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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Chemical Bonding and Molecular Structure

The correct order of ONO bond angle in the given species is:

(A) $NO_{2}^{+} < NO_{2} < NO_{3}^{-} < NO_{2}^{-}$
(B) $NO_{2}^{-} < NO_{3}^{-} < NO_{2} < NO_{2}^{+}$ ✓ Correct
(C) $NO_{3}^{-} < NO_{2}^{-} < NO_{2} < NO_{2}^{+}$
(D) $NO_{2}^{-} < NO_{3}^{-} < NO_{2}^{+} < NO_{2}$

Answer: (B)

Solution

The bond angle depends on the number of electron pairs around the central nitrogen atom.

$NO_{2}^{+}$ : Linear structure (sp hybridization), bond angle = 180°
$NO_{2}$ : Bent structure with one unpaired electron, bond angle ≈ 134°
$NO_{3}^{-}$ : Trigonal planar (sp² hybridization), bond angle = 120°
$NO_{2}^{-}$ : Bent structure with lone pair, bond angle ≈ 115°

As lone pair repulsion increases, bond angle decreases. The order is:

NO_{2}^{-} (115°) < NO_{3}^{-} (120°) < NO_{2} (134°) < NO_{2}^{+} (180°)

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option

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