---
title: "A student analyzes the potential-energy curves for the bonds in \\(\\text{F}_2\\), \\(\\text{ClF}\\), and \\(\\text{Cl}_2\\). The covalent radius of \\(\\text{F}\\) is smaller than that of \\(\\text{Cl}\\), and the equilibrium bond length is approximated by the sum of the bonded atoms’ covalent radii. All three curves are referenced to separated atoms at \\(U=0\\).  Which statement correctly identifies the curve for \\(\\text{ClF}\\) and compares its bond-dissociation energy with that of \\(\\text{Cl}_2\\)?"
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/120082/"
date_modified: "2026-08-21T08:41:18+00:00"
---

# A student analyzes the potential-energy curves for the bonds in \(\text{F}_2\), \(\text{ClF}\), and \(\text{Cl}_2\). The covalent radius of \(\text{F}\) is smaller than that of \(\text{Cl}\), and the equilibrium bond length is approximated by the sum of the bonded atoms’ covalent radii. All three curves are referenced to separated atoms at \(U=0\).

Which statement correctly identifies the curve for \(\text{ClF}\) and compares its bond-dissociation energy with that of \(\text{Cl}_2\)?

A student analyzes the potential-energy curves for the bonds in \(\text{F}_2\), \(\text{ClF}\), and \(\text{Cl}_2\). The covalent radius of \(\text{F}\) is smaller than that of \(\text{Cl}\), and the equilibrium bond length is approximated by the sum of the bonded atoms’ covalent radii. All three curves are referenced to separated atoms at \(U=0\).

Which statement correctly identifies the curve for \(\text{ClF}\) and compares its bond-dissociation energy with that of \(\text{Cl}_2\)?

![Draw a strictly grayscale potential-energy graph. Label the horizontal axis “Internuclear distance, \(r\) (\(\text{pm}\))” and the vertical axis “Potential energy, \(U\) (\(\text{kJ mol}^{-1}\))”. Use bare axes with no gridlines. Place horizontal tick marks at \(140\), \(165\), and \(200\), and vertical tick marks at \(0\), \(-160\), \(-240\), and \(-255\). A legend maps \(\text{P}\) to a solid line, \(\text{Q}\) to a dashed line, and \(\text{R}\) to a dotted line. The solid curve has its minimum at \((140,-160)\), the dashed curve at \((165,-255)\), and the dotted curve at \((200,-240)\). Mark each minimum with a small open circle. Each curve rises steeply at shorter distances and approaches the horizontal zero-energy level asymptotically from below at larger distances. No other labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787301678-653wHV.jpg)

- **A.** Curve \(\text{Q}\) represents \(\text{ClF}\), and dissociating \(\text{ClF}\) requires more energy per mole than dissociating \(\text{Cl}_2\), because \(\text{Q}\) has an intermediate equilibrium distance and a deeper well than \(\text{R}\).
- **B.** Curve \(\text{Q}\) represents \(\text{ClF}\), and dissociating \(\text{ClF}\) requires more energy per mole than dissociating \(\text{Cl}_2\), because \(\text{Q}\) has a shorter equilibrium distance than \(\text{R}\), and shorter bonds are necessarily stronger.
- **C.** Curve \(\text{Q}\) represents \(\text{ClF}\), and dissociating \(\text{ClF}\) requires less energy per mole than dissociating \(\text{Cl}_2\), because the minimum of \(\text{Q}\) is at a lower potential energy than the minimum of \(\text{R}\).
- **D.** Curve \(\text{P}\) represents \(\text{ClF}\), and dissociating \(\text{ClF}\) requires less energy per mole than dissociating \(\text{Cl}_2\), because the presence of \(\text{F}\) gives \(\text{ClF}\) the shortest equilibrium distance and the well for \(\text{P}\) is shallower than that for \(\text{R}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120082/*
