---
title: "The central atoms in \\(\\text{CH}_4\\), \\(\\text{NH}_3\\), and \\(\\text{H}_2\\text{O}\\) each have four electron domains in their respective Lewis diagrams. Which of the following correctly ranks the bond angles in these molecules from greatest to least, with the correct justification?"
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url: "https://nerd-notes.com/ubq/119610/"
date_modified: "2026-08-21T17:17:19+00:00"
---

# The central atoms in \(\text{CH}_4\), \(\text{NH}_3\), and \(\text{H}_2\text{O}\) each have four electron domains in their respective Lewis diagrams. Which of the following correctly ranks the bond angles in these molecules from greatest to least, with the correct justification?

The central atoms in \(\text{CH}_4\), \(\text{NH}_3\), and \(\text{H}_2\text{O}\) each have four electron domains in their respective Lewis diagrams. Which of the following correctly ranks the bond angles in these molecules from greatest to least, with the correct justification?

- **A.** \(\text{H}_2\text{O} > \text{NH}_3 > \text{CH}_4\), because oxygen has the highest electronegativity among the central atoms, which draws bonding electron pairs closer to the nucleus and widens the bond angle.
- **B.** \(\text{H}_2\text{O} > \text{NH}_3 > \text{CH}_4\), because having more lone pairs on the central atom causes greater electron repulsion that pushes the hydrogen atoms farther apart from one another.
- **C.** \(\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}\), because carbon has a larger atomic radius than nitrogen and oxygen, allowing the bonded hydrogen atoms to be positioned farther apart.
- **D.** \(\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}\), because nonbonding lone pairs exert greater repulsive force than bonding pairs, compressing the bond angles more as the number of lone pairs on the central atom increases.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119610/*