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

3 past-year The 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 complex)</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. $V_{2}O_{5}$</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 $H_{2}SO_{4}$ and $SO_{2}$</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. A-III, B-IV, C-I, D-II
  2. A-II, B-I, C-IV, D-III
  3. A-IV, B-I, C-III, D-II
  4. A-III, B-I, C-IV, D-II
### Core Logic (A) $V_2O_5$ $\rightarrow$ Catalyses the oxidation of $SO_2$ into $SO_3$ in the Contact process for the manufacture of $H_2SO_4$. (Matches III) (B) Fe $\rightarrow$ Acts as a catalyst in Haber's process for the preparation of ammonia from $N_2 / H_2$ mixture. (Matches I) (C) $PdCl_2$ $\rightarrow$ Wacker process for the oxidation of ethyne to ethanal. (Matches IV) (D) Ni complex $\rightarrow$ Catalyses the polymerisation of alkynes. (Matches II) ### Step 1: Final Match A $\rightarrow$ III B $\rightarrow$ I C $\rightarrow$ IV D $\rightarrow$ II ### Pattern Recognition Classic industrial catalysts match: Iron = Haber (Ammonia). Vanadium Pentoxide = Contact (Sulfuric Acid). Palladium Chloride = Wacker (Acetaldehyde). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

Q57 (2024)

The calculated 'spin-only' magnetic moment of $Ti^{2+}(3d^{2})$ is :
  1. 2.84 BM
  2. 5.92 BM
  3. 4.90 BM
  4. 3.87 BM
### Related Formula $$\mu = \sqrt{n(n + 2)} \text{ B.M.}$$ where $n$ is the number of unpaired electrons. ### Core Logic Electronic configuration of Titanium ($Ti, Z=22$): $[Ar] \, 4s^2 3d^2$. For $Ti^{2+}$, two electrons are removed from the 4s orbital first. Electronic configuration of $Ti^{2+} \implies [Ar] \, 4s^0 3d^2$. Number of unpaired electrons ($n$) = 2. ### Step 1: Calculate Magnetic Moment $$\mu = \sqrt{2(2 + 2)}$$ $$\mu = \sqrt{2 \times 4} = \sqrt{8}$$ $$\mu = 2.828 \text{ B.M.} \approx 2.84 \text{ B.M.}$$ ### Pattern Recognition The spin-only magnetic moment value is always slightly greater than the number of unpaired electrons ($n$). For $n=2$, $\mu$ is approx $2.8$. For $n=3$, $\mu \approx 3.87$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

Q63 (2024)

Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
  1. Its nearest inert gas is Radon.
  2. After losing one more electron, it acquires $4f^{14}$ electronic configuration.
  3. Its atomic number is 61.
  4. After losing one more electron, it acquires $4f^{0}$ electronic configuration.
### Core Logic Cerium ($Ce$, Atomic Number = 58) has the electronic configuration $[Xe] \, 4f^1 5d^1 6s^2$. In the +3 oxidation state, $Ce^{3+}$ has the configuration $[Xe] \, 4f^1$. Although +3 is the most common state for lanthanoids, Cerium readily loses one more electron from its $4f$ subshell to form $Ce^{4+}$. The electronic configuration of $Ce^{4+}$ becomes $[Xe] \, 4f^0$, which is the highly stable configuration of the nearest noble gas, Xenon. ### Step 1: Final Conclusion Cerium shows a +4 oxidation state because it attains a stable empty f-subshell ($4f^0$) configuration. ### Pattern Recognition Anomalous oxidation states in f-block elements (+2, +4) usually occur when the ion can achieve a stable $f^0$ (empty), $f^7$ (half-filled), or $f^{14}$ (fully filled) electronic configuration. For Ce (+4), it's $f^0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The 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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