---
title: "A student compares the bonding, hybridization, and geometry of the central carbon atom in three common molecules: \\(\\text{CO}_2\\), \\(\\text{CH}_2\\text{O}\\), and \\(\\text{CH}_4\\). Which of the following lists the molecules in order of decreasing bond angle around the central carbon atom, along with the correct justification?"
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url: "https://nerd-notes.com/ubq/120004/"
date_modified: "2026-08-21T14:21:31+00:00"
---

# A student compares the bonding, hybridization, and geometry of the central carbon atom in three common molecules: \(\text{CO}_2\), \(\text{CH}_2\text{O}\), and \(\text{CH}_4\). Which of the following lists the molecules in order of decreasing bond angle around the central carbon atom, along with the correct justification?

A student compares the bonding, hybridization, and geometry of the central carbon atom in three common molecules: \(\text{CO}_2\), \(\text{CH}_2\text{O}\), and \(\text{CH}_4\). Which of the following lists the molecules in order of decreasing bond angle around the central carbon atom, along with the correct justification?

- **A.** \(\text{CO}_2 > \text{CH}_2\text{O} > \text{CH}_4\), because the central carbon atom is \(sp^3\) hybridized in \(\text{CO}_2\), \(sp^2\) hybridized in \(\text{CH}_2\text{O}\), and \(sp\) hybridized in \(\text{CH}_4\).
- **B.** \(\text{CO}_2 > \text{CH}_2\text{O} > \text{CH}_4\), because the central carbon atom has \(2\) electron domains in \(\text{CO}_2\) (\(sp\), \(180^\circ\)), \(3\) electron domains in \(\text{CH}_2\text{O}\) (\(sp^2\), \(120^\circ\)), and \(4\) electron domains in \(\text{CH}_4\) (\(sp^3\), \(109.5^\circ\)).
- **C.** \(\text{CH}_4 > \text{CH}_2\text{O} > \text{CO}_2\), because increasing the number of electron domains from \(2\) to \(4\) increases repulsion among electron pairs, forcing the bonding domains farther apart and creating larger bond angles.
- **D.** \(\text{CH}_4 > \text{CH}_2\text{O} > \text{CO}_2\), because the central carbon atom is bonded to \(4\) terminal atoms in \(\text{CH}_4\), \(3\) terminal atoms in \(\text{CH}_2\text{O}\), and \(2\) terminal atoms in \(\text{CO}_2\), and molecules with more surrounding atoms have larger bond angles.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120004/*