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MCQEasyJEE 2026VSEPR Theory & Shapes of Molecules

JEE Chemistry 2026 Question with Solution

Identify the molecule (X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO3\mathrm{HNO_3}, H2SO4\mathrm{H_2SO_4}, NF3\mathrm{NF_3}, and O3\mathrm{O_3}. Choose the correct bond angle made by the central atom of the molecule (X).

  • A

    116∘116^\circ

  • B

    120∘120^\circ

  • C

    107∘107^\circ

  • D

    102∘102^\circ

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The molecules are HNO3\mathrm{HNO_3}, H2SO4\mathrm{H_2SO_4}, NF3\mathrm{NF_3}, and O3\mathrm{O_3}.

Find: The molecule with the maximum total number of lone pairs and the bond angle made by its central atom.

Approach: Count the total lone pairs in the Lewis dot structures of all given molecules, then use VSEPR theory for the identified molecule.

From the provided explanation:

  • HNO3\mathrm{HNO_3}: total 88 lone pairs
  • H2SO4\mathrm{H_2SO_4}: total 88 lone pairs
  • O3\mathrm{O_3}: total 66 lone pairs
  • NF3\mathrm{NF_3}: total 1010 lone pairs

Thus, molecule (X) is NF3\mathrm{NF_3}.

In NF3\mathrm{NF_3}, nitrogen has 33 bond pairs and 11 lone pair, so it is sp3sp^3 hybridized. Because fluorine is highly electronegative, the bonding electron pairs are pulled away from nitrogen, which reduces bond-pair repulsion near the central atom. Hence, the F−N−F\mathrm{F-N-F} bond angle becomes smaller than the ideal tetrahedral angle 109.5∘109.5^\circ.

The bond angle is approximately 102∘102^\circ.

Therefore, the correct option is D.

Common mistakes

  • Counting lone pairs only on the central atom is incorrect because the question asks for the maximum number of lone pairs in the whole molecule using the Lewis structure. Count lone pairs on both central and peripheral atoms.

  • Choosing 107∘107^\circ by comparing NF3\mathrm{NF_3} directly with NH3\mathrm{NH_3} is incorrect. In NF3\mathrm{NF_3}, fluorine pulls bonding electrons away, so the bond angle decreases further to about 102∘102^\circ.

  • Assuming the angle must be close to the tetrahedral value 109.5∘109.5^\circ is incorrect. VSEPR gives the basic geometry, but electronegativity effects modify the actual bond angle.

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