JEE Main Chemistry · Chapter 11

d- and f-Block Elements

Transition metal properties, magnetic moment, colour, KMnO₄ and K₂Cr₂O₇, lanthanoid contraction and actinoids.

NCERT Class 12JEE Main: 1 Q per shift 8 min read

1. Chapter at a Glance

The d-block (groups 3–12) contains the transition elements, which have partly filled d orbitals in the atom or in a common ion. The f-block contains the lanthanoids (4f) and actinoids (5f), called inner-transition elements. These metals show variable oxidation states, coloured ions, magnetism, catalytic activity and complex formation. JEE Main often asks spin-only magnetic moment, colour, oxidation states, lanthanoid contraction and the chemistry of KMnO₄ and K₂Cr₂O₇.

What JEE asks most
  • Electronic configurations (Cr and Cu exceptions) and number of unpaired electrons.
  • Spin-only magnetic moment μ = √(n(n+2)) BM.
  • Trends in E° values and the stability of oxidation states.
  • Preparation and reactions of KMnO₄ and K₂Cr₂O₇; lanthanoid contraction and its results.

2. Electronic Configuration and General Properties

General configuration: (n−1)d1–10 ns1–2. Two important exceptions in the 3d series come from the extra stability of half-filled and fully filled d subshells:

Cr: [Ar] 3d⁵ 4s¹    Cu: [Ar] 3d¹⁰ 4s¹
  • When transition atoms form ions, 4s electrons are removed before 3d electrons (Fe²⁺ = [Ar] 3d⁶).
  • Zn, Cd and Hg (d¹⁰ in atom and ions) are not regarded as true transition elements; they are soft, low-melting metals.
  • Strong metallic bonding (many unpaired d electrons) gives high melting points; the maximum is near the middle of each series (Cr, Mo, W). Mn and Tc are unexpectedly low.
  • Atomic radii decrease slowly across the series and are nearly constant in the middle; 4d and 5d radii are almost the same because of lanthanoid contraction.

3. Characteristic Properties

Variable oxidation states

Because ns and (n−1)d energies are close, both sets can be lost. Mn shows the most oxidation states in the 3d series (+2 to +7). The highest oxidation state of any element is +8 (Os in OsO₄, Ru in RuO₄). High oxidation states are stabilised by O and F (Mn₂O₇, CrO₃, VF₅).

ObservationReason
Mn²⁺ is very stable; Mn³⁺ is a strong oxidantMn²⁺ is d⁵ (half-filled)
Fe³⁺ more stable than Fe²⁺Fe³⁺ is d⁵
Cr²⁺ is a reducing agentCr³⁺ (t₂g³) is more stable
Cu⁺ disproportionates in waterHigh hydration enthalpy of Cu²⁺ outweighs its second IE
E°(Cu²⁺/Cu) is positive (+0.34 V)High atomisation enthalpy and low hydration enthalpy of Cu
Sc³⁺ and Zn²⁺ are colourlessd⁰ and d¹⁰: no d–d transition possible

Magnetic properties

Unpaired electrons make a substance paramagnetic. The spin-only formula is:

μ = √(n(n + 2)) BM   (n = number of unpaired electrons)
1.732.833.874.905.92 n=1n=2n=3n=4n=5 μ
Spin-only magnetic moment (BM) against the number of unpaired electrons. Mn²⁺ and Fe³⁺ (d⁵) have the highest value, 5.92 BM.

Colour, catalysis and other features

  • Colour arises from d–d transitions: the ion absorbs part of visible light and shows the complementary colour. Ti³⁺ (d¹) is purple, Cu²⁺ (d⁹) is blue.
  • Catalytic activity: variable oxidation states and a surface for adsorption — Fe in the Haber process, V₂O₅ in the contact process, Ni in hydrogenation.
  • Interstitial compounds: small atoms (H, C, N) trapped in the metal lattice (e.g. TiC, Fe₃H); they are hard, high-melting and non-stoichiometric.
  • Alloys form easily because atomic radii are similar (brass Cu–Zn, bronze Cu–Sn).

4. Potassium Dichromate and Potassium Permanganate

K₂Cr₂O₇ (orange)

  • Made from chromite ore FeCr₂O₄: fused with Na₂CO₃ in air → Na₂CrO₄ (yellow) → acidified to Na₂Cr₂O₇ → treated with KCl.
  • Chromate–dichromate equilibrium depends on pH: 2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O. Acid favours dichromate, base favours chromate.
  • Strong oxidising agent in acid: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ (green) + 7H₂O. Oxidises I⁻ to I₂, Fe²⁺ to Fe³⁺, H₂S to S.

KMnO₄ (purple)

  • Made from pyrolusite MnO₂: fused with KOH and an oxidant → K₂MnO₄ (green manganate), which disproportionates or is oxidised electrolytically to KMnO₄.
  • Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ (colourless) + 4H₂O. It acts as a self-indicator in titrations.
  • Neutral or weakly basic: MnO₄⁻ → MnO₂ (3e⁻). Oxidises I⁻ to IO₃⁻ in this medium.
  • HCl is not used to acidify KMnO₄ because it is oxidised to Cl₂.

5. The f-Block: Lanthanoids and Actinoids

  • Lanthanoids (Ce to Lu): common oxidation state +3. Ce⁴⁺ is a good oxidant (it gains the f⁰ state); Eu²⁺ and Yb²⁺ are reductants (f⁷, f¹⁴ in +2).
  • Lanthanoid contraction: a steady decrease in atomic and ionic radii from La to Lu, due to the poor shielding of 4f electrons. Consequences: Zr ≈ Hf in size, difficulty in separating lanthanoids, basicity of hydroxides decreases from La(OH)₃ to Lu(OH)₃.
  • Actinoids (Th to Lr): all radioactive; show a wider range of oxidation states (up to +7 for Np, Pu) because 5f, 6d and 7s energies are close. Actinoid contraction is larger than lanthanoid contraction.
  • Misch metal (a lanthanoid alloy with iron) is used in lighter flints.

6. Formula and Fact Master Table

ItemKey fact / formula
Spin-only momentμ = √(n(n+2)) BM
n-factor of KMnO₄acid 5, neutral 3, strongly basic 1
n-factor of K₂Cr₂O₇ (acid)6
Exceptional configurationsCr 3d⁵4s¹, Cu 3d¹⁰4s¹
Max oxidation state (3d)Mn +7
Max oxidation state (any)+8 (OsO₄, RuO₄)
Colourless ionsd⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Zn²⁺, Cu⁺)

7. Mind Map

d- and f-Block
Configuration
  • Cr, Cu exceptions
  • 4s lost first
Properties
  • Variable oxidation states
  • Colour, magnetism
  • Catalysis, alloys
Compounds
  • K₂Cr₂O₇
  • KMnO₄
  • Chromate ⇌ dichromate
f-block
  • Lanthanoid contraction
  • +3 common
  • Actinoids radioactive

8. Tips and Tricks

Shortcuts
  • Memorise μ values: n = 1, 2, 3, 4, 5 → 1.73, 2.83, 3.87, 4.90, 5.92 BM. Reverse-lookup n from a given μ.
  • To count unpaired electrons in an ion, first remove the 4s electrons, then fill the d orbitals by Hund's rule.
  • dn and d10−n have the same number of unpaired electrons (for high spin free ions).
  • Colour check: if the ion is d⁰ or d¹⁰, it is colourless (unless charge-transfer, as in MnO₄⁻ and Cr₂O₇²⁻).
Mistakes students make
  • Thinking KMnO₄'s colour is from d–d transition — Mn⁷⁺ is d⁰; the colour is due to charge transfer.
  • Removing 3d electrons before 4s when writing ionic configurations.
  • Believing Zn is a typical transition metal.

9. Important One-liners

  • Transition metals form coloured ions due to d–d transitions.
  • Mn²⁺ and Fe³⁺ both have 5 unpaired electrons.
  • Acidified K₂Cr₂O₇ turns from orange to green when it is reduced.
  • Cu⁺ is unstable in water and disproportionates to Cu²⁺ and Cu.
  • Lanthanoid contraction is caused by the poor shielding effect of 4f electrons.
  • All actinoids are radioactive.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

The spin-only magnetic moment of Mn²⁺ (Z of Mn = 25) is:

  1. 1.73 BM
  2. 3.87 BM
  3. 4.90 BM
  4. 5.92 BM
Show solution

Mn²⁺ = [Ar] 3d⁵: 5 unpaired electrons. μ = √(5 × 7) = √35 ≈ 5.92 BM.

Answer: (D)

Q2Easy

Which of the following ions is colourless in aqueous solution?

  1. Ti³⁺
  2. Cu²⁺
  3. Zn²⁺
  4. Fe³⁺
Show solution

Zn²⁺ is 3d¹⁰: no d–d transition is possible, so it is colourless. Ti³⁺ (d¹), Cu²⁺ (d⁹) and Fe³⁺ (d⁵) are coloured.

Answer: (C)

Q3Numerical

A 3d-metal ion has a spin-only magnetic moment of 3.87 BM. How many unpaired electrons does it have?

Show solution

√(n(n+2)) = 3.87 ⇒ n(n+2) ≈ 15 ⇒ n = 3.

Answer: 3

Q4Medium

When an acidified K₂Cr₂O₇ solution oxidises Fe²⁺, the colour changes from:

  1. Yellow to orange
  2. Orange to green
  3. Purple to colourless
  4. Green to orange
Show solution

Cr₂O₇²⁻ (orange, Cr⁺⁶) is reduced to Cr³⁺ (green). Purple to colourless is the KMnO₄ change; yellow to orange is chromate to dichromate on adding acid.

Answer: (B)

Q5Medium

Among the 3d series, the element showing the largest number of oxidation states is:

  1. Sc
  2. Cr
  3. Mn
  4. Zn
Show solution

Mn (3d⁵4s²) can use all seven electrons and shows every state from +2 to +7. Sc shows only +3 and Zn only +2.

Answer: (C)

Q6Hard

The main reason why Zr and Hf have almost identical radii is:

  1. Inert pair effect
  2. Lanthanoid contraction
  3. Both are in group 4
  4. Actinoid contraction
Show solution

The 14 lanthanoids between them add 4f electrons that shield poorly. The resulting contraction cancels the size increase expected on moving from period 5 to period 6.

Answer: (B)

Q7Hard

Why is Mn³⁺ a strong oxidising agent while Cr²⁺ is a strong reducing agent?

  1. Mn³⁺ → Mn²⁺ gives d⁵; Cr²⁺ → Cr³⁺ gives stable t₂g³
  2. Both have d⁴ configuration and are unstable in the same way
  3. Mn is more electronegative than Cr
  4. Cr²⁺ has a d⁵ configuration
Show solution

Both are d⁴ ions. Mn³⁺ gains an electron to become d⁵ (half-filled, very stable), so it oxidises other species. Cr²⁺ loses an electron to become Cr³⁺ with a stable half-filled t₂g³ set, so it reduces other species. Option (B) is incomplete because it does not explain the opposite behaviour; (D) is wrong since Cr²⁺ is d⁴.

Answer: (A)

Q8Hard

Adding NaOH to an orange solution of K₂Cr₂O₇ changes its colour to yellow because:

  1. Cr is reduced to Cr³⁺
  2. Dichromate changes to chromate
  3. Cr is oxidised to Cr⁷⁺
  4. A precipitate of Cr(OH)₃ forms
Show solution

Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O. This is an acid–base change; chromium stays +6 in both ions, so there is no redox reaction.

Answer: (B)

11. Quick Revision Summary

Remember before the exam
  • Cr = 3d⁵4s¹, Cu = 3d¹⁰4s¹; ions lose 4s electrons first.
  • μ = √(n(n+2)) BM; d⁵ gives 5.92 BM.
  • Colour from d–d transitions; d⁰ and d¹⁰ ions are colourless.
  • KMnO₄: purple → colourless (acid), n = 5; K₂Cr₂O₇: orange → green, n = 6.
  • Chromate (yellow) ⇌ dichromate (orange) with pH.
  • Lanthanoid contraction: poor 4f shielding; Zr ≈ Hf.
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