d- and f-Block Elements Practice

Take timed practice tests on d- and f-Block Elements for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

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d- and f-Block Elements Practice Tests

Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.

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JEE Main Session 1

3Questions

60s/QTimer

+4 / -1Scoring

Preview all 12 questions in JEE Main Session 1 (no login required)

Electronic configuration
1. Which of the following is NOT an exception to Aufbau principle for 3d metals?
- A. Ni (d⁸s²) (Correct)
- B. Cr (d⁵s¹)
- C. Cu (d¹⁰s¹)
- D. Mo (d⁵s¹)
Explanation: Cr: [Ar]3d⁵4s¹ (not 3d⁴4s²) — extra stability of half-filled d. Cu: [Ar]3d¹⁰4s¹ (not 3d⁹4s²) — extra stability of fully-filled d. Ni: [Ar]3d⁸4s² — no exception, follows Aufbau as expected.
Variable oxidation states
2. Transition metals show variable oxidation states because:
- A. Both (n−1)d and ns electrons participate in bonding (Correct)
- B. Their d-orbitals are empty
- C. They have large atomic radii
- D. They are metals
Explanation: Transition metals can lose electrons from both the ns and (n−1)d subshells, which are close in energy. This allows multiple stable oxidation states. E.g., Mn: +2 (d⁵), +4 (MnO₂), +6, +7 (KMnO₄).
Colour of compounds
3. Transition metal compounds are often coloured because:
- A. d-d transitions absorb visible light (complementary colour observed) (Correct)
- B. They reflect all light
- C. They are ionic
- D. Large lattice energy
Explanation: In a crystal field, d-orbitals split into different energy levels. Electrons absorb photons of visible light for d-d transitions → the complementary colour is observed. Zn²⁺ (d¹⁰), Sc³⁺ (d⁰) have no d-d transitions → colourless.
Catalytic properties
4. Transition metals and their compounds are excellent catalysts because of:
- A. Variable oxidation states allowing electron transfer AND ability to adsorb reactants on surfaces (Correct)
- B. High melting points
- C. Being noble metals
- D. Large atomic radii
Explanation: Transition metal catalysts work by: (1) changing oxidation state (homogeneous catalysis: Fe²⁺/Fe³⁺ in Fenton's reagent), (2) adsorbing reactants on d-orbital-rich surfaces (heterogeneous: Fe in Haber, Pt in Ostwald).
Magnetic properties
5. The spin-only magnetic moment formula is:
- A. μ = √(n(n+2)) BM, where n = number of unpaired electrons (Correct)
- B. μ = n BM
- C. μ = 2n BM
- D. μ = n(n+1) BM
Explanation: Spin-only magnetic moment: μ = √(n(n+2)) BM (Bohr magnetons), where n = number of unpaired electrons. For Fe³⁺ (d⁵, 5 unpaired): μ = √(5×7) = √35 ≈ 5.92 BM.
Lanthanide contraction
6. Lanthanide contraction causes Zr and Hf to have:
- A. Nearly identical atomic radii and properties (Correct)
- B. Very different radii
- C. Different oxidation states
- D. Hf to be more reactive than Zr
Explanation: Lanthanide contraction: the 14 lanthanides (4f filling) cause a gradual size decrease. By the time we reach 5d transition metals (Hf, Ta, W...), their radii are nearly equal to the 4d counterparts (Zr, Nb, Mo...). Zr and Hf have almost identical radii → very similar chemistry → hardest pair to separate.
JEE Main
7. KMnO₄ acts as oxidising agent in acidic, neutral, and basic media but gives different products. In basic medium, Mn goes from +7 to:
- A. +6 (manganate MnO₄²⁻) (Correct)
- B. +2 (Mn²⁺)
- C. +4 (MnO₂)
- D. 0 (Mn)
Explanation: KMnO₄ reductions: Acidic → Mn²⁺ (+2, 5e⁻ per Mn). Neutral/weakly basic → MnO₂ (+4, 3e⁻). Strongly basic → MnO₄²⁻ (+6, 1e⁻). Acidic conditions give strongest oxidation, used most in volumetric analysis.
Interstitial compounds
8. Interstitial compounds of transition metals (e.g., TiC, Fe₃C) are:
- A. Hard, high-melting, chemically inert, with non-stoichiometric compositions (Correct)
- B. Soft and reactive
- C. Ionic compounds
- D. Molecular compounds
Explanation: Interstitial compounds: small atoms (C, N, H) fit into interstitial voids of the transition metal lattice. Properties: very hard (reinforced structure), high melting points, retain metallic conductivity, chemically less reactive. TiN (gold colour), WC (cutting tools).
JEE Advanced
9. Among 3d transition metals, the one with the highest number of unpaired electrons is:
- A. Cr (3d⁵4s¹, 6 unpaired total) (Correct)
- B. Mn (3d⁵4s², 5 unpaired in d)
- C. Fe (3d⁶4s², 4 unpaired)
- D. Cu (3d¹⁰4s¹, 1 unpaired)
Explanation: Cr: [Ar]3d⁵4s¹ → 5 unpaired in 3d + 1 in 4s = 6 total unpaired electrons. Mn has 5d + 2s electrons but the 4s are paired → 5 unpaired. So Cr has most unpaired e⁻ among 3d metals.
Mixed
10. The f-block elements (lanthanides and actinides) are so called because:
- A. Their differentiating electrons fill f-orbitals (Correct)
- B. They have f in their name
- C. They all have f-shaped nuclei
- D. They are in the last f group
Explanation: In lanthanides (Ce to Lu), the 4f orbitals are being filled. In actinides (Th to Lr), the 5f orbitals are filled. The 'differentiating' (last added) electron enters the f subshell.
Mixed
11. Transition metals in higher oxidation states form:
- A. Covalent/oxo compounds (like CrO₃, Mn₂O₇) (Correct)
- B. Simple ionic compounds only
- C. Insoluble precipitates
- D. No compounds
Explanation: In higher oxidation states (+6, +7), the metal ion has high charge density and polarises anions significantly → covalent bonds form. CrO₃ (Cr +6), Mn₂O₇ (Mn +7, an oily liquid), OsO₄ (Os +8) are covalent. Lower states (+2) → ionic (FeCl₂, MnCl₂).
Mixed
12. The most stable oxidation state of cerium (Ce) is:
- A. +4 (Correct)
- B. +3
- C. +2
- D. +6
Explanation: Ce can exist in both +3 (4f¹) and +4 (4f⁰, empty f-shell = extra stability due to noble-gas-like configuration) states. CeO₂ (Ce⁴⁺) is a strong oxidising agent. +4 is the most stable for Ce, unlike most lanthanides which prefer +3.

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JEE Main Session 1

3Questions

60s/QTimer

+4 / -1Scoring

Preview all 12 questions in JEE Main Session 1 (no login required)

Electronic configuration
1. Which of the following is NOT an exception to Aufbau principle for 3d metals?
- A. Ni (d⁸s²) (Correct)
- B. Cr (d⁵s¹)
- C. Cu (d¹⁰s¹)
- D. Mo (d⁵s¹)
Explanation: Cr: [Ar]3d⁵4s¹ (not 3d⁴4s²) — extra stability of half-filled d. Cu: [Ar]3d¹⁰4s¹ (not 3d⁹4s²) — extra stability of fully-filled d. Ni: [Ar]3d⁸4s² — no exception, follows Aufbau as expected.
Variable oxidation states
2. Transition metals show variable oxidation states because:
- A. Both (n−1)d and ns electrons participate in bonding (Correct)
- B. Their d-orbitals are empty
- C. They have large atomic radii
- D. They are metals
Explanation: Transition metals can lose electrons from both the ns and (n−1)d subshells, which are close in energy. This allows multiple stable oxidation states. E.g., Mn: +2 (d⁵), +4 (MnO₂), +6, +7 (KMnO₄).
Colour of compounds
3. Transition metal compounds are often coloured because:
- A. d-d transitions absorb visible light (complementary colour observed) (Correct)
- B. They reflect all light
- C. They are ionic
- D. Large lattice energy
Explanation: In a crystal field, d-orbitals split into different energy levels. Electrons absorb photons of visible light for d-d transitions → the complementary colour is observed. Zn²⁺ (d¹⁰), Sc³⁺ (d⁰) have no d-d transitions → colourless.
Catalytic properties
4. Transition metals and their compounds are excellent catalysts because of:
- A. Variable oxidation states allowing electron transfer AND ability to adsorb reactants on surfaces (Correct)
- B. High melting points
- C. Being noble metals
- D. Large atomic radii
Explanation: Transition metal catalysts work by: (1) changing oxidation state (homogeneous catalysis: Fe²⁺/Fe³⁺ in Fenton's reagent), (2) adsorbing reactants on d-orbital-rich surfaces (heterogeneous: Fe in Haber, Pt in Ostwald).
Magnetic properties
5. The spin-only magnetic moment formula is:
- A. μ = √(n(n+2)) BM, where n = number of unpaired electrons (Correct)
- B. μ = n BM
- C. μ = 2n BM
- D. μ = n(n+1) BM
Explanation: Spin-only magnetic moment: μ = √(n(n+2)) BM (Bohr magnetons), where n = number of unpaired electrons. For Fe³⁺ (d⁵, 5 unpaired): μ = √(5×7) = √35 ≈ 5.92 BM.
Lanthanide contraction
6. Lanthanide contraction causes Zr and Hf to have:
- A. Nearly identical atomic radii and properties (Correct)
- B. Very different radii
- C. Different oxidation states
- D. Hf to be more reactive than Zr
Explanation: Lanthanide contraction: the 14 lanthanides (4f filling) cause a gradual size decrease. By the time we reach 5d transition metals (Hf, Ta, W...), their radii are nearly equal to the 4d counterparts (Zr, Nb, Mo...). Zr and Hf have almost identical radii → very similar chemistry → hardest pair to separate.
JEE Main
7. KMnO₄ acts as oxidising agent in acidic, neutral, and basic media but gives different products. In basic medium, Mn goes from +7 to:
- A. +6 (manganate MnO₄²⁻) (Correct)
- B. +2 (Mn²⁺)
- C. +4 (MnO₂)
- D. 0 (Mn)
Explanation: KMnO₄ reductions: Acidic → Mn²⁺ (+2, 5e⁻ per Mn). Neutral/weakly basic → MnO₂ (+4, 3e⁻). Strongly basic → MnO₄²⁻ (+6, 1e⁻). Acidic conditions give strongest oxidation, used most in volumetric analysis.
Interstitial compounds
8. Interstitial compounds of transition metals (e.g., TiC, Fe₃C) are:
- A. Hard, high-melting, chemically inert, with non-stoichiometric compositions (Correct)
- B. Soft and reactive
- C. Ionic compounds
- D. Molecular compounds
Explanation: Interstitial compounds: small atoms (C, N, H) fit into interstitial voids of the transition metal lattice. Properties: very hard (reinforced structure), high melting points, retain metallic conductivity, chemically less reactive. TiN (gold colour), WC (cutting tools).
JEE Advanced
9. Among 3d transition metals, the one with the highest number of unpaired electrons is:
- A. Cr (3d⁵4s¹, 6 unpaired total) (Correct)
- B. Mn (3d⁵4s², 5 unpaired in d)
- C. Fe (3d⁶4s², 4 unpaired)
- D. Cu (3d¹⁰4s¹, 1 unpaired)
Explanation: Cr: [Ar]3d⁵4s¹ → 5 unpaired in 3d + 1 in 4s = 6 total unpaired electrons. Mn has 5d + 2s electrons but the 4s are paired → 5 unpaired. So Cr has most unpaired e⁻ among 3d metals.
Mixed
10. The f-block elements (lanthanides and actinides) are so called because:
- A. Their differentiating electrons fill f-orbitals (Correct)
- B. They have f in their name
- C. They all have f-shaped nuclei
- D. They are in the last f group
Explanation: In lanthanides (Ce to Lu), the 4f orbitals are being filled. In actinides (Th to Lr), the 5f orbitals are filled. The 'differentiating' (last added) electron enters the f subshell.
Mixed
11. Transition metals in higher oxidation states form:
- A. Covalent/oxo compounds (like CrO₃, Mn₂O₇) (Correct)
- B. Simple ionic compounds only
- C. Insoluble precipitates
- D. No compounds
Explanation: In higher oxidation states (+6, +7), the metal ion has high charge density and polarises anions significantly → covalent bonds form. CrO₃ (Cr +6), Mn₂O₇ (Mn +7, an oily liquid), OsO₄ (Os +8) are covalent. Lower states (+2) → ionic (FeCl₂, MnCl₂).
Mixed
12. The most stable oxidation state of cerium (Ce) is:
- A. +4 (Correct)
- B. +3
- C. +2
- D. +6
Explanation: Ce can exist in both +3 (4f¹) and +4 (4f⁰, empty f-shell = extra stability due to noble-gas-like configuration) states. CeO₂ (Ce⁴⁺) is a strong oxidising agent. +4 is the most stable for Ce, unlike most lanthanides which prefer +3.