---
title: "The ground-state valence electron configurations for four elements in Period \\(3\\) are shown in the table below.  | Element | Valence electron configuration | | :— | :— | | \\(\\text{Mg}\\) | \\([\\text{Ne}]\\,3s^2\\) | | \\(\\text{Al}\\) | \\([\\text{Ne}]\\,3s^2\\,3p^1\\) | | \\(\\text{P}\\) | \\([\\text{Ne}]\\,3s^2\\,3p^3\\) | | \\(\\text{S}\\) | \\([\\text{Ne}]\\,3s^2\\,3p^4\\) |  Which of the following correctly ranks these elements in order of decreasing first ionization energy, and provides the valid electronic justification for the observed ranking?"
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url: "https://nerd-notes.com/ubq/123514/"
date_modified: "2026-09-28T11:59:59+00:00"
---

# The ground-state valence electron configurations for four elements in Period \(3\) are shown in the table below.

| Element | Valence electron configuration |
| :— | :— |
| \(\text{Mg}\) | \([\text{Ne}]\,3s^2\) |
| \(\text{Al}\) | \([\text{Ne}]\,3s^2\,3p^1\) |
| \(\text{P}\) | \([\text{Ne}]\,3s^2\,3p^3\) |
| \(\text{S}\) | \([\text{Ne}]\,3s^2\,3p^4\) |

Which of the following correctly ranks these elements in order of decreasing first ionization energy, and provides the valid electronic justification for the observed ranking?

The ground-state valence electron configurations for four elements in Period \(3\) are shown in the table below.

| Element | Valence electron configuration |
| :--- | :--- |
| \(\text{Mg}\) | \([\text{Ne}]\,3s^2\) |
| \(\text{Al}\) | \([\text{Ne}]\,3s^2\,3p^1\) |
| \(\text{P}\) | \([\text{Ne}]\,3s^2\,3p^3\) |
| \(\text{S}\) | \([\text{Ne}]\,3s^2\,3p^4\) |

Which of the following correctly ranks these elements in order of decreasing first ionization energy, and provides the valid electronic justification for the observed ranking?

- **A.** \(\text{P} > \text{S} > \text{Al} > \text{Mg}\), because removing an electron from a doubly occupied \(3p\) orbital in \(\text{S}\) requires less energy due to electron-electron repulsion, and removing an electron from \(\text{Al}\) requires more energy than from \(\text{Mg}\) due to a greater nuclear charge.
- **B.** \(\text{P} > \text{S} > \text{Mg} > \text{Al}\), because removing an electron from a doubly occupied \(3p\) orbital in \(\text{S}\) requires less energy due to electron-electron repulsion, and removing an electron from \(\text{Al}\) requires less energy than from \(\text{Mg}\) because the \(3p\) electron is in a higher-energy subshell shielded by the \(3s\) electrons.
- **C.** \(\text{S} > \text{P} > \text{Mg} > \text{Al}\), because \(\text{S}\) has a higher effective nuclear charge than \(\text{P}\) across Period \(3\), and removing an electron from \(\text{Al}\) requires less energy than from \(\text{Mg}\) because the \(3p\) electron is in a higher-energy subshell shielded by the \(3s\) electrons.
- **D.** \(\text{S} > \text{P} > \text{Al} > \text{Mg}\), because first ionization energy increases strictly from left to right across Period \(3\) as the number of protons and effective nuclear charge increase.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123514/*