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title: "A student considers the following possible Lewis contributors for gas-phase \\(\\text{CO}\\), which has \\(10\\) valence electrons.  | Contributor | Bonding and lone pairs | Formal charges | Electron-count feature | |—|—|—|—| | \\(\\mathrm{I}\\) | One \\(\\text{C}\\equiv\\text{O}\\) triple bond and \\(1\\) lone pair on each atom | \\(\\text{C}=-1\\), \\(\\text{O}=+1\\) | \\(8\\) electrons around each atom | | \\(\\mathrm{II}\\) | One \\(\\text{C}=\\text{O}\\) double bond, \\(1\\) lone pair on \\(\\text{C}\\), and \\(2\\) lone pairs on \\(\\text{O}\\) | \\(\\text{C}=0\\), \\(\\text{O}=0\\) | \\(6\\) electrons around \\(\\text{C}\\) and \\(8\\) around \\(\\text{O}\\) | | \\(\\mathrm{III}\\) | One \\(\\text{C}-\\text{O}\\) single bond, \\(1\\) lone pair on \\(\\text{C}\\), and \\(3\\) lone pairs on \\(\\text{O}\\) | \\(\\text{C}=+1\\), \\(\\text{O}=-1\\) | \\(4\\) electrons around \\(\\text{C}\\) and \\(8\\) around \\(\\text{O}\\) |  The measured dipole moment is only \\(0.12\\ \\text{D}\\), but its negative end is at \\(\\text{C}\\). Which statement best reconciles the dipole-moment data with the Lewis contributors?"
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url: "https://nerd-notes.com/ubq/120080/"
date_modified: "2026-08-21T08:41:18+00:00"
---

# A student considers the following possible Lewis contributors for gas-phase \(\text{CO}\), which has \(10\) valence electrons.

| Contributor | Bonding and lone pairs | Formal charges | Electron-count feature |
|—|—|—|—|
| \(\mathrm{I}\) | One \(\text{C}\equiv\text{O}\) triple bond and \(1\) lone pair on each atom | \(\text{C}=-1\), \(\text{O}=+1\) | \(8\) electrons around each atom |
| \(\mathrm{II}\) | One \(\text{C}=\text{O}\) double bond, \(1\) lone pair on \(\text{C}\), and \(2\) lone pairs on \(\text{O}\) | \(\text{C}=0\), \(\text{O}=0\) | \(6\) electrons around \(\text{C}\) and \(8\) around \(\text{O}\) |
| \(\mathrm{III}\) | One \(\text{C}-\text{O}\) single bond, \(1\) lone pair on \(\text{C}\), and \(3\) lone pairs on \(\text{O}\) | \(\text{C}=+1\), \(\text{O}=-1\) | \(4\) electrons around \(\text{C}\) and \(8\) around \(\text{O}\) |

The measured dipole moment is only \(0.12\ \text{D}\), but its negative end is at \(\text{C}\). Which statement best reconciles the dipole-moment data with the Lewis contributors?

A student considers the following possible Lewis contributors for gas-phase \(\text{CO}\), which has \(10\) valence electrons.

| Contributor | Bonding and lone pairs | Formal charges | Electron-count feature |
|---|---|---|---|
| \(\mathrm{I}\) | One \(\text{C}\equiv\text{O}\) triple bond and \(1\) lone pair on each atom | \(\text{C}=-1\), \(\text{O}=+1\) | \(8\) electrons around each atom |
| \(\mathrm{II}\) | One \(\text{C}=\text{O}\) double bond, \(1\) lone pair on \(\text{C}\), and \(2\) lone pairs on \(\text{O}\) | \(\text{C}=0\), \(\text{O}=0\) | \(6\) electrons around \(\text{C}\) and \(8\) around \(\text{O}\) |
| \(\mathrm{III}\) | One \(\text{C}-\text{O}\) single bond, \(1\) lone pair on \(\text{C}\), and \(3\) lone pairs on \(\text{O}\) | \(\text{C}=+1\), \(\text{O}=-1\) | \(4\) electrons around \(\text{C}\) and \(8\) around \(\text{O}\) |

The measured dipole moment is only \(0.12\ \text{D}\), but its negative end is at \(\text{C}\). Which statement best reconciles the dipole-moment data with the Lewis contributors?

- **A.** The resonance hybrid is only slightly electron-rich at \(\text{C}\). Contributor \(\mathrm{I}\) is important enough to overcome polarization toward \(\text{O}\) because it gives both atoms octets, while contributor \(\mathrm{II}\) and the greater electronegativity of \(\text{O}\) limit the dipole magnitude.
- **B.** The resonance hybrid is only slightly electron-rich at \(\text{C}\). Contributor \(\mathrm{I}\) is important because a negative formal charge is inherently more stable on less electronegative \(\text{C}\), while contributor \(\mathrm{II}\) limits the dipole magnitude by having no formal charges.
- **C.** The resonance hybrid is slightly electron-rich at \(\text{O}\). Contributor \(\mathrm{III}\) should outweigh contributor \(\mathrm{I}\) because negative formal charge is better accommodated by the more electronegative \(\text{O}\), while the incomplete octet at \(\text{C}\) has little effect.
- **D.** The small dipole indicates that the resonance hybrid is represented almost entirely by contributor \(\mathrm{II}\). Because its atoms have zero formal charge, the \(\text{C}=\text{O}\) bond is nonpolar, and contributors \(\mathrm{I}\) and \(\mathrm{III}\) can be neglected.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120080/*