JEE Main 2026 — Jan 23, Shift 1 — Chemistry solution

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°.

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