Chapter 4 Group d and f block elements Previous Year Questions

Chapter 8 – d- and f-Block Elements questions, structured neatly for exam preparation.

🎯 VERY SHORT ANSWER TYPE QUESTIONS (1 Mark)

Electronic configuration of Cr³⁺ (Z = 24):
Cr: [Ar] 3d⁵ 4s¹ → Cr³⁺: [Ar] 3d³
CuSO₄·5H₂O is blue but ZnSO₄ and anhydrous CuSO₄ are colourless:
Due to the presence of unpaired d-electrons in Cu²⁺ causing d–d transitions, giving blue colour. Zn²⁺ has a 3d¹⁰ configuration — no d–d transitions.
Mn third ionisation energy is high because:
The third electron is removed from the stable half-filled 3d⁵ configuration (Mn²⁺), requiring high energy.
Highest oxidation state in 3d series:
Manganese (Mn) shows +7 oxidation state (e.g., in KMnO₄).
Why Ag is a transition metal despite 4d¹⁰ configuration:
In some compounds, Ag exhibits +2 oxidation state (Ag²⁺ = 4d⁹), involving d-orbitals in bonding.
Low atomisation enthalpy of Zn:
Zn has a full d¹⁰ configuration → poor metallic bonding → low ΔₐH.
Element that doesn't show variable oxidation state among Cr, Co, Zn:
Zn (only +2) — due to stable d¹⁰ configuration.
Stable +3 oxidation states of La, Gd, Lu:
Due to extra stability from empty, half-filled, or completely filled f-orbitals respectively.
Consequence of Lanthanoid Contraction:
Poor shielding by 4f-electrons → similar atomic radii → difficulty in separation (lanthanoid resemblance).
Why 5d series elements have higher 1st ionisation enthalpy:
Due to increased effective nuclear charge from lanthanoid contraction.
Why Mn²⁺ is more stable than Fe²⁺ in oxidation to +3 state:
Mn²⁺ has a stable 3d⁵ half-filled configuration → more stable.
Magnetic moment of Cu²⁺ (Z = 29):
Cu²⁺ = 3d⁹ → 1 unpaired e⁻
μ = √(n(n+2)) = √3 = 1.73 BM
Shape of chromate ion (CrO₄²⁻):
Tetrahedral
Why V₂O₅ acts as catalyst:
Due to variable oxidation states of vanadium → easily accept/lose electrons.
What are interstitial compounds:
Compounds formed when small atoms (H, C, N) occupy interstitial spaces in metal lattices.
Two reasons transition metals are good catalysts:
(i) Variable oxidation states
(ii) Provide surface for reactants to adsorb and react.
Equation: Thiosulphate + alkaline KMnO₄:
8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 2OH⁻ + 6SO₄²⁻
Ore for K₂Cr₂O₇ preparation:
Chromite, FeCr₂O₄
Electronic configuration of Lu³⁺ (Z = 71):
Lu: [Xe] 4f¹⁴ 5d¹ 6s² → Lu³⁺ = [Xe] 4f¹⁴
Most common oxidation state of actinoids:
+3
Catalysts used in processes:
(a) Contact Process (H₂SO₄) → V₂O₅
(b) Polythene → Ziegler-Natta catalyst
Lanthanoid showing +4 oxidation state:
Cerium (Ce)
One ore each of Mn and Cr:
Mn → Pyrolusite (MnO₂)
Cr → Chromite (FeCr₂O₄)
Why Cd²⁺ ion is white:
Cd²⁺ has d¹⁰ configuration → no d–d transitions → colourless.
Structure of dichromate ion (Cr₂O₇²⁻):

(Sketch required in actual exam — V-shape with shared oxygen)
Basic character of monoxides (descending):
TiO > VO > CrO > FeO

✅ Short Answer Type Questions (3 Marks)

Q1. Account for the following:

(a) La(OH)₃ is more basic than Lu(OH)₃
(b) Zn²⁺ salts are white
(c) Cu(I) compounds are unstable in aqueous solution and undergo disproportionation

Answer:
(a) La³⁺ has a larger ionic size than Lu³⁺ due to lanthanoid contraction. Larger size means weaker hold on OH⁻ → more basic hydroxide.
(b) Zn²⁺ has a 3d¹⁰ configuration (no unpaired electrons) → no d–d transitions → salts are white.
(c) Cu⁺ disproportionates:
2Cu⁺ → Cu²⁺ + Cu
Because Cu⁺ is unstable in aqueous medium due to high hydration energy of Cu²⁺.

Q2. Describe the oxidising action of potassium dichromate with following. Write ionic equations for its reactions with:

(a) Iodide ion
(b) Iron (II)
(c) H₂S

Answer:
(a) 6I⁻ + Cr₂O₇²⁻ + 14H⁺ → 3I₂ + 2Cr³⁺ + 7H₂O
(b) 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
(c) 3H₂S + Cr₂O₇²⁻ + 8H⁺ → 3S + 2Cr³⁺ + 7H₂O

Q3. Deduce the number of 3d electrons in the following ions:

(a) Fe³⁺
(b) Cu²⁺
(c) Sc³⁺

Answer:
Fe (Z = 26) → Fe³⁺ = [Ar] 3d⁵ → 5 electrons
Cu (Z = 29) → Cu²⁺ = [Ar] 3d⁹ → 9 electrons
Sc (Z = 21) → Sc³⁺ = [Ar] → 0 electrons

Q4. Why do transition metals form alloys? Write any two characteristics of interstitial compounds.

Answer:
Transition metals have similar atomic sizes and crystal structures → atoms can replace each other to form alloys.

Characteristics of interstitial compounds:

Hard and rigid due to non-metal atoms fitting into metal lattice.
High melting and boiling points; often retain metallic conductivity.

Q5. In the reaction:

3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
(a) Why Mn(VI) changes to Mn(VII) and Mn(IV)?
(b) What special name is given to such type of reactions?

Answer:
(a) Mn shows variable oxidation states. Here, Mn(VI) undergoes simultaneous oxidation to Mn(VII) and reduction to Mn(IV).
(b) This is a disproportionation reaction.

Q6. What happens when:

(a) Thiosulphate ions react with alkaline KMnO₄
(b) Ferrous oxalate reacts with acidified KMnO₄
(c) Sulphurous acid reacts with acidified KMnO₄
Write balanced equations.

Answer:
(a) 3S₂O₃²⁻ + 8KMnO₄ + 5H₂O → 6SO₄²⁻ + 8MnO₂ + 2OH⁻
(b) 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O
(c) 5H₂SO₃ + 2KMnO₄ + 3H₂SO₄ → 5H₂SO₄ + 2MnSO₄ + K₂SO₄ + 3H₂O

Q7. Name the catalysts used in the following processes:

(a) Manufacture of ammonia by Haber’s Process
(b) Oxidation of ethyne to ethanol
(c) Photographic industry

Answer:
(a) Iron (Fe) with molybdenum as promoter
(b) Acidified KMnO₄
(c) AgBr with light (acts as light-sensitive catalyst)

Q8. Among TiCl₄, VCl₃, and FeCl₂, which one will be drawn more strongly into a magnetic field and why?

Answer:
Fe²⁺ has 3d⁶ → 4 unpaired electrons → maximum magnetic moment (4.9 BM)
Thus, FeCl₂ is drawn most strongly due to highest number of unpaired electrons.

Q9. Complete the following equations:

(a) MnO₄²⁻ + H⁺ → .............
(b) KMnO₄ → (on heating) ............
(c) MnO₄⁻ + Fe²⁺ + C₂O₄²⁻ + H⁺ → ..........

Answer:
(a) 2MnO₄²⁻ + 4H⁺ → MnO₄⁻ + MnO₂ + 2H₂O
(b) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
(c) 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ + 10Fe²⁺ → 2Mn²⁺ + 10CO₂ + 10Fe³⁺ + 8H₂O

Q10. How do you account for the following?

(a) Cr²⁺ is a reducing agent while Mn³⁺ is an oxidising agent
(b) Actinoids show more oxidation states than lanthanoids
(c) Most transition metal ions show colours in aqueous solution

Answer:
(a) Cr²⁺ has d⁴ → tends to lose 1e⁻ to reach stable d³ (Cr³⁺)
Mn³⁺ has d⁴ → tends to gain 1e⁻ to reach stable d⁵ (Mn²⁺)
(b) Due to small energy gap between 5f, 6d, and 7s orbitals in actinoids
(c) Unpaired d-electrons undergo d–d transitions → absorb visible light → coloured

Absolutely! Below are the detailed answers to the Long Answer Type Questions (5-MARKS) from Chapter 8 – d- and f-Block Elements, with questions followed by accurate, exam-oriented answers:

✍️ LONG ANSWER TYPE QUESTIONS (5 MARKS)

Q1. A green compound ‘A’ on fusion with NaOH in the presence of air forms yellow compound ‘B’. Acidification of ‘B’ gives orange compound ‘C’. ‘C’ reacts with ammonium salt to form compound ‘D’, which on heating gives compound ‘A’ and nitrogen gas. Identify A to D and write the chemical equations.

Answer:

A = Cr₂O₃ (green)
B = Na₂CrO₄ (yellow)
C = Na₂Cr₂O₇ (orange)
D = (NH₄)₂Cr₂O₇

Reactions:

Cr₂O₃ + NaOH + O₂ → Na₂CrO₄ (B) + H₂O
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ (C) + H₂O
Na₂Cr₂O₇ + 2NH₄⁺ → (NH₄)₂Cr₂O₇ (D)
(NH₄)₂Cr₂O₇ → Cr₂O₃ (A) + N₂↑ + 4H₂O

Q2. Assign reasons for the following:

(a) No regular trend in E° values of M²⁺/M in 3d-series
(b) Gradual decrease in ionic radii of M²⁺ in 3d-series
(c) Transition metals form complexes
(d) Ce³⁺ is easily oxidised to Ce⁴⁺
(e) Tantalum and Palladium are used in electroplating

Answer:

(a) Due to varying enthalpies of atomisation, hydration, and ionisation energies → no regular trend in electrode potentials.
(b) Due to increasing nuclear charge → stronger pull on electrons → smaller ionic radii (d-block contraction).
(c) Due to small size, high charge, and availability of vacant d-orbitals for bonding.
(d) Ce³⁺ ([Xe] 4f¹) → easily loses 4f electron to form stable Ce⁴⁺ ([Xe])
(e) Tantalum and Palladium are corrosion-resistant and provide smooth, lustrous coatings → ideal for electroplating.

Q3. Account for the following:

(a) Actinoids display a variety of oxidation states
(b) Yb²⁺ is a good reductant
(c) Cerium (IV) is a good analytical reagent
(d) Transition metal fluorides are ionic, chlorides/bromides are covalent
(e) Hydrochloric acid attacks all actinoids

Answer:

(a) Small energy gap between 5f, 6d, and 7s orbitals → multiple oxidation states.
(b) Yb²⁺ → [Xe]4f¹⁴ → stable → tends to lose electron → good reductant.
(c) Ce⁴⁺ oxidises easily to Ce³⁺ → used in redox titrations.
(d) Fluoride is small and highly electronegative → forms ionic bonds. Cl⁻, Br⁻ → larger, polarizable → covalent character.
(e) Actinoids are highly electropositive → react with dilute HCl to form chlorides and H₂ gas.

Q4. Explain the following with suitable reasons:

(a) Co²⁺ is stable in aqueous solution but oxidised in presence of complexing agent
(b) Eu²⁺, Yb²⁺ are good reductants but Tb⁴⁺ is an oxidant
(c) AgCl dissolves in NH₃
(d) Cr²⁺ is a stronger reducing agent than Fe²⁺
(e) Highest oxidation state of transition metals appears in oxoanions

Answer:

(a) Co²⁺ (aq) stable due to hydration energy, but with ligands, Co³⁺ is stabilised by ligand field → gets oxidised.
(b) Eu²⁺ = 4f⁷, Yb²⁺ = 4f¹⁴ → stable → prefer to donate electrons → good reductants. Tb⁴⁺ → oxidises due to stability of 4f⁷.
(c) AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻ (complex formation increases solubility)
(d) Cr²⁺ (d⁴) reduces to stable d³ (Cr³⁺); Fe²⁺ (d⁶) less reducing.
(e) Oxoanions (e.g., MnO₄⁻, Cr₂O₇²⁻) stabilise high oxidation states by π bonding with O.

Q5. A white crystalline compound ‘A’ when heated with K₂Cr₂O₇ and conc. H₂SO₄ evolves reddish brown gas ‘B’. ‘B’ turns NaOH yellow, giving solution ‘C’. Adding CH₃COOH and lead acetate gives yellow ppt ‘D’. Heating ‘A’ with NaOH evolves gas forming brown ppt ‘E’ with K₂HgI₄. Identify A–E with reactions.

Answer:

A = NH₄Cl
B = CrO₂Cl₂ (reddish brown gas)
C = Na₂CrO₄ (yellow)
D = PbCrO₄ (yellow ppt)
E = HgNH₂Cl (brown ppt)

Reactions:

NH₄Cl + K₂Cr₂O₇ + H₂SO₄ → CrO₂Cl₂ (B) + NH₄HSO₄ + KHSO₄ + H₂O
CrO₂Cl₂ + 4NaOH → Na₂CrO₄ (C) + 2NaCl + 2H₂O
Na₂CrO₄ + Pb(CH₃COO)₂ + CH₃COOH → PbCrO₄ (D) + 2CH₃COONa
NH₄Cl + NaOH → NH₃↑ + NaCl + H₂O
NH₃ + K₂HgI₄ → HgNH₂I (E) + KI + NH₄I (brown ppt)

Q6. (a) Describe the preparation of potassium dichromate (K₂Cr₂O₇) from chromite ore. (b) Chromates and dichromates are interconvertible with change in pH. Explain with equations.

Answer:

(a) Preparation from chromite ore:

Fusion:
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Acidification:
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + H₂O

(b) Interconversion:

Chromate to dichromate (acidic pH):
2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O
Dichromate to chromate (basic pH):
Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O

Q7. Explain giving reasons:

(a) Transition metals are less reactive than alkali & alkaline earth metals
(b) Cu²⁺/Cu has positive E°
(c) Middle elements in transition series have high melting points
(d) Small decrease in atomic size across a series
(e) Variability in oxidation states of transition vs non-transition metals

Answer:

(a) Due to high ionisation energy, strong metallic bonding, and partial d-orbital shielding.
(b) Due to high hydration energy of Cu²⁺ → stabilised → positive E° value.
(c) Maximum unpaired electrons → stronger metallic bonding → high melting point.
(d) d-electrons shield poorly → increased nuclear charge → small atomic contraction.
(e) Transition metals have variable oxidation states due to similar energy of (n–1)d and ns orbitals. Non-transition metals show fixed oxidation states.

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