NEET · Chemistry —

d- and f-Block Elements appeared 3 times across 1 year — 6.7% of Chemistry. This question is from Catalytic Properties of Transition Metals.

Year 2024 Total
Questions 3 3
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Match List I with List II:
List-I (Transition metal/compound)List-II (Catalytic Role)
A. V₂O₅(I) Preparation of ammonia from N₂/H₂ mixture
B. Fe(II) Polymerisation of alkynes
C. PdCl₂(III) Preparation of H₂SO₄ and SO₃
D. Ni complex(IV) Oxidation of ethyne to ethanal
Choose the correct answer from the options given below:

Solution & Explanation

Related Formula
2SO₂ + O₂ V₂O₅ 2SO₃ N₂ + 3H₂ Fe 2NH₃
Core Logic

A. V₂O₅ catalyses SO₂ arrow SO₃ in H₂SO₄ manufacture (III). B. Fe is catalyst in Haber's process for NH₃ (I). C. PdCl₂ is catalyst in Wacker process (IV). D. Ni complex is catalyst in polymerization of alkynes (II).

Step 1: Final Matching

A-III, B-I, C-IV, D-II.

Pattern Recognition

Direct NCERT inorganic chemistry catalytic applications.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Reference Study Guides

D And F Block Elements

More d- and f-Block Elements Previous-Year Questions

Q57 neet_2024_05_may_morning Spin-Only Magnetic Moment
The calculated ‘spin-only’ magnetic moment of Ti²⁺ (3d²) is:
  • A. (1) 2.84 BM
  • B. (2) 5.92 BM
  • C. (3) 4.90 BM
  • D. (4) 3.87 BM

Solution

Related Formula
μ = √(n(n + 2)) B.M.
Core Logic

Electronic configuration of Ti²⁺ = [Ar] 3d². Number of unpaired electrons n = 2.

μ = √(2(2 + 2)) = √(8) ≈ 2.84 B.M.
Step 1: Final Selection

Calculated spin-only magnetic moment is 2.84 BM.

Pattern Recognition

n=1 arrow 1.73, n=2 arrow 2.84, n=3 arrow 3.87, n=4 arrow 4.90, n=5 arrow 5.92.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Q63 neet_2024_05_may_morning Lanthanoid Oxidation States
Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
  • A. (1) Its nearest inert gas is Radon.
  • B. (2) After losing one more electron, it acquires 4f¹⁴ electronic configuration.
  • C. (3) Its atomic number is 61.
  • D. (4) After losing one more electron, it acquires 4f⁰ electronic configuration.

Solution

Related Formula
Ce = [Xe] 4f¹ 5d¹ 6s² Ce⁴⁺ = [Xe] 4f⁰
Core Logic

Cerium (Z=58) has electronic configuration [Xe]4f¹ 5d¹ 6s². Upon losing 4 electrons, it acquires extra stability of noble gas configuration (4f⁰).

Step 1: Conclusion

Reasoning matches statement (4).

Pattern Recognition

Extra stability of empty (4f⁰), half-filled (4f⁷), and fully-filled (4f¹⁴) subshells drives unusual oxidation states in f-block elements.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

More d- and f-Block Elements Questions — neet_2024_05_may_morning

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)