The d- and f-Block Elements NEET PYQs (2005–2026) with Solutions

10 solved The d- and f-Block Elements PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumGeneral Properties

   Match List-I with List-II: Choose the correct answer from the options given below.

   1. AOption A: A–III, B–IV, C–I, D–II
   2. BOption B: A–IV, B–I, C–III, D–II
   3. COption C: A–II, B–I, C–IV, D–III
   4. DOption D: A–III, B–I, C–IV, D–II

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   Correct answer: (D) A–III, B–I, C–IV, D–II

   Explanation

   V₂O₅ → Contact process (SO₂ to SO₃); Fe → Haber process; PdCl₂ → Oxidation of ethyne to ethanal; Ni complex → Polymerisation of alkynes. Hence option (D) is correct.

2. Question 2 (NEET 2026)

   NEET 2026EasyMelting points of transition metals

   Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Generally, $3d$ transition metals have high melting points. Reason R: Involvement of $3d$-electrons in addition to $4s$-electrons in the interatomic metallic bonding. In light of the above statements, choose the most appropriate answer from the options given below:

   1. AOption A: Both A and R are correct and R is the correct explanation of A
   2. BOption B: Both A and R are correct and R is NOT the correct explanation of A
   3. COption C: A is correct but R is not correct.
   4. DOption D: A is not correct but R is correct

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   Correct answer: (A) Both A and R are correct and R is the correct explanation of A

   Explanation

   Transition metals (except Zn, Cd, Hg) are hard and have high melting and boiling points, so Assertion A is correct. NCERT explains that the high melting points are due to the involvement of a greater number of $(n-1)d$ electrons, in addition to the $ns$ electrons, in interatomic metallic bonding. For the $3d$ series, the $3d$ and $4s$ electrons together make strong metallic bonds. So R is correct and is the correct explanation of A: option (A).

3. Question 3 (NEET 2026)

   NEET 2026MediumProperties of nitrogen family elements

   Identify the incorrect statement from the following:

   1. AOption A: Nitrogen can form pπ–pπ multiple bonds with itself
   2. BOption B: P(C₂H₅)₃ and As(C₆H₅)₃ form dπ–dπ bond with transition metals
   3. COption C: Phosphorus, arsenic and antimony show catenation property
   4. DOption D: Nitrogen can form dπ–pπ bond with oxygen

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   Correct answer: (D) Nitrogen can form dπ–pπ bond with oxygen

   Explanation

   Nitrogen and oxygen do not have vacant d orbitals, so dπ–pπ bonding is not possible. Nitrogen forms pπ–pπ bonds with itself (N≡N). Therefore statement (4) is incorrect.

4. Question 4 (NEET 2026)

   NEET 2026EasyLanthanoids & Actinoids

   Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:

   1. AOption A: After losing one more electron, it acquires 4f¹⁴ electronic configuration
   2. BOption B: Its nearest inert gas is Radon
   3. COption C: Its atomic number is 61
   4. DOption D: After losing one more electron, it acquires 4f⁰ electronic configuration

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   Correct answer: (D) After losing one more electron, it acquires 4f⁰ electronic configuration

   Explanation

   Cerium (Z=58) has configuration [Xe]4f¹5d¹6s². In +4 oxidation state it attains stable empty 4f⁰ configuration, making +4 state possible. Hence option (D) is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumElectronic configuration of lanthanide ions

   The lanthanide ion having four unpaired electrons is (Given : Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)

   1. AOption A: $\mathrm{Nd^{3+}}$
   2. BOption B: $\mathrm{Ce^{3+}}$
   3. COption C: $\mathrm{Tb^{3+}}$
   4. DOption D: $\mathrm{Ho^{3+}}$

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   Correct answer: (D) $\mathrm{Ho^{3+}}$

   Explanation

   $\mathrm{Ln^{3+}}$ ions have the configuration $[\mathrm{Xe}]4f^n$, where $n = Z - 57$. $\mathrm{Ce^{3+}}$ ($Z = 58$): $4f^1$, 1 unpaired electron. $\mathrm{Nd^{3+}}$ ($Z = 60$): $4f^3$, 3 unpaired electrons. $\mathrm{Tb^{3+}}$ ($Z = 65$): $4f^8$, 6 unpaired electrons (7 singly filled, one paired). $\mathrm{Ho^{3+}}$ ($Z = 67$): $4f^{10}$. Seven orbitals hold 10 electrons, so 3 are paired and 4 are unpaired. So the answer is $\mathrm{Ho^{3+}}$, option (D).

6. Question 6 (NEET 2026)

   NEET 2026MediumCarbon family properties and oxidation states

   Identify the incorrect statement from the following:

   1. AOption A: Carbon has the ability to form pπ–pπ multiple bond with itself
   2. BOption B: ECl₃ (E = B and Al) is a monomer when E = B and dimer when E = Al
   3. COption C: The order of catenation property of Group 14 elements is C >> Si > Ge ≈ Sn
   4. DOption D: Oxygen exhibits only −2 oxidation state

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   Correct answer: (D) Oxygen exhibits only −2 oxidation state

   Explanation

   Oxygen shows oxidation states of 0 (O₂), −1 (peroxides), −2 (oxides), and positive values in compounds with fluorine. Therefore statement (4) is incorrect.

7. Question 7 (NEET 2026)

   NEET 2026EasyThermal decomposition of potassium permanganate

   The green paramagnetic species formed by heating $\mathrm{KMnO_4}$ at 513 K is

   1. AOption A: $\mathrm{K_2MnO_4}$
   2. BOption B: $\mathrm{Mn_3O_4}$
   3. COption C: $\mathrm{MnO}$
   4. DOption D: $\mathrm{KO_2}$

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   Correct answer: (A) $\mathrm{K_2MnO_4}$

   Explanation

   On heating at 513 K, potassium permanganate decomposes: $$\mathrm{2KMnO_4 \xrightarrow{513\ K} K_2MnO_4 + MnO_2 + O_2}$$ Potassium manganate, $\mathrm{K_2MnO_4}$, is dark green. In the manganate ion$\mathrm{MnO_4^{2-}}$, Mn is in the$+6$state ($3d^1$) with one unpaired electron, so it is paramagnetic.$\mathrm{Mn_3O_4}$and$\mathrm{MnO}$are not formed in this reaction, and$\mathrm{KO_2}$ is a potassium superoxide, not a manganese product. So the answer is option (A).

8. Question 8 (NEET 2025)

   NEET 2025EasyNitrogen Family

   Given below are two statements : Statement I : Like nitrogen that can form ammonia, arsenic can form arsine. Statement II : Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below.

   1. AOption A: Both Statement I and Statement II are correct
   2. BOption B: Both Statement I and Statement II are incorrect
   3. COption C: Statement I is correct but Statement II is incorrect
   4. DOption D: Statement I is incorrect but Statement II is correct

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   Correct answer: (C) Statement I is correct but Statement II is incorrect

   Explanation

   Arsenic forms arsine (AsH₃), analogous to ammonia. Antimony can form antimony pentoxide (Sb₂O₅), so Statement II is incorrect.

9. Question 9 (NEET 2025)

   NEET 2025MediumElectronic Configuration

   Which among the following electronic configurations belong to main group elements? A. [Ne]3s¹ B. [Ar]3d³4s² C. [Kr]4d¹⁰5s²5p⁵ D. [Ar]3d¹⁰4s¹ E. [Rn]5f⁰6d²7s² Choose the correct answer from the options given below.

   1. AOption A: B and E only
   2. BOption B: A and C only
   3. COption C: D and E only
   4. DOption D: A, C and D only

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   Correct answer: (B) A and C only

   Explanation

   Main group elements belong to s-block and p-block. [Ne]3s¹ corresponds to sodium (s-block) and [Kr]4d¹⁰5s²5p⁵ corresponds to iodine (p-block). The remaining configurations belong to transition or inner transition elements.

10. Question 10 (NEET 2025)

    NEET 2025MediumCatalysts and Coordination Compounds

    Match List - I with List - II Choose the correct answer from the options given below :

    1. AOption A: A-I, B-II, C-IV, D-III
    2. BOption B: A-II, B-III, C-I, D-IV
    3. COption C: A-I, B-II, C-III, D-IV
    4. DOption D: A-I, B-IV, C-III, D-II

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    Correct answer: (C) A-I, B-II, C-III, D-IV

    Explanation

    Haber process uses Fe catalyst. Wacker oxidation uses PdCl₂ catalyst. Wilkinson catalyst is [(PPh₃)₃RhCl]. Ziegler catalyst consists of TiCl₄ and trialkyl aluminium compounds such as Al(CH₃)₃. Hence the correct matching is A-I, B-II, C-III, D-IV.