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CBSE Class 12 Chemistry Chapter 4 The d- and f-Block Elements Worksheet with Answers

Chapter 4 of NCERT Class 12 Chemistry explores the fascinating chemistry of d-block and f-block elements — transition metals, lanthanoids and actinoids. This printable worksheet is designed for rigorous board exam preparation, offering a mix of objective and subjective questions that mirror the 2025 CBSE Class 12 Chemistry paper pattern. Complete this in 90 minutes under timed conditions for maximum benefit.

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Key takeaways

✓Printable 90-minute worksheet covering transition metals, lanthanoids and actinoids with complete answer key
✓Section A contains 6 MCQs testing fundamental concepts of d-block and f-block element properties
✓Section D features 5 three-mark short-answer questions on electronic configuration and oxidation states
✓Section E includes 3 five-mark HOTS questions demanding analytical thinking about trends and reactions
✓Case-study question integrates real-world applications of transition metal compounds in catalysis and industry
✓Answer key provides step-by-step explanations aligned with NCERT Class 12 Chemistry terminology and marking scheme

Quick Chapter Recap: The d- and f-Block Elements

Before attempting the worksheet, revisit these core concepts from NCERT Class 12 Chemistry Chapter 4. Transition metals occupy the d-block of the periodic table, exhibiting variable oxidation states, coloured compounds, paramagnetism and catalytic activity due to partially filled d-orbitals. The first transition series spans Sc (Z=21) to Zn (Z=30). Lanthanoids (4f series, Ce to Lu) and actinoids (5f series, Th to Lr) constitute the f-block, showing lanthanoid contraction and actinoid contraction respectively. Key trends include atomic radii, ionisation enthalpy, metallic character and stability of oxidation states. Compounds like K₂Cr₂O₇ (orange dichromate) and KMnO₄ (purple permanganate) are vital oxidising agents. Understanding electronic configurations using the Aufbau principle and Hund's rule is essential for explaining magnetic properties and colour.

Transition metals: elements with partially filled d-orbitals in ground state or common oxidation states
Lanthanoid contraction: steady decrease in atomic and ionic radii from La to Lu due to poor shielding by 4f electrons
Actinoid contraction: similar radius decrease in 5f series, more pronounced than lanthanoid contraction
Variable oxidation states arise because (n-1)d and ns electrons have comparable energies
Coloured ions result from d-d electronic transitions absorbing visible light
Interstitial compounds form when small atoms (H, C, N) occupy voids in metal lattices

Section A: Multiple Choice Questions (1 mark each)

Attempt all six MCQs. Each question carries 1 mark. Choose the most appropriate answer from the four options provided. These questions test your grasp of fundamental concepts like electronic configuration, oxidation states, magnetic properties and characteristic reactions of transition metals, lanthanoids and actinoids. Refer to NCERT Class 12 Chemistry Chapter 4 for standard terminology. Pay close attention to exceptions such as the electronic configurations of Cr and Cu, which deviate from Aufbau predictions due to stability of half-filled and fully-filled d-orbitals. Remember that Zn, Cd and Hg are not considered true transition metals because their d-orbitals are completely filled in the ground state and common oxidation states.

Q1. Which of the following is NOT a characteristic property of transition metals? (a) High melting points (b) Variable oxidation states (c) Coloured compounds (d) Always diamagnetic
Q2. The electronic configuration of Cu²⁺ ion is: (a) [Ar]3d⁹ (b) [Ar]3d⁷4s² (c) [Ar]3d⁸4s¹ (d) [Ar]3d¹⁰
Q3. Lanthanoid contraction is caused by: (a) Poor shielding effect of 4f electrons (b) Increase in nuclear charge (c) Both (a) and (b) (d) Decrease in nuclear charge
Q4. Which actinoid element is used in nuclear reactors? (a) Thorium (b) Uranium (c) Plutonium (d) All of these
Q5. The highest oxidation state exhibited by transition elements is generally shown by: (a) First member of the series (b) Middle members (c) Last member (d) Random distribution
Q6. Identify the diamagnetic ion among the following: (a) Fe²⁺ (b) Cu²⁺ (c) Zn²⁺ (d) Ni²⁺

Section B: Fill in the Blanks (1 mark each)

Complete each statement with the correct term or value. These questions focus on important facts, definitions and nomenclature from NCERT Class 12 Chemistry Chapter 4. Write your answers precisely — spelling and chemical formulae matter. Remember that transition elements are defined by IUPAC as elements with an incomplete d sub-shell in the metallic state or in any of their common oxidation states. Lanthanoids and actinoids show a progressive filling of 4f and 5f orbitals respectively. Use standard IUPAC conventions for oxidation states and chemical formulae. Pay attention to the distinction between lanthanoids (generally +3 oxidation state) and actinoids (show greater range of oxidation states from +3 to +7).

Q7. The element ________ is the first member of the 3d transition series.
Q8. Potassium dichromate (K₂Cr₂O₇) is a powerful ________ agent in acidic medium.
Q9. The most common oxidation state of lanthanoids is ________.
Q10. Interstitial compounds are formed when small atoms like H, C, N occupy ________ in the metal lattice.
Q11. The magnetic moment of a substance with three unpaired electrons is approximately ________ B.M. (Bohr Magneton).

Section C: Match the Following (5 marks)

Match the items in Column A with the most appropriate items in Column B. Write the correct pairs (e.g., 1-d, 2-a) in your answer sheet. This section tests your ability to connect transition metal compounds with their characteristic colours, uses or properties. Many d-block compounds display striking colours due to d-d transitions; for example, hydrated copper(II) salts are blue, while anhydrous salts are white. Potassium permanganate is deep purple, potassium dichromate is orange. Understanding these visual identifiers helps in qualitative analysis. Similarly, knowing industrial applications — such as TiO₂ in paints, V₂O₅ as catalyst in contact process, and MnO₂ in dry cells — is crucial for Class 12 Chemistry board exams and competitive tests.

Column A: (1) KMnO₄ solution (2) K₂Cr₂O₇ solution (3) Anhydrous CuSO₄ (4) Hydrated CuSO₄ (5) TiO₂
Column B: (a) White powder (b) Blue crystals (c) Purple colour (d) Orange colour (e) White pigment in paints

Section D: Short Answer Questions (3 marks each)

Answer any five questions in 50-80 words each. Allocate approximately 4 minutes per question. These three-mark Class 12 Chemistry solutions require you to explain concepts, write equations or compare properties. Use chemical equations wherever applicable and balance them correctly. When discussing electronic configurations, clearly indicate the orbitals involved and apply Hund's rule and Pauli exclusion principle. For questions on trends (atomic radius, ionisation enthalpy, oxidation states), provide reasoning based on nuclear charge, shielding effect and penetration of orbitals. Remember that transition metals form coloured compounds because their d-orbitals are partially filled, allowing d-d transitions in the visible region of the spectrum. Lanthanoid contraction affects the chemistry of the third transition series, making their atomic radii nearly identical to the second series elements.

Q12. Explain why Zn, Cd and Hg are not considered typical transition elements. (3 marks)
Q13. Write the electronic configuration of Fe²⁺ and Fe³⁺ ions. Which is more stable and why? (3 marks)
Q14. What is lanthanoid contraction? State two consequences of lanthanoid contraction. (3 marks)
Q15. Why do transition metals exhibit variable oxidation states? Illustrate with examples from the first transition series. (3 marks)
Q16. Calculate the magnetic moment (spin-only) for Mn²⁺ ion. (Given: Atomic number of Mn = 25) (3 marks)

Section E: Long Answer and HOTS Questions (5 marks each)

Answer all three questions in 100-150 words each. Allocate 8-10 minutes per question. These five-mark problems demand deep conceptual clarity, analytical reasoning and the ability to integrate multiple topics from NCERT Class 12 Chemistry Chapter 4. HOTS (Higher Order Thinking Skills) questions may ask you to compare trends across periods or groups, predict properties of unknown compounds, or explain anomalies. When discussing trends in ionisation enthalpy or atomic radius, consider factors like effective nuclear charge, electron-electron repulsion and shielding. For questions on compounds like K₂Cr₂O₇ or KMnO₄, write balanced redox equations in acidic or alkaline medium as specified. Distinguish clearly between lanthanoids (all radioactive except Pm, mostly +3 state) and actinoids (all radioactive, wider oxidation state range +3 to +7). Use diagrams or electron-box notation where helpful to illustrate electronic configurations or bonding.

Q17. (a) Why do transition elements form coloured compounds? (b) Explain why Sc³⁺ salts are colourless while Ti³⁺ salts are coloured. (5 marks)
Q18. (a) What are interstitial compounds? (b) Discuss three characteristic properties of interstitial compounds formed by transition metals. (5 marks)
Q19. Compare and contrast the properties of lanthanoids and actinoids under the following heads: (a) Electronic configuration (b) Oxidation states (c) Radioactivity (d) Magnetic properties (e) Chemical reactivity (5 marks)

Section F: Case-Study Based Question (4 marks)

Read the passage carefully and answer the sub-questions that follow. Case-study questions are a recent addition to the CBSE Class 12 Chemistry board exam pattern, testing your ability to apply theoretical knowledge to real-world scenarios. This question integrates concepts of catalysis, oxidation states, electronic configuration and industrial applications of transition metals. The passage below describes how transition metal compounds are used in heterogeneous catalysis for industrial processes such as the Haber process (Fe catalyst for ammonia synthesis), contact process (V₂O₅ for sulphuric acid production) and hydrogenation reactions (Ni catalyst for vegetable oil hardening). Understanding the role of variable oxidation states and availability of vacant d-orbitals in catalytic mechanisms is crucial for answering these sub-questions accurately and scoring full marks in your CBSE 12 Chemistry board exam.

Difficulty Level and Time Allocation

This CBSE Class 12 Chemistry Chapter 4 worksheet is classified as Moderate to Advanced difficulty, suitable for students aiming to score 90+ in their 2025 board exams. The distribution mirrors the actual CBSE Class 12 Chemistry marking scheme: MCQs test recall and basic application (6 marks), fill-in-the-blanks check factual accuracy (5 marks), matching assesses concept linkage (5 marks), short answers demand explanation and calculation (15 marks for five questions), long answers require synthesis and HOTS (15 marks for three questions), and the case-study integrates real-world application (4 marks). Total marks: 50. Suggested time: 90 minutes in one sitting under exam-like conditions. Keep a periodic table, calculator and rough paper handy. After completion, self-assess using the detailed answer key provided in the next section. Identify weak areas and revisit the corresponding NCERT Class 12 Chemistry Chapter 4 sections or Class 12 Chemistry notes before re-attempting those questions.

Section A (MCQs): 6 marks, 8-10 minutes — quick fact-based questions
Section B (Fill-in-the-blanks): 5 marks, 5-6 minutes — requires precise terminology
Section C (Matching): 5 marks, 5 minutes — visual and application-based recall
Section D (Short answers): 15 marks, 20-25 minutes — explanations and calculations
Section E (Long answers): 15 marks, 30-35 minutes — in-depth analysis and comparison
Section F (Case-study): 4 marks, 8-10 minutes — applied reasoning from passage

Answer Key with Explanations

Below is the complete answer key for the CBSE Class 12 Chemistry Chapter 4 worksheet. Each answer includes a brief explanation aligned with NCERT Class 12 Chemistry terminology and the CBSE marking scheme. For MCQs, the correct option is highlighted with reasoning. Fill-in-the-blank answers are exact terms from the chapter. Matching answers show correct pairings. Short-answer and long-answer model responses include key points, chemical equations and numerical workings where applicable. Use this key not just to check correctness but to understand the logic and presentation expected in board exams. If your answer differs in wording but conveys the same scientific meaning, award yourself credit. However, ensure chemical formulae, oxidation states and numerical values are accurate. For case-study sub-questions, compare your reasoning with the provided explanations. This self-assessment process is vital for effective revision and building confidence before the 2025 CBSE Class 12 Chemistry board exam.

Detailed Answer Key: Section A (MCQs)

A1. (d) Always diamagnetic — INCORRECT statement. Most transition metal ions possess unpaired d-electrons, making them paramagnetic. Only species with fully paired electrons (e.g., Zn²⁺, Cu⁺) are diamagnetic. A2. (a) [Ar]3d⁹ is correct. Copper (Z=29) has ground-state configuration [Ar]3d¹⁰4s¹. Removing two electrons to form Cu²⁺ means losing the 4s¹ electron and one from 3d¹⁰, leaving [Ar]3d⁹. A3. (c) Both (a) and (b). Lanthanoid contraction arises because 4f electrons shield nuclear charge poorly, and with increasing atomic number the effective nuclear charge increases, pulling electrons closer and reducing ionic radius progressively from La to Lu. A4. (d) All of these. Thorium, uranium and plutonium are all used in nuclear reactors; uranium-235 and plutonium-239 undergo fission, thorium-232 is a fertile material. A5. (b) Middle members of the 3d series (like Mn in +7 in KMnO₄, Cr in +6 in dichromate) exhibit the highest oxidation states because a greater number of unpaired d and s electrons are available for bonding. A6. (c) Zn²⁺ is [Ar]3d¹⁰ with all electrons paired, hence diamagnetic. Fe²⁺, Cu²⁺ and Ni²⁺ all have unpaired d-electrons and are paramagnetic.

Detailed Answer Key: Sections B, C, D, E and F

Section B Answers: A7. Scandium (Sc). A8. oxidising. A9. +3. A10. interstitial sites (or voids). A11. 3.87 B.M. (μ = √(3×5) = √15 ≈ 3.87). Section C Matching Answers: 1-c (KMnO₄ is purple), 2-d (K₂Cr₂O₇ is orange), 3-a (anhydrous CuSO₄ is white), 4-b (hydrated CuSO₄·5H₂O is blue), 5-e (TiO₂ is white pigment). Section D Model Answers: A12. Zn ([Ar]3d¹⁰4s²), Cd and Hg have completely filled d-orbitals in ground state and common +2 oxidation state ([Ar]3d¹⁰ for Zn²⁺), so no partially filled d-orbitals; hence not typical transition elements. A13. Fe²⁺: [Ar]3d⁶; Fe³⁺: [Ar]3d⁵. Fe³⁺ is more stable due to half-filled d⁵ configuration providing extra stability. A14. Lanthanoid contraction is the steady decrease in size from La³⁺ to Lu³⁺ due to poor shielding by 4f electrons. Consequences: (i) similarity in size of 2nd and 3rd transition series elements; (ii) difficulty in separation of lanthanoids. A15. (n-1)d and ns electrons have similar energies, so both participate in bonding. Example: Mn shows +2 (MnCl₂), +4 (MnO₂), +6 (K₂MnO₄), +7 (KMnO₄). A16. Mn²⁺ is [Ar]3d⁵ with 5 unpaired electrons; μ = √(5×7) = √35 ≈ 5.92 B.M. Section E Model Answers: A17. (a) Partially filled d-orbitals allow d-d transitions absorbing visible light, causing colour. (b) Sc³⁺ [Ar]3d⁰ has no d electrons, no transitions, colourless; Ti³⁺ [Ar]3d¹ has one d electron, d-d transitions occur, coloured. A18. (a) Interstitial compounds form when small non-metal atoms (H, C, N, B) occupy interstitial voids in metal lattices. (b) Properties: (i) Harder than pure metal, (ii) Higher melting point, (iii) Retain metallic conductivity, (iv) Chemically inert. A19. Lanthanoids: 4f¹⁻¹⁴5d⁰⁻¹6s²; mostly +3; one radioactive (Pm); less reactive; smaller magnetic moments. Actinoids: 5f¹⁻¹⁴6d⁰⁻¹7s²; +3 to +7; all radioactive; highly reactive; larger moments. Section F Answers: (i) Vanadium pentoxide (V₂O₅). (ii) Variable oxidation states allow formation of intermediates; vacant d-orbitals adsorb reactants. (iii) Zero oxidation state (elemental Fe).

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Frequently asked questions

What is the recommended time to complete this CBSE Class 12 Chemistry Chapter 4 worksheet?+

Allocate 90 minutes in one sitting under timed, exam-like conditions. This mirrors the pace required in the actual CBSE board exam. Section-wise: MCQs 8-10 min, fill-blanks 5-6 min, matching 5 min, short answers 20-25 min, long answers 30-35 min, case-study 8-10 min. Practice under time pressure to build speed and accuracy.

How many marks does this worksheet carry and how does it align with the CBSE board exam pattern?+

This worksheet totals 50 marks, structured to reflect the actual CBSE Class 12 Chemistry paper: 6 MCQs (6 marks), 5 fill-in-the-blanks (5 marks), matching (5 marks), 5 short answers (15 marks), 3 long answers (15 marks) and one case-study (4 marks). The distribution and difficulty match recent board exam trends, making it ideal for realistic self-assessment.

Why are Zn, Cd and Hg not classified as true transition metals?+

According to IUPAC, transition elements must have a partially filled d sub-shell in the ground state or in common oxidation states. Zn, Cd and Hg have electronic configurations ending in (n-1)d¹⁰ ns², and their common +2 ions have fully filled d¹⁰ configurations, leaving no unpaired d-electrons. Hence they lack typical transition metal properties like variable oxidation states, coloured compounds and paramagnetism.

What is lanthanoid contraction and why is it important in the chemistry of the third transition series?+

Lanthanoid contraction is the gradual decrease in atomic and ionic radii across the lanthanoid series (La to Lu) despite increasing atomic number, caused by poor shielding of nuclear charge by 4f electrons. This contraction results in the third transition series elements (Hf to Hg) having nearly the same atomic radii as their second-series counterparts (Zr to Cd), leading to similar chemical properties and difficulties in separating these pairs.

How do I calculate the magnetic moment (spin-only formula) for transition metal ions?+

First write the electronic configuration of the ion. Count the number of unpaired electrons (n). Apply the spin-only formula: μ = √(n(n+2)) Bohr Magneton (B.M.). For example, Fe²⁺ is [Ar]3d⁶ with 4 unpaired electrons, so μ = √(4×6) = √24 ≈ 4.90 B.M. This formula is valid when orbital contribution to magnetic moment is negligible, which is generally the case for first-row transition metals.

Why do transition metals form coloured compounds while s-block and p-block elements generally do not?+

Transition metals have partially filled d-orbitals. When white light falls on their compounds, electrons absorb specific wavelengths to undergo d-d transitions (promotion from lower-energy to higher-energy d-orbitals). The complementary colour of the absorbed wavelength is transmitted or reflected, giving the compound its characteristic colour. s- and p-block ions typically have completely filled or empty d-orbitals, so d-d transitions are not possible, making them colourless.

What are interstitial compounds and how do they differ from typical ionic or covalent compounds?+

Interstitial compounds form when small non-metal atoms (like H, C, N, B) occupy the interstitial voids (spaces) in the crystal lattice of transition metals. Unlike ionic compounds, they retain metallic properties such as electrical conductivity and lustre. Unlike typical covalent compounds, they are very hard, have high melting points and are chemically inert. Examples include TiH₁.₇, Fe₃C (cementite in steel) and VH₀.₅₆.

How can I effectively use the answer key provided in this worksheet for self-assessment?+

After completing the worksheet under timed conditions, check each answer against the key. Award yourself marks as per the marking scheme. For MCQs, note why the correct option is right and others are wrong. For numerical problems, verify every step of your working. For descriptive answers, compare your reasoning, equations and key points with the model answer. Identify patterns in mistakes (e.g., calculation errors, incomplete equations) and revise those NCERT Class 12 Chemistry sections before re-attempting similar questions.

Which topics from Chapter 4 are most heavily weighted in CBSE Class 12 Chemistry board exams?+

Recent CBSE board papers emphasise electronic configurations, calculation of magnetic moments, explanation of lanthanoid contraction, properties and reactions of compounds like K₂Cr₂O₇ and KMnO₄, variable oxidation states, and reasons for colour and catalytic activity. Case-study questions often integrate industrial applications (catalysts, alloys). Focus on these high-weightage areas and practice numerical problems involving oxidation-state determination and magnetic moment calculations.

Can I use this worksheet for quick revision before the board exam or should I treat it as a full practice test?+

Treat it as a full 90-minute practice test at least twice: once during initial revision (after completing NCERT Chapter 4) and again 10-15 days before the board exam. For quick revision, attempt only Section A (MCQs), Section B (fill-blanks) and Section D (short answers) in 30 minutes. The case-study and HOTS questions in Section E are best tackled when you have time to think analytically, so save them for comprehensive practice sessions rather than last-minute cramming.

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