d- and f-Block Elements Previous Year Questions — NEET Chemistry

3 past-year d- and f-Block Elements questions from NEET (Chemistry).

Q53 (2024)

Match List I with List II: <div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I (Transition metal/compound)</th><th style="border: 1px solid #888; padding: 8px;">List-II (Catalytic Role)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. $V_2O_5$</td><td style="border: 1px solid #888; padding: 8px;">(I) Preparation of ammonia from $N_2/H_2$ mixture</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Fe</td><td style="border: 1px solid #888; padding: 8px;">(II) Polymerisation of alkynes</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. $PdCl_2$</td><td style="border: 1px solid #888; padding: 8px;">(III) Preparation of $H_2SO_4$ and $SO_3$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Ni complex</td><td style="border: 1px solid #888; padding: 8px;">(IV) Oxidation of ethyne to ethanal</td></tr></tbody></table></div> Choose the correct answer from the options given below:
  1. (1) A-III, B-IV, C-I, D-II
  2. (2) A-II, B-I, C-IV, D-III
  3. (3) A-IV, B-I, C-III, D-II
  4. (4) A-III, B-I, C-IV, D-II
### Related Formula $$2SO_2 + O_2 \xrightarrow{V_2O_5} 2SO_3$$ $$N_2 + 3H_2 \xrightarrow{Fe} 2NH_3$$ ### Core Logic A. $V_2O_5$ catalyses $SO_2 \rightarrow SO_3$ in $H_2SO_4$ manufacture (III). B. Fe is catalyst in Haber's process for $NH_3$ (I). C. $PdCl_2$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements

Q57 (2024)

The calculated ‘spin-only’ magnetic moment of $Ti^{2+} (3d^2)$ is:
  1. (1) 2.84 BM
  2. (2) 5.92 BM
  3. (3) 4.90 BM
  4. (4) 3.87 BM
### Related Formula $$\mu = \sqrt{n(n + 2)} \text{ B.M.}$$ ### Core Logic Electronic configuration of $Ti^{2+} = [Ar] 3d^2$. Number of unpaired electrons $n = 2$. $$\mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.84 \text{ B.M.}$$ ### Step 1: Final Selection Calculated spin-only magnetic moment is 2.84 BM. ### Pattern Recognition $n=1 \rightarrow 1.73, n=2 \rightarrow 2.84, n=3 \rightarrow 3.87, n=4 \rightarrow 4.90, n=5 \rightarrow 5.92$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements

Q63 (2024)

Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
  1. (1) Its nearest inert gas is Radon.
  2. (2) After losing one more electron, it acquires $4f^{14}$ electronic configuration.
  3. (3) Its atomic number is 61.
  4. (4) After losing one more electron, it acquires $4f^0$ electronic configuration.
### Related Formula $$Ce = [Xe] 4f^1 5d^1 6s^2 \implies Ce^{4+} = [Xe] 4f^0$$ ### Core Logic Cerium ($Z=58$) has electronic configuration $[Xe]4f^1 5d^1 6s^2$. Upon losing 4 electrons, it acquires extra stability of noble gas configuration ($4f^0$). ### Step 1: Conclusion Reasoning matches statement (4). ### Pattern Recognition Extra stability of empty ($4f^0$), half-filled ($4f^7$), and fully-filled ($4f^{14}$) subshells drives unusual oxidation states in f-block elements. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements
Rankbit System
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%)

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