---
title: "A student compares the molecular structures of the nitrosyl halides \\(\\text{NOF}\\), \\(\\text{NOCl}\\), and \\(\\text{NOBr}\\). For each molecule, the most favorable Lewis structure has the connectivity \\(\\text{X}-\\text{N}=\\text{O}\\), with one lone pair on the central \\(\\text{N}\\) atom. Selected electronegativity values are shown below.  | \\(\\text{X}\\) | Electronegativity, \\( \\chi_{\\text{X}} \\) | |—|—| | \\(\\text{F}\\) | \\(4.0\\) | | \\(\\text{Cl}\\) | \\(3.2\\) | | \\(\\text{Br}\\) | \\(3.0\\) |  Let \\(\\theta_{\\text{X}}\\) represent the \\(\\text{X}-\\text{N}-\\text{O}\\) bond angle. Which ranking from greatest to least and accompanying explanation are correct?"
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url: "https://nerd-notes.com/ubq/120079/"
date_modified: "2026-08-21T09:20:32+00:00"
---

# A student compares the molecular structures of the nitrosyl halides \(\text{NOF}\), \(\text{NOCl}\), and \(\text{NOBr}\). For each molecule, the most favorable Lewis structure has the connectivity \(\text{X}-\text{N}=\text{O}\), with one lone pair on the central \(\text{N}\) atom. Selected electronegativity values are shown below.

| \(\text{X}\) | Electronegativity, \( \chi_{\text{X}} \) |
|—|—|
| \(\text{F}\) | \(4.0\) |
| \(\text{Cl}\) | \(3.2\) |
| \(\text{Br}\) | \(3.0\) |

Let \(\theta_{\text{X}}\) represent the \(\text{X}-\text{N}-\text{O}\) bond angle. Which ranking from greatest to least and accompanying explanation are correct?

A student compares the molecular structures of the nitrosyl halides \(\text{NOF}\), \(\text{NOCl}\), and \(\text{NOBr}\). For each molecule, the most favorable Lewis structure has the connectivity \(\text{X}-\text{N}=\text{O}\), with one lone pair on the central \(\text{N}\) atom. Selected electronegativity values are shown below.

| \(\text{X}\) | Electronegativity, \( \chi_{\text{X}} \) |
|---|---|
| \(\text{F}\) | \(4.0\) |
| \(\text{Cl}\) | \(3.2\) |
| \(\text{Br}\) | \(3.0\) |

Let \(\theta_{\text{X}}\) represent the \(\text{X}-\text{N}-\text{O}\) bond angle. Which ranking from greatest to least and accompanying explanation are correct?

- **A.** \(\theta_{\text{F}} > \theta_{\text{Cl}} > \theta_{\text{Br}}\), because a more electronegative \(\text{X}\) draws the \(\text{N}-\text{X}\) bonding pair farther from \(\text{N}\), reducing lone-pair–bonding-pair repulsion and allowing a larger angle.
- **B.** \(\theta_{\text{Cl}} > \theta_{\text{Br}} > \theta_{\text{F}}\), because the shorter \(\text{N}-\text{Cl}\) bond places its bonding pair closer to \(\text{N}\) than does the \(\text{N}-\text{Br}\) bond, while \(\text{F}\) withdraws the most bonding density.
- **C.** \(\theta_{\text{Br}} > \theta_{\text{Cl}} > \theta_{\text{F}}\), because as \(\text{X}\) becomes less electronegative, more \(\text{N}-\text{X}\) bonding density remains near \(\text{N}\), strengthening bonding-domain repulsions and opening the \(\text{X}-\text{N}-\text{O}\) angle.
- **D.** \(\theta_{\text{Br}} > \theta_{\text{F}} > \theta_{\text{Cl}}\), because \(\text{Br}\) withdraws the least bonding density, but lone pairs on \(\text{Cl}\) repel the lone pair on \(\text{N}\) more strongly than lone pairs on \(\text{F}\) do.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120079/*