---
title: "A student investigates the thermodynamic stability and bond strengths of hydrogen halide molecules. The atomic radii of selected halogen elements are shown in the table below.  | Halogen | Atomic radius (pm) | | :— | :— | | \\(\\text{F}\\) | \\(42\\) | | \\(\\text{Cl}\\) | \\(79\\) | | \\(\\text{Br}\\) | \\(94\\) |  Which of the following correctly ranks the \\(\\text{H}-\\text{X}\\) bond dissociation energies from greatest to least, and provides the best justification?"
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url: "https://nerd-notes.com/ubq/123776/"
date_modified: "2026-09-28T12:30:16+00:00"
---

# A student investigates the thermodynamic stability and bond strengths of hydrogen halide molecules. The atomic radii of selected halogen elements are shown in the table below.

| Halogen | Atomic radius (pm) |
| :— | :— |
| \(\text{F}\) | \(42\) |
| \(\text{Cl}\) | \(79\) |
| \(\text{Br}\) | \(94\) |

Which of the following correctly ranks the \(\text{H}-\text{X}\) bond dissociation energies from greatest to least, and provides the best justification?

A student investigates the thermodynamic stability and bond strengths of hydrogen halide molecules. The atomic radii of selected halogen elements are shown in the table below.

| Halogen | Atomic radius (pm) |
| :--- | :--- |
| \(\text{F}\) | \(42\) |
| \(\text{Cl}\) | \(79\) |
| \(\text{Br}\) | \(94\) |

Which of the following correctly ranks the \(\text{H}-\text{X}\) bond dissociation energies from greatest to least, and provides the best justification?

- **A.** \(\text{HBr} > \text{HCl} > \text{HF}\), because the larger halogen atoms have more total electrons, resulting in greater polarizability and stronger covalent bonds.
- **B.** \(\text{HBr} > \text{HCl} > \text{HF}\), because the larger valence electron clouds of the halogen atoms allow for a greater volume of orbital overlap with hydrogen.
- **C.** \(\text{HF} > \text{HCl} > \text{HBr}\), because the greater electronegativity of fluorine draws valence electron density inward, increasing electron-electron shielding of the nuclei.
- **D.** \(\text{HF} > \text{HCl} > \text{HBr}\), because the smaller atomic radius of fluorine results in a shorter bond length and stronger Coulombic attraction between the nuclei and the shared electron pair.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123776/*