ONO bond angles in NO₂⁺, NO₂, NO₃⁻ and NO₂⁻

01Original question — verbatim

02Molecular geometry viewer

Every species is drawn to scale at its experimental ONO angle. Yellow dots are the non-bonding electrons that live on nitrogen — a full lone pair in NO₂⁻, a single odd electron in NO₂, none at all in NO₂⁺ and NO₃⁻. That one difference is the whole question.

03VSEPR domain builder

Pick the electron domains around nitrogen and the builder names the species and its geometry. Every one of the four is reachable — that is the point: the four species differ only in how many σ-bonded oxygens there are and what, if anything, occupies the remaining domain.

04Measured ONO bond angles

Experimental gas-phase / crystallographic values. The ideal VSEPR angle for three electron domains is 120° (grey line) — NO₃⁻ sits exactly on it, the lone pair in NO₂⁻ pulls below it, and the lone electron in NO₂ lets the angle open above it.

05Ordering exercise

Click the species in order of increasing ONO bond angle.

06Step-by-step solution

1. Count the electron domains on nitrogen

   Only σ-bonded atoms and non-bonding electrons on the central atom count as domains; π bonds do not.

   NO₂⁺: 16 e⁻ → 2 σ + nothing = SN 2  ·  NO₂: 17 e⁻ → 2 σ + 1 electron = SN 3  ·  NO₃⁻: 24 e⁻ → 3 σ + nothing = SN 3  ·  NO₂⁻: 18 e⁻ → 2 σ + 1 lone pair = SN 3

2. NO₂⁺ is the odd one out — it is linear

   With only two domains and no non-bonding electrons on N (isoelectronic with CO₂, 16 valence electrons), the nitronium ion is sp hybridised and strictly linear.

   O=N⁺=O  ⟹  ∠ONO = 180° — the largest, so it must sit last.

   That alone kills (A) and (D).

3. The remaining three all have three domains — rank them by repulsion

   Three domains give an ideal 120°. What moves a species off 120° is the character of the third domain:

   repulsion:  lone pair  >  bonding pair  >  single (odd) electron
   • NO₃⁻ — three identical N–O bonding domains, nothing else on N. Perfect D_3h symmetry ⟹ exactly 120°.
   • NO₂⁻ — the third domain is a full lone pair, which repels harder than a bonding pair and squeezes the two N–O bonds together ⟹ 115.4°, below 120°.
   • NO₂ — the third domain holds only one electron, which repels less than a bonding pair, so the bonds spread apart ⟹ 134.1°, above 120°.

4. Assemble the order

   NO₂⁻ (115.4°) < NO₃⁻ (120°) < NO₂ (134.1°) < NO₂⁺ (180°)

   Read the options against this list rather than against a half-remembered rule: (C) is the tempting near-miss because it has NO₂ and NO₂⁺ in the right places but swaps the first two. The list matches (B) exactly.

5. Answer

   (B) NO₂⁻ < NO₃⁻ < NO₂ < NO₂⁺

   A useful memory hook: along NO₂⁺ → NO₂ → NO₂⁻ you are adding one electron at a time to the same skeleton, and the angle closes 180° → 134° → 115° each time. NO₃⁻ is simply pinned at 120° by symmetry, and 115.4 < 120 < 134.1 slots it into second place.

07Concepts and classification