---
title: "The atomic numbers and valence electron configurations for three halogens are provided in the table below.  | Element | Atomic number | Valence electron configuration | | :— | :— | :— | | \\(\\text{F}\\) | \\(9\\) | \\(2s^2\\,2p^5\\) | | \\(\\text{Cl}\\) | \\(17\\) | \\(3s^2\\,3p^5\\) | | \\(\\text{Br}\\) | \\(35\\) | \\(4s^2\\,4p^5\\) |  Which of the following lists the elements in order of decreasing electronegativity and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/119615/"
date_modified: "2026-08-21T08:11:47+00:00"
---

# The atomic numbers and valence electron configurations for three halogens are provided in the table below.

| Element | Atomic number | Valence electron configuration |
| :— | :— | :— |
| \(\text{F}\) | \(9\) | \(2s^2\,2p^5\) |
| \(\text{Cl}\) | \(17\) | \(3s^2\,3p^5\) |
| \(\text{Br}\) | \(35\) | \(4s^2\,4p^5\) |

Which of the following lists the elements in order of decreasing electronegativity and provides the correct justification?

The atomic numbers and valence electron configurations for three halogens are provided in the table below.

| Element | Atomic number | Valence electron configuration |
| :--- | :--- | :--- |
| \(\text{F}\) | \(9\) | \(2s^2\,2p^5\) |
| \(\text{Cl}\) | \(17\) | \(3s^2\,3p^5\) |
| \(\text{Br}\) | \(35\) | \(4s^2\,4p^5\) |

Which of the following lists the elements in order of decreasing electronegativity and provides the correct justification?

- **A.** \(\text{F} > \text{Cl} > \text{Br}\), because \(\text{Br}\) has the greatest number of occupied electron shells, creating electron-electron repulsions that push bonding electrons away.
- **B.** \(\text{F} > \text{Cl} > \text{Br}\), because the valence electrons in \(\text{F}\) occupy a shell closer to the nucleus, resulting in a stronger Coulombic attraction for shared electrons.
- **C.** \(\text{Br} > \text{Cl} > \text{F}\), because \(\text{Br}\) has the greatest nuclear charge (\(35\) protons), giving it the strongest attraction for bonding electrons.
- **D.** \(\text{Br} > \text{Cl} > \text{F}\), because \(\text{Br}\) has the largest atomic radius, which allows it to more easily accommodate additional shared electrons.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119615/*