---
title: "A student draws the Lewis diagram shown below for xenon tetrafluoride (\\(\\text{XeF}_4\\)). Which of the following claims best describes the predicted \\(\\text{F}-\\text{Xe}-\\text{F}\\) bond angles in \\(\\text{XeF}_4\\) and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/119174/"
date_modified: "2026-08-19T12:32:49+00:00"
---

# A student draws the Lewis diagram shown below for xenon tetrafluoride (\(\text{XeF}_4\)). Which of the following claims best describes the predicted \(\text{F}-\text{Xe}-\text{F}\) bond angles in \(\text{XeF}_4\) and provides the correct justification?

A student draws the Lewis diagram shown below for xenon tetrafluoride (\(\text{XeF}_4\)). Which of the following claims best describes the predicted \(\text{F}-\text{Xe}-\text{F}\) bond angles in \(\text{XeF}_4\) and provides the correct justification?

![A Lewis structure centered on a Xe atom. Exactly four single bonds extend orthogonally from the central Xe atom to four surrounding F atoms arranged at 90-degree intervals (top, bottom, left, right). Each of the four F atoms has exactly three lone pairs of dots drawn on its outer perimeter, with no lone pairs drawn between the F and Xe atoms. The central Xe atom carries exactly two lone pairs of dots, with one pair positioned in the upper-left quadrant between the top and left bonds, and the other pair positioned in the lower-right quadrant between the bottom and right bonds. The diagram contains exactly 36 valence electrons total (8 bonding electrons and 28 nonbonding electrons). No formal charges, brackets, or other annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787142769-X8wvB1.jpg)

- **A.** The \(\text{F}-\text{Xe}-\text{F}\) bond angles are equal to \(90^\circ\) because the two lone pairs on the central \(\text{Xe}\) atom occupy positions \(180^\circ\) apart, exerting symmetrical repulsions above and below the square plane.
- **B.** The \(\text{F}-\text{Xe}-\text{F}\) bond angles are less than \(90^\circ\) because lone pair-bonding pair repulsions are stronger than bonding pair-bonding pair repulsions, compressing the bonds toward each other.
- **C.** The \(\text{F}-\text{Xe}-\text{F}\) bond angles are less than \(90^\circ\) because six electron domains around the central \(\text{Xe}\) atom create steric crowding that distorts the square planar geometry.
- **D.** The \(\text{F}-\text{Xe}-\text{F}\) bond angles are equal to \(90^\circ\) because the central \(\text{Xe}\) atom forms four equivalent bonds with zero lone pairs in its valence shell.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119174/*