JEE Main 2026 Jan 22 Shift 2, Chemistry Q11: Valence Shell Electron Pair Repulsion (VSEPR) Theory

JEE Main2026Jan 22, Shift 2Chemistry

Among $H_{2} S, H_{2} O, N F_{3}, N H_{3}$ and $C H C l_{3}$ , identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is :

Solution

Approximate dipole moments (Debye): $H_{2} S \approx 0.95$ , $H_{2} O \approx 1.85$ , $N F_{3} \approx 0.23$ , $N H_{3} \approx 1.47$ , $C H C l_{3} \approx 1.04$ .

$N F_{3}$ has the lowest dipole moment because the bond dipoles ( $N \to F$ ) partially cancel the lone-pair dipole on nitrogen.

Nitrogen in $N F_{3}$ is $s p^{3}$ with three N–F bonds and one lone pair.

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JEE Main 2026 Jan 22 Shift 2, Chemistry Q11: Valence Shell Electron Pair Repulsion (VSEPR) Theory

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