is 3d5 4s1 and Cu is 3d10 4s1, due to the stability of half-filled and fully-filled d subshells.
WATCH OUT
Assuming all transition metal ions are coloured
Colour needs unpaired d-electrons (d-d transitions); Sc3+ (d0) and Zn2+ (d10) are colourless.
WATCH OUT
Calling Zn a transition metal
Zn has a full d10 configuration in its ground and +2 states, so it is not a true transition metal.
WATCH OUT
Using the wrong n for KMnO4 in different media
In acidic medium MnO4- gains 5 electrons (to Mn2+), in neutral 3 (to MnO2), and in alkaline 1 (to MnO4 2-).

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Configuration
Write the electronic configurations of Cr (Z = 24) and Cu (Z = 29).
Show solution
Cr: [Ar] 3d5 4s1 (not 3d4 4s2). Cu: [Ar] 3d10 4s1 (not 3d9 4s2).
Q2MEDIUM· Colour
Why are transition metal ions coloured?
Show solution
Because of d-d transitions: unpaired d-electrons absorb part of visible light and jump to higher d-orbitals; the complementary colour is transmitted, giving the ion its colour.
Q3MEDIUM· Balancing
Balance in acidic medium: Cr2O7 2- + Fe2+ -> Cr3+ + Fe3+.
Show solution
Cr2O7 2- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+.
Q4MEDIUM· Reasoning
Why is Zn not considered a transition metal?
Show solution
Zn is [Ar] 3d10 4s2; its d-orbitals are completely filled in the ground state and in its only common oxidation state (+2). It shows no variable oxidation states or coloured ions, so it is not a transition metal.
Q5MEDIUM· Lanthanoid Contraction
What is lanthanoid contraction and give two consequences.
Show solution
It is the steady decrease in atomic/ionic radius from La to Lu due to poor shielding by 4f electrons. Consequences: lanthanoids have very similar properties and are hard to separate; post-lanthanoid pairs like Zr/Hf and Nb/Ta have nearly identical radii and properties.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • d-block: (n-1)d^1-10 ns^1-2; Cr and Cu are exceptions.
  • Variable oxidation states arise from close (n-1)d and ns energies.
  • Colour is due to d-d transitions; needs unpaired d-electrons.
  • Magnetism: paramagnetic (unpaired e-), diamagnetic (none), ferromagnetic (Fe, Co, Ni).
  • Good catalysts due to variable oxidation states and surface adsorption (Fe, Ni, Pt, V2O5).
  • K2Cr2O7 (orange) and KMnO4 (purple) are strong oxidising agents.
  • Lanthanoid contraction (poor 4f shielding) makes Zr/Hf and Nb/Ta nearly identical.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks across the chapter

Question typeMarks eachTypical countWhat it tests
K2Cr2O7 / KMnO43-51Preparation and oxidising action
General properties31Oxidation states, colour, magnetism, catalysis
f-block / lanthanoid contraction2-31Lanthanoid contraction and actinoid comparison
Prep strategy
  • Learn the Cr and Cu configuration exceptions
  • Connect colour and magnetism to unpaired d-electrons
  • Memorise the dichromate and permanganate half-reactions
  • Master lanthanoid contraction and its consequences

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Catalysis in industry

Transition metal catalysts drive ammonia, sulfuric acid, and hydrogenation processes worldwide.

Analytical chemistry

K2Cr2O7 and KMnO4 are standard oxidising agents in volumetric titrations.

Magnets and alloys

Ferromagnetic Fe, Co, Ni and lanthanoid magnets are used in motors, data storage, and electronics.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Quote the Cr and Cu exceptions explicitly
  2. Link colour/magnetism to unpaired electrons
  3. Write balanced half-reactions for the oxidants
  4. Explain lanthanoid contraction with its cause and consequences

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Calculate magnetic moments using the spin-only formula sqrt(n(n+2)).
  • Explore the variable oxidation states and actinide chemistry of the f-block in depth.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Chemistry examHigh
JEE Main and Advanced (d- and f-block)High
NEET ChemistryMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Transition metals have variable oxidation states because their (n-1)d and ns orbitals are close in energy, so they can readily accept and donate electrons during a reaction, providing low-energy pathways. They also offer surfaces on which reactant molecules can adsorb, weaken bonds, and react more easily. These properties make metals like iron (Haber process), nickel (hydrogenation), platinum (catalytic converters), and vanadium pentoxide (Contact process) excellent catalysts.

Potassium permanganate is deep purple. In acidic medium MnO4- is reduced by gaining 5 electrons to nearly colourless Mn2+. In neutral or faintly alkaline medium it gains 3 electrons to form a brown precipitate of MnO2. In strongly alkaline medium it gains 1 electron to form the green manganate ion MnO4 2-. The medium therefore controls both the extent of reduction and the colour observed.
Verified by the tuition.in editorial team
Last reviewed on 30 May 2026. Written and reviewed by subject-matter experts — read about our process.
Editorial process →
Header Logo