---
title: "| Element | Atomic number | Valence electron configuration | | :— | :— | :— | | \\(\\text{O}\\) | \\(8\\) | \\(2s^2\\,2p^4\\) | | \\(\\text{F}\\) | \\(9\\) | \\(2s^2\\,2p^5\\) | | \\(\\text{Cl}\\) | \\(17\\) | \\(3s^2\\,3p^5\\) |  The atomic numbers and valence electron configurations for three nonmetal atoms are provided in the table above.  Which of the following correctly ranks the atoms in order of decreasing atomic radius, and provides the correct justification based on Coulomb’s law and atomic structure?"
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url: "https://nerd-notes.com/ubq/120177/"
date_modified: "2026-08-21T16:02:48+00:00"
---

# | Element | Atomic number | Valence electron configuration |
| :— | :— | :— |
| \(\text{O}\) | \(8\) | \(2s^2\,2p^4\) |
| \(\text{F}\) | \(9\) | \(2s^2\,2p^5\) |
| \(\text{Cl}\) | \(17\) | \(3s^2\,3p^5\) |

The atomic numbers and valence electron configurations for three nonmetal atoms are provided in the table above.

Which of the following correctly ranks the atoms in order of decreasing atomic radius, and provides the correct justification based on Coulomb’s law and atomic structure?

| Element | Atomic number | Valence electron configuration |
| :--- | :--- | :--- |
| \(\text{O}\) | \(8\) | \(2s^2\,2p^4\) |
| \(\text{F}\) | \(9\) | \(2s^2\,2p^5\) |
| \(\text{Cl}\) | \(17\) | \(3s^2\,3p^5\) |

The atomic numbers and valence electron configurations for three nonmetal atoms are provided in the table above.

Which of the following correctly ranks the atoms in order of decreasing atomic radius, and provides the correct justification based on Coulomb's law and atomic structure?

- **A.** \(\text{Cl} > \text{O} > \text{F}\), because the valence electrons in \(\text{Cl}\) occupy an energy level with a larger principal quantum number (\(n = 3\)) than those in \(\text{O}\) and \(\text{F}\) (\(n = 2\)), and \(\text{F}\) has a greater effective nuclear charge than \(\text{O}\).
- **B.** \(\text{Cl} > \text{F} > \text{O}\), because the valence electrons in \(\text{Cl}\) occupy the \(n = 3\) energy level, and \(\text{F}\) has more valence electrons than \(\text{O}\), which increases electron-electron repulsions and expands the valence shell.
- **C.** \(\text{O} > \text{F} > \text{Cl}\), because \(\text{O}\) has fewer protons than \(\text{F}\) so its valence electrons experience less Coulombic attraction, while \(\text{Cl}\) has a much higher nuclear charge (\(Z = 17\)) that draws its electron shells closest to the nucleus.
- **D.** \(\text{O} > \text{Cl} > \text{F}\), because \(\text{O}\) has the lowest effective nuclear charge among the three elements, while \(\text{F}\) has the highest electronegativity and therefore holds its electrons most tightly.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120177/*