Identify the molecule ( X ) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO 3 , H 2 SO 4 , NF 3 and O 3 . Choose the correct bond angle made by the central atom of the molecule (X).

Chemistry · Chemical Bonding

JEE Main 2026 — 23 January, Morning Shift — Question 62

Identify the molecule $(\mathrm{X})$ with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{NF}_{3}$ and $\mathrm{O}_{3}$ . Choose the correct bond angle made by the central atom of the molecule (X).

Option A:
$120^{\circ}$
Option B:
$107^{\circ}$
Option C:
$102^{\circ}$
Correct
Option D:
$116^{\circ}$

Answer: C

Step-by-step solution

$\mathrm{NF}_{3}$ Number of lone pair in $\mathrm{NF}_{3}=10$ Bond angle in $\mathrm{NF}_{3} \approx 102^{\circ}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Bonding
Topic: Introduction to Bonding and Lewis Octet Theory