NEET The p-Block Elements PYQs & Practice Questions

10 free NEET Chemistry questions on The p-Block Elements, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Halogens · Hydrides and Oxides of Nitrogen Family · Group 15 – Oxoacids of Nitrogen · Group 16 – Allotropy of Sulphur · Group 17 – Interhalogen Compounds · Group 18 – Noble Gases · Group 15 – Phosphorus Oxoacids · Group 16 – Sulphuric Acid · Group 17 – Trends in Halogens · Group 15 – Nitrogen Family Hydrides · Group 18 – Structures of Xenon Compounds · Group 16 – Ozone

1. Q1NEET 2026Oxidizing Action of MnO2 with Halide Salts and Conc. H2SO4

   Given below are two statements: Statement-I: Heating NaCl with concentrated H2SO4 and MnO2 results in oxidation of Mn. Statement-II: Heating NaI with concentrated H2SO4 and MnO2 results in reduction of Mn. In light of the above statements, choose the most appropriate answer from the options given below:

   • A. Both Statement-I and Statement-II are incorrect.
   • B. Statement-I is correct but Statement-II is incorrect.
   • C. Statement-I is incorrect but Statement-II is correct.
   • D. Both Statement-I and Statement-II are correct.
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   Answer: B — Statement-I is correct but Statement-II is incorrect.

   In both reactions, NaX + H2SO4 generates HX, which is then oxidized to X2 by MnO2; in this process MnO2 acts as the oxidizing agent and Mn is itself REDUCED from +4 to +2 (not oxidized). So Statement-I, which claims 'oxidation of Mn', is incorrect. Statement-II correctly states that heating NaI with conc. H2SO4 and MnO2 results in reduction of Mn (as MnO2 oxidizes I⁻ to I2), which is correct. So Statement-I is incorrect but Statement-II is correct.

2. Q2NEET 2026Valence Orbitals and Maximum Covalency

   The correct statement is:

   • A. Beryllium has three valence orbitals
   • B. Magnesium has a maximum covalency of four.
   • C. Aluminium has five valence orbitals
   • D. Boron has a maximum covalency of four.
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   Answer: C — Aluminium has five valence orbitals

   Boron (period 2) has only four valence orbitals available (2s, 2px, 2py, 2pz) with no accessible d-orbitals, so its maximum covalency is limited to four (e.g., in BF4⁻, BH4⁻). Beryllium also has four valence orbitals (2s + three 2p), not three. Magnesium and aluminium (period 3 elements) have access to 3d orbitals, allowing expanded octets and covalency greater than four (e.g., [AlF6]3⁻), so statements about them being limited to four or having only a few valence orbitals are incorrect.

3. Q3NEET 2025Hydrides and Oxides of Nitrogen 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:

   • A. Both Statement I and Statement II are correct
   • B. Both Statement I and Statement II are incorrect
   • C. Statement I is correct but Statement II is incorrect
   • D. Statement I is incorrect but Statement II is correct
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   Answer: B — Both Statement I and Statement II are incorrect

   Statement I is correct: all group 15 elements (N, P, As, Sb, Bi) form hydrides (NH3, PH3, AsH3, SbH3, BiH3). Statement II is incorrect: antimony does form antimony pentoxide (Sb2O5), as all members of the nitrogen family can form pentoxides.

4. Q4NEET 2025Xenon Compounds - Shapes and Hybridization

   Match List - I with List - II. List-I: A. XeO3, B. XeF2, C. XeOF4, D. XeF6. List-II: I. sp³d; linear, II. sp³; pyramidal, III. sp³d³; distorted octahedral, IV. sp³d²; square Pyramidal. Choose the correct answer from the options given below:

   • A. A-II, B-I, C-IV, D-III
   • B. A-II, B-I, C-III, D-IV
   • C. A-IV, B-II, C-III, D-I
   • D. A-IV, B-II, C-I, D-III
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   Answer: —

   XeO3 is sp³ hybridized, pyramidal (II). XeF2 is sp³d, linear (I). XeOF4 is sp³d², square pyramidal (IV). XeF6 is sp³d³, distorted octahedral (III). So A-II, B-I, C-IV, D-III.

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5. Q5NEET 2023Inert Pair Effect

   Taking stability as the factor, which one of the following represents the correct relationship?

   • A. InI3 > InI
   • B. AlCl > AlCl3
   • C. TlI > TlI3
   • D. TlCl3 > TlCl
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   Answer: B — AlCl > AlCl3

   Due to the inert pair effect, heavier p-block elements (like thallium) show increasing stability of the lower oxidation state (+1) relative to the higher oxidation state (+3) down the group. Thus TlI (Tl in +1 state) is more stable than TlI3 (Tl in +3 state), making TlI > TlI3 the correct relationship.

6. Q6NEET 2023Uses of Noble Gases

   Given below are two statements: one is labelled as Assertion A and the other labelled as Reason R: Assertion A: Helium is used to dilute oxygen in diving apparatus. Reason R: Helium has high solubility in O2. In the light of the above statements, choose the correct answer from the options given below.

   • A. Both A and R are true and R is NOT the correct explanation of A.
   • B. A is true but R is false.
   • C. A is false but R is true.
   • D. Both A and R are true and R is the correct explanation of A.
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   Answer: A — Both A and R are true and R is NOT the correct explanation of A.

   Assertion A is true — helium is used to dilute oxygen in diving apparatus (NCERT reference). However, Reason R is false — helium is actually used because it has much lower solubility in blood (not high solubility), which is why it prevents the bends (decompression sickness) in scuba divers. So A is true but R is false.

7. Q7NEET 2023Allotropes of Carbon

   Match List-I with List-II: List-I: A. Coke, B. Diamond, C. Fullerene, D. Graphite. List-II: I. Carbon atoms are sp3 hybridised. II. Used as a dry lubricant. III. Used as a reducing agent. IV. Cage like molecules. Choose the correct answer from the options given below.

   • A. A-IV, B-I, C-II, D-III
   • B. A-III, B-I, C-IV, D-II
   • C. A-III, B-IV, C-I, D-II
   • D. A-II, B-IV, C-I, D-III
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   Answer: A — A-IV, B-I, C-II, D-III

   Coke is used as a reducing agent in metallurgical processes (A-III). Diamond has a 3D network structure where each carbon atom is sp3 hybridised, bonded to 4 other carbons (B-I). Fullerene has a cage-like molecular structure (C-IV). Graphite is used as a dry lubricant due to its layered structure held by weak van der Waals forces that allow layers to slide (D-II). So the correct match is A-III, B-I, C-IV, D-II.

8. Q8NEET 2022Interhalogen Compounds

   Given below are two statements: one is labelled as Assertions (A) and the other is labelled as Reason(R). Assertion (A): ICl is more reactive than I2. Reason (R): I-Cl bond is weaker than I-I bond. In the light of the above statements, choose the most appropriate answer from the options given below:

   • A. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
   • B. (A) is correct but (R) is not correct.
   • C. (A) is not correct but (R) is correct.
   • D. Both (A) and (R) are correct but (R) is the correct explanation of (A)
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   Answer: C — (A) is not correct but (R) is correct.

   ICl is more reactive than I2 because the I-Cl bond, being a bond between two different atoms (polar bond), is weaker than the I-I bond (bond between identical atoms), making it easier to break. Both A and R are correct and R correctly explains A.

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9. Q9NEET 2022Hydrides of Group 16

   Given below are two statements: Statement I: The boiling points of the following hydrides of group 16 elements increases in the order - H2O < H2S < H2Se < H2Te Statement II: The boiling points of these hydrides increase with increase in molar mass. In the light of the above statements, choose the most appropriate answer from the options given below:

   • A. Both Statement I and Statement II are incorrect.
   • B. Statement I is correct but Statement II is incorrect.
   • C. Statement I is incorrect but Statement II is correct.
   • D. Both Statement I and Statement II are correct.
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   Answer: —

   The correct order of boiling points for group 16 hydrides is H2S < H2Se < H2Te < H2O, because water has exceptionally high boiling point due to hydrogen bonding, breaking the general trend of increasing boiling point with molar mass. So Statement I's order is incorrect, and Statement II's generalization is also incorrect due to this H-bonding exception.

10. Q10NEET 2022Hydrides

    Match List-I with List-II. List-I (Hydrides): (a) MgH2 (b) GeH4 (c) B2H6 (d) HF List-II (Nature): (i) Electron precise (ii) Electron deficient (iii) Electron rich (iv) Ionic Choose the correct answer from the options given below:

    • A. a-(iii), b-(i), c-(ii), d-(iv)
    • B. a-(i), b-(ii), c-(iv), d-(iii)
    • C. a-(ii), b-(iii), c-(iv), d-(i)
    • D. a-(iv), b-(i), c-(ii), d-(iii)
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    Answer: C — a-(ii), b-(iii), c-(iv), d-(i)

    MgH2 is an ionic hydride (iv); GeH4 is electron precise (i) since Ge forms 4 bonds using all valence electrons; B2H6 is electron deficient (ii) since boron doesn't have enough electrons for normal 2-centre bonds in all positions; HF is electron rich (iii) since fluorine has more electrons than needed for bonding (lone pairs).