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NEET P-Block Elements MCQs with answers

155 questions on P-Block Elements for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

mediumNEET 2021

1. Strongest hydrohalic acid:

  1. HF
  2. HCl
  3. HBr
  4. HI
Show answer and explanation

Answer: (D) HI

Acid strength ∝ 1/(H-X bond strength). Bond strength F>Cl>Br>I. So HI > HBr > HCl > HF.

Quick tip: HI strongest acid. HF weakest acid but strongest bond. Electronegativity ≠ acid strength.

NCERT Class 12, Chapter 7, page 210

easy

2. Which of the following Group 14 elements exhibits the maximum catenation property?

  1. Silicon
  2. Germanium
  3. Carbon
  4. Lead
Show answer and explanation

Answer: (C) Carbon

Catenation ability depends primarily on the M-M bond strength. The C-C bond enthalpy is exceptionally high (348 kJ/mol) compared to Si-Si (297 kJ/mol), Ge-Ge (260 kJ/mol), and Sn-Sn (240 kJ/mol), giving carbon the greatest tendency to catenate.

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3. The basicity of H₄P₂O₇ is:

  1. 3
  2. 1
  3. 2
  4. 4
Show answer and explanation

Answer: (D) 4

Basicity counts the ionisable (O−H) hydrogens. H₄P₂O₇ has 4 such hydrogens.

hard

4. Unlike other halogens which form interhalogen compounds of formulas AX, AX₃, AX₅, and AX₇, fluorine never acts as the central atom in any interhalogen compound. What is the fundamental reason for this behaviour?

  1. Fluorine has the lowest standard electrode potential among the halogens
  2. Fluorine is the most electronegative element and lacks vacant d-orbitals in its valence shell
  3. The F-F single bond energy is unusually small
  4. Fluorine cannot form covalent bonds with other p-block elements
Show answer and explanation

Answer: (B) Fluorine is the most electronegative element and lacks vacant d-orbitals in its valence shell

In interhalogen compounds (e.g., IF₇, ClF₃), the central atom must be less electronegative than the surrounding halogens and must expand its octet using vacant d-orbitals. Because fluorine is the most electronegative element and possesses no d-orbitals, it cannot act as the central atom.

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5. Given below are two statements: Statement I: The maximum coordination number of boron in aqueous solutions is 4, whereas aluminium achieves a coordination number of 6. Statement II: Aluminium possesses vacant 3d-orbitals of suitable energy that allow it to expand its coordination sphere, which are absent in boron. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Boron belongs to period 2 and has only four valence orbitals (2s, 2p), limiting its coordination number to 4 (e.g., [B(OH)₄]⁻). Aluminium, belonging to period 3, has low-lying vacant 3d-orbitals, allowing it to form octahedral complexes such as [Al(H₂O)₆]³⁺ with a coordination number of 6.

easy

6. Which of the following elements has the highest density at standard temperature and pressure?

  1. Silicon (Si)
  2. Germanium (Ge)
  3. Lead (Pb)
  4. Tin (Sn)
Show answer and explanation

Answer: (C) Lead (Pb)

Density generally increases down the group in p-block elements due to a substantial increase in atomic mass relative to the increase in atomic volume. Therefore, lead (Pb) has the highest density among the Group 14 elements.

hard

7. An aqueous solution of boric acid, B(OH)₃, behaves as a weak monobasic acid. Its acidic nature originates from:

  1. Acting as a Lewis acid by accepting an OH⁻ ion from water: B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺
  2. Direct release of a proton from the B-OH unit: B(OH)₃ → [BO(OH)₂]⁻ + H⁺
  3. Auto-ionisation forming H⁺ and B(OH)₄⁻ ions
  4. Release of three protons in successive stages
Show answer and explanation

Answer: (A) Acting as a Lewis acid by accepting an OH⁻ ion from water: B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺

Because boron has only six valence electrons in B(OH)₃ and lacks octet completion, it acts as a Lewis acid rather than a Brønsted acid. It accepts an electron pair from the oxygen of a water molecule (abstracting an OH⁻ ion), thereby releasing a hydronium ion into solution.

hard

8. Which of the following trends represents the correct order of the second ionisation enthalpy (ΔᵢH₂) across the elements of Period 2?

  1. F > O > N > C > B > Be > Li
  2. O > F > N > C > Li > B > Be
  3. Li > O > F > N > C > B > Be
  4. Li > F > O > N > C > B > Be
Show answer and explanation

Answer: (C) Li > O > F > N > C > B > Be

Removal of a second electron from Li⁺ involves breaking the exceptionally stable 1s² noble-gas shell, giving Li the highest ΔᵢH₂. Oxygen has higher ΔᵢH₂ than fluorine and nitrogen because O⁺ has a half-filled 2p³ configuration.

hard

9. Given below are two statements: Statement I: Nitrogen dioxide (NO₂) readily dimerizes to dinitrogen tetroxide (N₂O₄), whereas chlorine dioxide (ClO₂) does not dimerize. Statement II: The odd electron in NO₂ is localized on nitrogen's 2p orbital, whereas in ClO₂ the odd electron is extensively delocalized over vacant d-orbitals. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true
Show answer and explanation

Answer: (A) Both Statement I and Statement II are true

In NO₂, the odd unpaired electron resides in an sp² hybrid orbital primarily on the small nitrogen atom with no d-orbitals, causing high spin density that drives dimerization to N₂O₄. In contrast, ClO₂ does not dimerize because the odd electron is extensively delocalized across the molecule using chlorine's d-orbitals.

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10. The ground state electronic configuration of the newly synthesized superheavy element Oganesson (Z = 118, Group 18) terminates in:

  1. 6d¹⁰ 7s² 7p⁶
  2. 7s² 7p⁶
  3. 5f¹⁴ 6d¹⁰ 7s² 7p⁵
  4. 5f¹⁴ 6d¹⁰ 7s² 7p⁶
Show answer and explanation

Answer: (D) 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Oganesson (Z = 118) completes the 7th period. Its electron configuration is [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶, having 8 valence electrons in the 7th shell preceded by filled 5f and 6d subshells.

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11. The number of lone pairs on the xenon atom in XeF₆ is:

  1. 2
  2. 3
  3. 0
  4. 1
Show answer and explanation

Answer: (D) 1

Xe has 8 valence electrons; after forming the bonds in XeF₆, 1 lone pair remains.

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12. Match List-I (Group 15 Oxides) with List-II (Acid-Base Character) and select the correct option: List-I: (A) N₂O₃ (B) As₂O₃ (C) Sb₂O₃ (D) Bi₂O₃ List-II: (I) Predominantly basic (II) Strongly acidic (III) Amphoteric (moderately acidic) (IV) Amphoteric (weakly acidic)

  1. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  2. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  3. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
  4. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Show answer and explanation

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

Down Group 15, the metallic character increases, causing the oxides to transition from acidic to basic. N₂O₃ and P₄O₆ are strongly acidic, As₂O₃ and Sb₂O₃ are amphoteric (with As₂O₃ being slightly more acidic than Sb₂O₃), and Bi₂O₃ is predominantly basic.

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13. Which of the following hydride pairs shows the largest difference in boiling point primarily due to extensive intermolecular hydrogen bonding present only in the first member?

  1. PH₃ and AsH₃
  2. CH₄ and SiH₄
  3. H₂S and H₂Se
  4. H₂O and H₂S
Show answer and explanation

Answer: (D) H₂O and H₂S

Due to the small size and exceptionally high electronegativity of oxygen, H₂O exhibits extensive intermolecular hydrogen bonding, raising its boiling point to 373 K. H₂S molecules lack hydrogen bonding and have a much lower boiling point (213 K), giving a difference of 160 K, which is the largest among the given pairs.

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14. Given below are two statements regarding the relative stability of dihalides and tetrahalides of Group 14: Statement I: GeCl₂ is more stable than GeCl₄, whereas PbCl₄ is more stable than PbCl₂. Statement II: The stability of the +2 oxidation state increases down Group 14 due to the inert pair effect. In light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

Due to the inert pair effect, the stability of the +2 oxidation state increases down Group 14 (+4 is more stable for Ge, whereas +2 is more stable for Pb). Thus, GeCl₄ is more stable than GeCl₂, and PbCl₂ is more stable than PbCl₄. Statement I is incorrect while Statement II is correct.

easy

15. How many degenerate orbitals contain at least one electron in the valence shell of a neutral ground-state Tin atom (Sn, Z = 50)?

  1. 2
  2. 1
  3. 3
  4. 5
Show answer and explanation

Answer: (A) 2

The valence shell configuration of Tin (Group 14, Period 5) is 5s² 5p². According to Hund's rule of maximum multiplicity, the two 5p electrons occupy two separate degenerate 5p orbitals (e.g., 5pₓ¹ 5pᵧ¹), leaving one 5p orbital vacant. Thus, exactly 2 degenerate p-orbitals contain at least one electron.

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