Class 12 Chemistry · NCERT Chapter 4 (Part 1)
~14 min readd and f Block Elements Class 12 Chemistry Notes
The d and f block elements give transition metals their colour, magnetism and catalytic power. This chapter covers the electronic configuration of the 3d series, variable oxidation states, magnetic moments, potassium dichromate, potassium permanganate and the lanthanoid and actinoid series.
Why are transition metal compounds so often coloured?
Their ions have partially filled d orbitals. When visible light falls on them, electrons undergo d–d transitions and absorb a part of the visible spectrum; the transmitted light shows the complementary colour. Zn²⁺ (d¹⁰) has no such transition and is colourless.
Position and General Configuration of d-Block Elements
The d-block spans groups 3–12, where the (n−1)d subshell is being filled. The elements scandium to zinc (3d¹–3d¹⁰4s²) form the first transition series; the term transition element strictly means an element whose atom or ion in the common oxidation state has an incomplete d subshell.
- Exceptions: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹ — half-filled and filled d subshells add exchange and symmetry stability.
- Zn, Cd and Hg (d¹⁰) are d-block but not transition elements by the strict definition.
- Four series: 3d (Sc–Zn), 4d (Y–Cd), 5d (Lu–Hg), 6d (incomplete).
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MCQ Questions
Atomic, Ionic and Ionisation Trends in the 3d Series
Across the 3d series, atomic radii first decrease, then stay nearly constant because the added d electrons shield each other imperfectly while the nuclear charge grows. Ionic radii fall steadily with the charge on the ion.
- Atomic radius: falls slightly from Sc to Mn, then roughly constant to Zn.
- Ionic radii: r(M³⁺) < r(M²⁺) < r(M); isoelectronic ions shrink as the charge rises.
- Ionisation enthalpies increase across the series but rise slowly: the shielding by d electrons keeps them comparable to s-block values initially.
- Cr and Cu show smaller first ionisation enthalpies than expected from half- and fully-filled d subshells.
Standard electrode potentials
Most M²⁺/M values are negative, telling us metals are stronger reducing agents than hydrogen, but the values do not change smoothly; complexation and hydration energies cause irregularities (e.g. Mn, Co, Zn).
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Oxidation States and Their Variation
Transition metals show many oxidation states because the (n−1)d and ns electrons have comparable energies and can all participate in bonding. The maximum oxidation state parallels the number of (n−1)d + ns electrons available for bonding.
- Highest states increase left to right: Sc +3, Ti +4, V +5, Cr +6, Mn +7 — then Cu +2, Zn +2.
- +2 and +3 are the common aquarium states; higher states appear in oxo-anions (MnO₄⁻, Cr₂O₇²⁻).
- Stable states are favoured when the ion has a half-filled or empty d subshell: Mn²⁺ (d⁵), Zn²⁺ (d¹⁰).
- In higher states the metals are strong oxidising agents, e.g. KMnO₄ and K₂Cr₂O₇.
Remember
MCQ Questions
Colour, Magnetic and Catalytic Properties
Colour from d–d transitions
An ion with partially filled d orbitals absorbs a portion of visible light to promote an electron between the split d levels; the transmitted light is the complementary colour. Thus Cu²⁺ (aq) blue, Cr³⁺ green, Ni²⁺ green, Mn²⁺ pale pink, and Zn²⁺/Sc³⁺ (d⁰ or d¹⁰) colourless.
- Unpaired d electrons give paramagnetism; many unpaired electrons → ferromagnetic in the solid (Fe, Co, Ni).
- Cu⁺ (d¹⁰) diamagnetic, Cu²⁺ (d⁹) paramagnetic.
- Catalysis: transition metals adsorb reactants and change oxidation state to form intermediates — Fe (Haber), V₂O₅ (contact), Pt/Ni (hydrogenation).
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Potassium Dichromate and Permanganate
- K₂Cr₂O₇: prepared from chromite ore (FeCr₂O₄) via roasting, fusion with Na₂CO₃, leaching and acidification; oxidised from Na₂CrO₄. Orange crystals, strong oxidant in acid: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
- KMnO₄: prepared from pyrolusite (MnO₂) by fusion with KOH/air to K₂MnO₄, then electrolytic oxidation. Purple crystals.
- MnO₄⁻ reduces differently by medium: acid → Mn²⁺ (n = 5), neutral → MnO₂ (n = 3), alkaline → MnO₄²⁻ (n = 1).
- Uses: K₂Cr₂O₇ in volumetric analysis and chrome tanning; KMnO₄ as oxidant, disinfectant and in titrations.
Why they are strong oxidants
In acidic medium K₂Cr₂O₇ oxidises ferrous salts, iodides and sulphides; KMnO₄ oxidises Fe²⁺, C₂O₄²⁻, and many organic groups. Both act as self-indicators and are primary/standard reagents in titrations.
MCQ Questions
The f-Block: Lanthanoids and Actinoids
The f-block houses the lanthanoids (4f, La–Lu) and actinoids (5f, Ac–Lr). Their common oxidation state is +3, with the inner f electrons poorly shielding the outer ones.
- Lanthanoid contraction: radii shrink steadily from Ce³⁺ to Lu³⁺ (~15 pm) because 4f shielding is weak — properties of successive elements become very similar.
- Lanthanoids: most stable state +3; Ce⁴⁺ (colourless, strong oxidant) and Eu²⁺ also known; separation by ion-exchange or fractional crystallisation.
- Actinoids: radioactive; show wider oxidation states (+3…+7); actinoid contraction is larger; uranium and plutonium power nuclear reactors.
- Both series form coloured ions (f–f transitions) but the colours are weaker than d–d colours.
Contraction consequences
MCQ Questions
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q1–Q3)
- 1Write the electronic configurations of the elements: (i) Cr, (ii) Cu, (iii) Zn2+.
- 2Why are the transition metals known to form coloured ions?
- 3Calculate the spin‑only magnetic moment (in BM) of Fe2+ (high spin) and Co3+ (low spin).
Answers at a glance
- 1Cr: [Ar] 3d5 4s1; Cu: [Ar] 3d10 4s1; Zn2+: [Ar] 3d10.
- 2Due to d‑d transitions arising from crystal field splitting of partially filled d orbitals in an ligand field.
- 3Fe2+ (high spin): 4.90 BM; Co3+ (low spin): 0 BM.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1What are transition elements?
- 2Give the electronic configuration of Cr atom.
- 3What is lanthanoid contraction?
2 Mark Questions
- 1Explain why Cu+ ion is not stable in aqueous solution.
- 2Why do transition metals exhibit variable oxidation states?
3 Mark Questions
- 1Describe the preparation and properties of potassium dichromate.
- 2Explain the origin of colour in transition metal complexes.
5 Mark Questions
- 1Compare the chemistry of actinoids with that of lanthanoids with reference to electronic configuration, oxidation states and chemical reactivity.
- 2An alloy of Cu, Zn and Ni is used in silver‑ware manufacture. Explain why such alloys are resistant to corrosion.
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Class 12 Chemistry — downloadable assignments and CBSE-pattern tests (PDF).
Topic-wise practice covering the first half of the chapter, with answer key.
Topic-wise practice covering the second half of the chapter, with answer key.
Full CBSE-pattern test (90 minutes, 40 marks) with answer key.
Second full CBSE-pattern test (90 minutes, 40 marks) with answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
General configuration
For d-block; Cu is an exception: 3d¹⁰4s¹.
Magnetic moment
n = number of unpaired electrons.
Maximum oxidation state
Highest for Mn (+7: MnO₄⁻).
Cation radii trend
Ionic radii decrease with increasing positive charge.
Lanthanoid contraction
Poor 4f shielding makes radii shrink slowly across the series.
Equivalent weight in redox
KMnO₄ in acid: n = 5; K₂Cr₂O₇ in acid: n = 6.
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- General electronic configuration + exceptions (Cu 3d¹⁰4s¹, Cr 3d⁵4s¹) are compulsory 1-mark/MCQ facts.
- Variable oxidation states and the reason — nearly equal energies of (n−1)d and ns orbitals.
- Magnetic moment μ = √(n(n+2)) BM — numericals on unpaired electrons (Sc³⁺ vs Fe³⁺).
- Coloured ions from d–d transitions; why Zn²⁺ and Sc³⁺ are colourless.
- Properties: catalytic action, formation of alloys and interstitial compounds, high melting points.
- Preparations and oxidising behaviour of K₂Cr₂O₇ and KMnO₄ — a standard 3/5-mark question.
- Lanthanoid contraction: cause, effect on properties and separation by ion exchange.
- Actinoids: radioactivity, wider range of oxidation states, actinoid contraction.
Practice Numericals
Step-by-step worked examples
Standard exam-level problems solved with clear conceptual breakdowns.
Magnetic moment of Fe³⁺
Step-by-Step Solution
- 1Fe: [Ar]3d⁶4s². Fe³⁺: [Ar]3d⁵.
- 2Number of unpaired electrons n = 5.
- 3μ = √(n(n+2)) = √(5 × 7) = √35.
Final Answer
μ ≈ 5.92 BM
Colourless Sc³⁺ vs violet Ti³⁺
Step-by-Step Solution
- 1Sc³⁺ = [Ar]3d⁰ — no d electrons, so no d–d transition is possible.
- 2Ti³⁺ = [Ar]3d¹ — one d electron absorbs visible light by a d–d transition, giving a violet colour.
Final Answer
Presence of a d electron (d–d transition) makes Ti³⁺ coloured.
FAQ
Frequently asked questions
Why do transition elements show variable oxidation states?
The (n−1)d and ns orbitals have nearly equal energies, so one or more electrons from each can participate in bonding. The maximum possible state rises from +3 in Sc to +7 in Mn, then falls as the d orbitals fill and the effective nuclear charge binds the electrons more tightly (Sc: +3…Mn: +7…Zn: +2).
What is the lanthanoid contraction and what does it cause?
Across the lanthanoid series the 4f electrons shield the nuclear charge poorly, so the effective nuclear charge rises and the ionic radii steadily shrink — a contraction of ~15 pm from Ce³⁺ to Lu³⁺. It makes neighbouring lanthanoids chemically similar, makes their separation by ion exchange possible, and tightens the radii of the following (5d) elements (e.g. Zr–Hf similarity).
Why do transition metals make good catalysts?
They adsorb reactants on their surface, and by showing variable oxidation states they form reaction intermediates at a lower activation energy. Examples: Pt in the contact process, V₂O₅ in the contact process for H₂SO₄, and Fe in the Haber process.
Why are transition metals hard and have high melting points?
Their atoms contribute both ns and (n−1)d electrons to a strong metallic bond — more bonding electrons per atom than typical s-block metals. The cohesive force peaks around V, Cr and Mo where the number of unpaired d electrons is highest.
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