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The d- and f-Block Elements – Class 12 Chemistry notes, important questions and MCQs

NCERT Class 12 Chemistry, chapter 4. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Transition (d-block) elements show variable oxidation states, coloured compounds, catalytic activity and complex formation. Lanthanoids and actinoids form the f-block.

d-block properties

General configuration (n − 1)d¹⁻¹⁰ ns¹⁻²; Zn, Cd, Hg are not typical transition elements (full d¹⁰).
Variable oxidation states (Mn shows +2 to +7); colour due to d–d transitions; paramagnetism from unpaired electrons, µ = √[n(n + 2)] BM.
Catalysts: Fe (Haber process), V₂O₅ (contact process), Ni (hydrogenation).

Important compounds and f-block

KMnO₄ (purple, strong oxidant) and K₂Cr₂O₇ (orange, oxidant in acid).
Lanthanoid contraction: steady decrease in size, making Zr and Hf nearly the same size.
Actinoids are all radioactive; many are man-made.

Key terms

Transition element: element with partly filled d-orbitals in the atom or common ions
Lanthanoid contraction: gradual decrease in atomic and ionic size across lanthanoids
Interstitial compounds: compounds formed when small atoms like H, C or N are trapped in the crystal lattice of transition metals

Formulas

µ = √[n(n + 2)] BM

Important questions with answers

Why do transition metals form alloys easily?2 marks

Their atomic sizes are similar, so atoms of one metal can easily replace atoms of another in the crystal lattice.

What is lanthanoid contraction? Give two consequences.3 marks

The steady decrease in atomic and ionic radii from La to Lu, due to poor shielding by 4f electrons. Consequences: Zr and Hf have almost the same radius and similar properties, making them hard to separate; basicity of lanthanoid hydroxides decreases from La(OH)₃ to Lu(OH)₃.

Describe the preparation of potassium dichromate from chromite ore with equations, and write the action of acidified K₂Cr₂O₇ on Fe²⁺ ions.5 marks

Chromite is fused with sodium carbonate in air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂. The sodium chromate is acidified: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O. Treating with KCl gives K₂Cr₂O₇: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. As an oxidant in acid: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O (orange to green).

Quick revision

Variable oxidation states.
Coloured ions from d–d transitions.
Zr and Hf are similar because of lanthanoid contraction.

Exam tips

Colourless ions like Zn²⁺, Sc³⁺ have d⁰ or d¹⁰ configuration (no d–d transition).
Colour and paramagnetism of transition metal ions come from unpaired d electrons; Sc³⁺ and Zn²⁺ (d⁰, d¹⁰) are colourless.

Practice MCQs

Which is used as a catalyst in the Haber process?
1. Ni
2. Fe
3. V₂O₅
4. Pt
Show answer
Answer: (b) Fe. Iron.
KMnO₄ solution is
1. green
2. orange
3. purple
4. colourless
Show answer
Answer: (c) purple. Purple.
Zn²⁺ salts are colourless because
1. its d-orbitals are completely filled
2. Zn is a metal
3. it has no electrons
4. it is reactive
Show answer
Answer: (a) its d-orbitals are completely filled. d¹⁰, no d–d transition.
Magnetic moment of an ion with 3 unpaired electrons is
1. 1.73 BM
2. 2.83 BM
3. 3.87 BM
4. 4.90 BM
Show answer
Answer: (c) 3.87 BM. √15.
Which ion is colourless
1. Ti³⁺
2. Zn²⁺
3. Cu²⁺
4. Fe²⁺
Show answer
Answer: (b) Zn²⁺. d¹⁰.
Zr and Hf have almost identical radii because of
1. inert pair effect
2. hydrogen bonding
3. d–d transition
4. lanthanoid contraction
Show answer
Answer: (d) lanthanoid contraction. Lanthanoid contraction.
The highest oxidation state shown by manganese is
1. +4
2. +6
3. +7
4. +8
Show answer
Answer: (c) +7. In KMnO₄.
Spin-only magnetic moment of Mn²⁺ (Z = 25) is about
1. 2.83 BM
2. 4.90 BM
3. 5.92 BM
4. 3.87 BM
Show answer
Answer: (c) 5.92 BM. 5 unpaired: √35.

All chapters of Class 12 Chemistry

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.