---
title: "A student models isolated \\(\\text{NO}_2^-\\), \\(\\text{NO}_3^-\\), and \\(\\text{NO}_2^+\\) ions produced in an atmospheric-ion mass spectrometer. The student constructs Lewis structures that satisfy the octet rule for nitrogen and assumes that equivalent major resonance contributors contribute equally to each resonance hybrid. Which choice correctly ranks the average \\(\\text{N}-\\text{O}\\) bond lengths from greatest to least and explains the ranking?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/120052/"
date_modified: "2026-08-21T08:40:59+00:00"
---

# A student models isolated \(\text{NO}_2^-\), \(\text{NO}_3^-\), and \(\text{NO}_2^+\) ions produced in an atmospheric-ion mass spectrometer. The student constructs Lewis structures that satisfy the octet rule for nitrogen and assumes that equivalent major resonance contributors contribute equally to each resonance hybrid. Which choice correctly ranks the average \(\text{N}-\text{O}\) bond lengths from greatest to least and explains the ranking?

A student models isolated \(\text{NO}_2^-\), \(\text{NO}_3^-\), and \(\text{NO}_2^+\) ions produced in an atmospheric-ion mass spectrometer. The student constructs Lewis structures that satisfy the octet rule for nitrogen and assumes that equivalent major resonance contributors contribute equally to each resonance hybrid. Which choice correctly ranks the average \(\text{N}-\text{O}\) bond lengths from greatest to least and explains the ranking?

- **A.** \(\text{NO}_2^+ > \text{NO}_2^- > \text{NO}_3^-\), because average \(\text{N}-\text{O}\) bond length increases as average bond order increases.
- **B.** \(\text{NO}_2^- > \text{NO}_3^- > \text{NO}_2^+\), because the greater number of equivalent resonance contributors for \(\text{NO}_3^-\) gives it a greater average bond order than \(\text{NO}_2^-\), while \(\text{NO}_2^+\) contains double bonds.
- **C.** \(\text{NO}_3^- > \text{NO}_2^+ > \text{NO}_2^-\), because the lone pair on nitrogen in \(\text{NO}_2^-\) contributes an additional bonding pair to the delocalized \(\text{N}-\text{O}\) bonding system.
- **D.** \(\text{NO}_3^- > \text{NO}_2^- > \text{NO}_2^+\), because their average \(\text{N}-\text{O}\) bond orders are \(\dfrac{4}{3}\), \(\dfrac{3}{2}\), and \(2\), respectively, and greater bond order corresponds to shorter bond length.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120052/*