The d- and f-Block Elements Class 12 Notes - CBSE Chemistry Chapter 8

Chapter summary

The d- and f-Block Elements covers the position and electronic configuration of transition elements, periodic trends in size, density and melting point, variable oxidation states, magnetic moment, colour and catalytic behaviour, the important oxidisers potassium permanganate and potassium dichromate, and the f-block lanthanoids and actinoids with lanthanoid contraction. It is a high-yield NEET chapter where questions test electronic configurations, the spin-only magnetic moment, the reasons for colour, and lanthanoid contraction.

Chapter notes
🃏 Flash Cards: The d- and f-Block Elements

Class 12 Chemistry · Chapter 8 – swipe through all 9 cards to understand the whole chapter.

🧲Start here1/9

Who Are the d-Block Elements?

Transition elements sit in the middle of the table (groups 3-12) where the inner (n-1)d orbitals fill up.

(n-1)d110 ns02

True transition element = partially filled d in element OR a common ion.

  • 4 series: 3d (Sc→Zn), 4d, 5d, 6d (incomplete)
  • Zn, Cd, Hg (d10s2) are NOT true transition metals
  • ns electrons leave before (n-1)d → Fe2⁺ is 3d6
⚛️Key exceptions2/9

Cr and Cu Configurations

Half-filled and fully-filled d-subshells are extra stable, so Cr and Cu break the simple filling rule.

Cr: [Ar]3d54s1 · Cu: [Ar]3d104s1

Not 3d44s2 / 3d94s2 — the lone 4s electron buys d5 / d10 stability.

  • Cr → half-filled 3d5 is stable
  • Cu → fully-filled 3d10 is stable
  • Unpaired d-electrons drive colour, magnetism, oxidation states
📏Core trends3/9

Size, Density & Melting Point

Added electrons enter an inner d-shell, so across a series atomic radius stays almost constant.

radius ≈ constant (mid) · MP peaks mid-series · density rises

Os and Ir are the densest metals; Mn, Tc dip in MP (stable d5).

  • Radius: small drop, flat middle, slight rise at Cu, Zn
  • Very high MP from strong ns + (n-1)d metallic bonding
  • High atomisation enthalpy → good catalysts and alloys
🔢Core concept4/9

Variable Oxidation States

ns and (n-1)d energies are very close, so a varying number of electrons can bond.

Mn: +2 to +7 (widest in 3d) · states differ by 1

Highest state lives in the oxide/fluoride: Mn2O7, KMnO4, OsO4 (+8).

  • +2 grows more stable across the series (Mn2⁺ = stable d5)
  • O and F stabilise the highest states
  • Sc shows only +3; Zn shows only +2
🧭Spin-only5/9

Magnetic Moment

Paramagnetism comes from unpaired d-electrons; count them and plug into the spin-only formula.

μ = √[n(n+2)] BM

n = unpaired e⁻. n=0 → diamagnetic. Unit is Bohr Magneton (BM).

  • More unpaired e⁻ → more paramagnetic
  • Fe3⁺ (3d5, n=5) → μ = √35 ≈ 5.9 BM
  • Ignores orbital contribution (spin-only approximation)
🎨Key concept6/9

Colour, Catalysis & Complexes

d-d electronic transitions absorb visible light, so most transition-metal ions are coloured.

d-d transition → colour · d0 and d10 ions are colourless

Sc3⁺, Ti4⁺ (d0) and Zn2⁺, Cu⁺ (d10) → no d-d jump → colourless.

  • Catalysts (V2O5, Fe, Ni, Pt) via variable states / adsorption
  • Form complexes and interstitial compounds (H, C, N in lattice)
  • Similar atomic sizes → form alloys (e.g. steel)
🌍f-Block7/9

Lanthanoids & Actinoids

Inner transition elements fill the deeper (n-2)f orbitals and sit below the main table.

Lanthanoids: 4f, Ce→Lu · Actinoids: 5f, Th→Lr (all radioactive)

Characteristic state +3; actinoids reach +6/+7 (close 5f, 6d, 7s).

  • Lanthanoids: dominant +3 state
  • Eu, Yb → +2 and Ce, Tb → +4 (half/full 4f link)
  • Ce4⁺ is a strong oxidising agent
📉High yield8/9

Lanthanoid Contraction

Across the lanthanoids, atomic and ionic size steadily shrinks because 4f electrons shield poorly.

poor 4f shielding → size ↓ across Ce→Lu → Zr ≈ Hf

Pairs Zr/Hf and Nb/Ta become almost identical and hard to separate.

  • Rising effective nuclear charge pulls electrons inward
  • 2nd and 3rd transition series end up similar in size
  • Basicity differences of later lanthanoids aid separation
⚗️Must know9/9

KMnO4 & K2Cr2O7

Two NEET-staple oxidisers — learn their colour, structure and acidic-medium half-reactions.

MnO4⁻ + 8H⁺ + 5e⁻ → Mn2⁺ + 4H2O · Cr2O72⁻ + 14H⁺ + 6e⁻ → 2Cr3⁺ + 7H2O

KMnO4: Mn(+7), purple, tetrahedral MnO4⁻. K2Cr2O7: Cr(+6), orange.

  • Acidic medium: MnO4⁻ takes 5e⁻, Cr2O72⁻ takes 6e⁻
  • Neutral/basic MnO4⁻ → 3e⁻ → MnO2 (brown)
  • CrO42⁻ (yellow, basic) ⇌ Cr2O72⁻ (orange, acidic)
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📝 Practice The d- and f-Block Elements — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
The incorrect statement among the following is:
Correct answer: B. Most trivalent lanthanoid ions are actually coloured in the solid state (due to f-f transitions), so the statement calling them colourless is incorrect. The other three statements are correct.
🔎 See the full step-by-step solution in the app →
Q2NEET 2021
Which one of the following reactions is a metal displacement reaction?
Correct answer: B. In Cr₂O₃ + 2Al → Al₂O₃ + 2Cr (the thermite/aluminothermy reaction) the more reactive metal Al displaces the less reactive metal Cr — a metal displacement reaction. Option C displaces hydrogen (a non-metal), and A and D are decomposition reactions.
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Q3NEET 2020
The calculated spin-only magnetic moment of the Cr²⁺ ion is:
Correct answer: A. Cr²⁺ is [Ar]3d⁴ with 4 unpaired electrons. μ = √(n(n+2)) = √(4(4+2)) = √(24) ≈ 4.90 BM.
🔎 See the full step-by-step solution in the app →
Q4NEET 2015
Which of the following processes does NOT involve oxidation of iron?
Correct answer: C. In Fe(CO)₅, CO is a neutral ligand and iron is in the 0 oxidation state, so forming it from Fe(0) involves no change in oxidation state. Rusting, displacing Cu²⁺, and liberating H₂ from steam all oxidise iron.
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Q5NEET 2015
A 3d-series ion shows a spin-only magnetic moment of 2.84 BM. The ion is (At. no. Ni = 28, Ti = 22, Cr = 24, Co = 27):
Correct answer: A. 2.84 = √(n(n+2)) ⇒ n(n+2)=8 ⇒ n=2. Ni²⁺ is [Ar]3d⁸ with exactly 2 unpaired electrons. (Ti³⁺ d¹ → 1, Cr³⁺ d³ → 3, Co²⁺ d⁷ → 3.)
🔎 See the full step-by-step solution in the app →
Q6NEET 2014
The pair of compounds that can exist together is:
Correct answer: C. FeCl₂ and SnCl₂ both contain metals in lower oxidation states that cannot oxidise one another, so they coexist. FeCl₃ would oxidise SnCl₂ (and KI), and HgCl₂ would oxidise SnCl₂, so those pairs react.
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Q7NEET 2010
Which one of the following ions has the electronic configuration [Ar]3d⁶? (At. no. Mn = 25, Fe = 26, Co = 27, Ni = 28)
Correct answer: D. Co (Z=27) is [Ar]3d⁷4s²; Co³⁺ loses 4s² and one 3d electron → [Ar]3d⁶. (Ni³⁺=3d⁷, Mn³⁺=3d⁴, Fe³⁺=3d⁵.)
🔎 See the full step-by-step solution in the app →
Q8NEET 2005
Four successive members of the first-row transition elements are listed with their atomic numbers. Which is expected to have the highest third ionisation enthalpy: V (Z=23), Cr (Z=24), Fe (Z=26), Mn (Z=25)?
Correct answer: D. The third electron is removed from the M²⁺ ion. For Mn, Mn²⁺ is [Ar]3d⁵, a stable half-filled shell; removing the third electron from this stable configuration needs the most energy, so Mn has the highest third ionisation enthalpy.
🔎 See the full step-by-step solution in the app →
Q9NEET 1995
Stainless steel contains iron and:
Correct answer: A. Stainless steel is corrosion-resistant because, besides iron (~73%), it contains chromium (~18%) and nickel (~8%). So the added constituents are Cr and Ni.
🔎 See the full step-by-step solution in the app →
Q10NEET 1991
The general electronic configuration of transition elements is:
Correct answer: C. In transition elements the (n-1)d subshell is being filled while the outermost ns holds one or two electrons, giving (n-1)d¹⁻¹⁰ ns¹⁻² (ns⁰ only in some ions). Among the options, (n-1)d¹⁻¹⁰ ns¹⁻² best represents the neutral atoms.
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Frequently Asked Questions

What is a transition element?

A transition element is one that has a partially filled d-subshell either in its elementary (ground) state or in one of its common oxidation states. By this definition zinc, cadmium and mercury are not regarded as true transition elements because their d-subshell is completely filled (d10).

How is the magnetic moment of a transition-metal ion calculated?

The spin-only magnetic moment equals the square root of n times (n plus 2) Bohr magnetons, where n is the number of unpaired electrons. More unpaired electrons means more paramagnetic, while an ion with no unpaired electrons is diamagnetic. For example, the Fe3+ ion with five unpaired electrons has a moment of about 5.9 Bohr magnetons.

Why are most transition-metal ions coloured?

Most transition-metal ions are coloured because of d-d transitions, where an electron jumps between d-orbitals split into different energy levels, absorbing part of visible light. Ions with empty d-orbitals (d0, like Sc3+) or completely filled d-orbitals (d10, like Zn2+) are colourless because no d-d transition is possible.

What is lanthanoid contraction and one of its consequences?

Lanthanoid contraction is the steady decrease in atomic and ionic size across the lanthanoid series, caused by the poor shielding of the nucleus by 4f electrons so that increasing nuclear charge pulls the electrons inward. A key consequence is that zirconium and hafnium have almost identical sizes and are very difficult to separate.

Why does manganese show the widest range of oxidation states in the 3d series?

Manganese has five 3d and two 4s electrons available, and because the energies of the 3d and 4s electrons are very close, all of them can take part in bonding. This allows manganese to show oxidation states from +2 up to +7, the widest range in the 3d series.

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