d- and f-block Elements (NCERT Class 12 Chemistry)

Short Answer Q&A – d- and f-block Elements (NCERT Class 12 Chemistry)

1. Q: What are transition elements?
   A: Elements whose atoms or ions (in a stable oxidation state) contain partially filled d-orbitals.
2. Q: Give the general outer electronic configuration of d-block elements.
   A: (n-1)d^1–10 ns^1–2.
3. Q: Why are Zn, Cd, and Hg not considered true transition elements?
   A: They have completely filled d-orbitals (d¹⁰) in both atomic and common oxidation states (+2).
4. Q: Why is Ag considered a transition element?
   A: It exhibits the +2 oxidation state (Ag²⁺), with partially filled 4d⁹ configuration.
5. Q: Write the electronic configuration of Cr (Z=24).
   A: [Ar] 3d⁵ 4s¹ (Half-filled stability).
6. Q: Write the electronic configuration of Cu (Z=29).
   A: [Ar] 3d¹⁰ 4s¹ (Fully-filled stability).
7. Q: Write the electronic configuration of Mn²⁺ (Z=25).
   A: [Ar] 3d⁵.
8. Q: Which 3d series element shows the maximum number of oxidation states?
   A: Manganese (Mn), from +2 to +7.
9. Q: What is the most common oxidation state of the 3d series elements?
   A: +2.
10. Q: Why do transition elements exhibit variable oxidation states?
    A: Due to the small energy difference between (n-1)d and ns orbitals; both participate in bonding.
11. Q: Why do transition metals have high melting points?
    A: Due to strong interatomic metallic bonding involving a large number of unpaired d-electrons.
12. Q: Explain the anomalous low melting points of Mn and Tc.
    A: Their stable d⁵ configuration restricts the number of electrons for metallic bonding.
13. Q: What is the general trend in atomic radii across a transition series?
    A: Decreases, then remains nearly constant, then increases slightly at the end.
14. Q: Define the Lanthanoid Contraction.
    A: The gradual decrease in atomic and ionic radii across the Lanthanoid series.
15. Q: What is the main cause of the Lanthanoid Contraction?
    A: Poor shielding effect of 4f electrons.
16. Q: What is a major consequence of Lanthanoid Contraction?
    A: 4d and 5d elements (e.g., Zr and Hf) have almost identical atomic radii.
17. Q: Why is ionization enthalpy of 5d elements higher than 3d and 4d?
    A: Due to the greater effective nuclear charge caused by Lanthanoid Contraction.
18. Q: Why is Cr²⁺ a strong reducing agent?
    A: It changes from d⁴ to stable half-filled d³ configuration (Cr³⁺).
19. Q: Why is Mn³⁺ a strong oxidizing agent?
    A: It changes from d⁴ to the stable half-filled d⁵ configuration (Mn²⁺).
20. Q: Spin-only magnetic moment for Fe³⁺ (Z=26)?
    A: Fe³⁺ is 3d⁵ (n=5), μ = √35 BM (≈ 5.92 BM).
21. Q: What is the source of colour in transition metal ions?
    A: d-d transitions of electrons after absorbing light energy.
22. Q: Why are Sc³⁺ and Ti⁴⁺ ions colourless?
    A: d⁰ configurations, so no d-d transitions are possible.
23. Q: Why are transition metals used as catalysts?
    A: Due to variable oxidation states and large surface area.
24. Q: What are interstitial compounds?
    A: Non-stoichiometric compounds formed by trapping small atoms (H, C, N) in crystal voids.
25. Q: One property of interstitial compounds?
    A: Very hard and chemically inert.
26. Q: Why do transition metals form alloys easily?
    A: Similar atomic sizes allow substitution in the crystal lattice.
27. Q: Which 3d series element has a positive standard electrode potential?
    A: Only Copper (Cu).
28. Q: Why can’t Cu liberate H₂ from acids?
    A: Positive standard electrode potential ($E^o_{\text{Cu}^{2+}/\text{Cu}}$); reaction non-spontaneous.
29. Q: Which oxide is more basic: CrO or CrO₃?
    A: CrO.
30. Q: Nature of oxide Mn₂O₇?
    A: Highly acidic (+7 oxidation state).
31. Q: Formula of the ore for potassium dichromate?
    A: Chromite (FeCr₂O₄).
32. Q: Oxidation state of Cr in K₂Cr₂O₇?
    A: +6.
33. Q: Colour of chromate ion (CrO₄^2–)?
    A: Yellow.
34. Q: How to convert chromate ion to dichromate ion?
    A: Add acid (H⁺).
35. Q: Colour of dichromate ion (Cr₂O₇^2–)?
    A: Orange.
36. Q: Why is K₂Cr₂O₇ a strong oxidizing agent in acidic medium?
    A: Cr(VI) readily reduces to stable Cr(III).
37. Q: Ore for potassium permanganate?
    A: Pyrolusite (MnO₂).
38. Q: Colour of permanganate ion (MnO₄⁻)?
    A: Deep purple.
39. Q: Product when KMnO₄ reacts in strongly acidic medium?
    A: Mn²⁺ (oxidation state: +2).
40. Q: Product when KMnO₄ reacts in neutral medium?
    A: MnO₂ (brown precipitate).
41. Q: Which is more stable in aqueous solution, Cu(II) or Cu(I)?
    A: Cu(II), due to higher hydration enthalpy.
42. Q: Why are Zn²⁺ salts generally white?
    A: Zn²⁺ has filled 3d¹⁰, so no d-d transitions.
43. Q: Reason for green colour change when acidified K₂Cr₂O₇ is reduced?
    A: Formation of Cr³⁺.
44. Q: Green species formed during preparation of KMnO₄?
    A: Manganate ion (MnO₄^2–).
45. Q: 3d series element with lowest enthalpy of atomisation?
    A: Zinc (Zn).
46. Q: Why is the E^o_Mn²⁺/Mn value more negative than expected?
    A: Extra stability of the half-filled Mn²⁺ (3d⁵).
47. Q: Name a transition element present in Vitamin B₁₂.
    A: Cobalt (Co).
48. Q: Type of oxide: Cr₂O₃?
    A: Amphoteric (Cr in +3 state).
49. Q: How does stability of higher oxidation states change down a group (e.g., Group 6)?
    A: Increases (e.g., W(VI) more stable than Cr(VI)).
50. Q: Example of a paramagnetic transition metal ion?
    A: Fe²⁺ or Ti³⁺.
51. Q: Why are f-block elements called inner transition elements?
    A: Because they involve filling of the inner (n–2)f subshell.
52. Q: General outer electronic configuration of f-block elements?
    A: (n–2)f^1–14 (n–1)d^0–1 ns².
53. Q: Common name for elements Ce (Z=58) to Lu (Z=71)?
    A: Lanthanoids (Lanthanides).
54. Q: Most common oxidation state of Lanthanoids?
    A: +3.
55. Q: Why is the +3 state most stable for Lanthanoids?
    A: Often corresponds to stable f⁰, f⁷, or f¹⁴ configurations.
56. Q: Lanthanoid with stable +2 oxidation state?
    A: Europium (Eu²⁺) or Ytterbium (Yb²⁺).
57. Q: Lanthanoid with stable +4 state?
    A: Cerium (Ce⁴⁺) or Terbium (Tb⁴⁺).
58. Q: Why does Ce⁴⁺ act as a strong oxidizing agent?
    A: It reverts easily to stable Ce³⁺.
59. Q: General trend in basic character of Lanthanoid hydroxides?
    A: Basic character decreases from La(OH)₃ to Lu(OH)₃.
60. Q: Reason for decreasing basic character of Lanthanoid hydroxides?
    A: Lanthanoid Contraction decreases ionic size and increases covalent character.
61. Q: Why are most Lanthanoid ions paramagnetic?
    A: Presence of unpaired f-electrons.
62. Q: Diamagnetic Lanthanoid ions?
    A: Lu³⁺ (4f¹⁴) and Ce⁴⁺ (4f⁰).
63. Q: What is Mischmetal?
    A: Alloy containing ~95% Lanthanoids and 5% iron.
64. Q: Main use of Mischmetal?
    A: Production of lighter flints and tracer bullets.
65. Q: Common name for elements Th (Z=90) to Lr (Z=103)?
    A: Actinoids (Actinides).
66. Q: Most common oxidation state of Actinoids?
    A: +3.
67. Q: Why do Actinoids show more oxidation states than Lanthanoids?
    A: 5f, 6d, and 7s have comparable energy.
68. Q: Key difference (radioactivity) between Lanthanoids and Actinoids?
    A: All Actinoids are radioactive; only Promethium (Pm) among Lanthanoids is radioactive.
69. Q: Why is Actinoid Contraction greater than Lanthanoid Contraction?
    A: 5f orbitals shield poorly compared to 4f.
70. Q: What are transuranium elements?
    A: Atomic number > 92 (Uranium).
71. Q: Common configuration for M²⁺ ions in the 3d series?
    A: 3dⁿ (Electrons lost from 4s first).
72. Q: Which 3d element forms only +3 oxidation state?
    A: Scandium (Sc).
73. Q: Why is separating Lanthanoids difficult?
    A: Close similarity in chemical properties due to Lanthanoid Contraction.
74. Q: Fe²⁺ solution + acidified K₂Cr₂O₇: What happens?
    A: Orange K₂Cr₂O₇ reduced to green Cr³⁺.
75. Q: SO₂ gas passed through acidified K₂Cr₂O₇ solution?
    A: Orange changes to green (Cr(VI) to Cr(III)).
76. Q: Nature of oxide MnO?
    A: Basic (+2 state).
77. Q: Common term for coloured Ti and V compounds?
    A: Coloured pigments for paints and dyes.
78. Q: Most acidic oxide of Chromium?
    A: CrO₃ (+6 state).
79. Q: Transition element essential in chlorophyll?
    A: Magnesium (Mg) (typically compared; Mg is s-block).
80. Q: Why do Actinoid ions form more stable complexes than Lanthanoids?
    A: Higher charge density and tendency for covalent bonding.
81. Q: Orbital contribution to magnetic moment is significant for?
    A: Actinoid ions and heavy transition metal ions.
82. Q: Main source of colour in Lanthanoid ions?
    A: f-f transitions.
83. Q: Electronic configuration of Lu³⁺ (Z=71)?
    A: [Xe] 4f¹⁴.
84. Q: Electronic configuration of Gd³⁺ (Z=64)?
    A: [Xe] 4f⁷.
85. Q: Element used in catalytic hydrogenation of vegetable oils?
    A: Nickel (Ni) or Palladium (Pd).
86. Q: Effect of oxidation state on covalent character?
    A: Covalent character increases with higher oxidation state.
87. Q: Most common Lanthanoid ion in optical devices (lasers)?
    A: Neodymium (Nd³⁺).
88. Q: Highest oxidation state in transition metals?
    A: +8 (OsO₄, RuO₄).
89. Q: Chemical nature of V₂O₅?
    A: Amphoteric (+5 state).
90. Q: Two metal oxides used as pigments?
    A: Cr₂O₃ (green), CoO (blue).
91. Q: Why is Ti³⁺ (3d¹) paramagnetic and coloured?
    A: One unpaired electron; d-d transition possible.
92. Q: Which 3d element has only one oxidation state besides zero?
    A: Scandium (+3).
93. Q: Why are Actinoids highly electropositive?
    A: Relatively low ionization energies.
94. Q: Example: Fe alloy used in magnetic materials?
    A: Alnico (Al, Ni, Co, Fe).
95. Q: Atomic volume trend from Sc to Zn?
    A: Decreases toward center and increases at end.
96. Q: Catalyst in Contact Process (H₂SO₄ production)?
    A: V₂O₅.
97. Q: Maximum oxidation state shown by Fe?
    A: +6 (FeO₄^2–).
98. Q: Why is TiO₂ an ideal white pigment?
    A: Ti⁴⁺ is d⁰, making it colourless and highly reflective.
99. Q: Radioactive Lanthanoid?
    A: Promethium (Pm).
100. Q: Actinoid used as nuclear fuel?
     A: Uranium (U) and Plutonium (Pu).
101. Q: Tendency of transition metals to form compounds with other metals?
     A: Alloys.
102. Q: Write the configuration of Cr³⁺.
     A: [Ar] 3d³.
103. Q: What determines the number of unpaired electrons in a transition metal ion?
     A: Oxidation state and surrounding ligands (Crystal Field Theory).
104. Q: Trend in first ionization enthalpy (3d elements)?
     A: Generally increases, with exceptions at Cr and Cu.
105. Q: Why is E^o_{M³⁺/M²⁺} difference small between Cr and Fe?
     A: Stable d³ and d⁵ configurations.
106. Q: Most stable oxidation state of Ni?
     A: +2.
107. Q: Configuration of Eu²⁺?
     A: [Xe] 4f⁷.
108. Q: Chemical similarity between Lanthanoids and Actinoids?
     A: Both form stable +3 ions.