JEE Main 2026 Jan 23 Shift 1, Chemistry Q15: Valence Shell Electron Pair Repulsion (VSEPR) Theory

JEE Main2026Jan 23, Shift 1Chemistry

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).

A
$120°$
B
$107°$
C
$102°$
D
$116°$

Solution

Count lone pairs in each Lewis structure:

• HNO $_{3}$ : 8 lone pairs (the three O atoms carry the bulk; N has 0).

• H $_{2}$ SO $_{4}$ : 8 lone pairs.

• O $_{3}$ : 5 lone pairs.

• NF $_{3}$ : 10 lone pairs (1 on N + 3 on each of the three F atoms).

$∴$ X = NF $_{3}$ .

Geometry of NF $_{3}$ : sp $^{3}$ central N with one lone pair, pyramidal. The bond angle is compressed below the tetrahedral value because the highly electronegative F atoms pull the bonding electrons further from N, allowing the lone pair to push the N–F bonds closer together. Observed F–N–F bond angle $\approx$ 102°.

Concept behind this question

Valence Shell Electron Pair Repulsion (VSEPR) Theory Notes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Jan 23 Shift 1, Chemistry Q15: Valence Shell Electron Pair Repulsion (VSEPR) Theory

Concept behind this question

More previous year questions on Valence Shell Electron Pair Repulsion (VSEPR) Theory

Practice more Chemistry