---
title: "A student investigating trends in ionic crystal lattices examines the data below for four isoelectronic ions with the ground-state electron configuration \\(1s^2 2s^2 2p^6\\).  | Ion | Number of protons | Number of electrons | | :— | :— | :— | | \\(\\text{O}^{2-}\\) | \\(8\\) | \\(10\\) | | \\(\\text{F}^-\\)| \\(9\\) | \\(10\\) | | \\(\\text{Na}^+\\) | \\(11\\) | \\(10\\) | | \\(\\text{Mg}^{2+}\\) | \\(12\\) | \\(10\\) |  Which of the following correctly ranks the ions in order of decreasing ionic radius, and provides the best explanation for this trend?"
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url: "https://nerd-notes.com/ubq/121459/"
date_modified: "2026-08-23T05:04:37+00:00"
---

# A student investigating trends in ionic crystal lattices examines the data below for four isoelectronic ions with the ground-state electron configuration \(1s^2 2s^2 2p^6\).

| Ion | Number of protons | Number of electrons |
| :— | :— | :— |
| \(\text{O}^{2-}\) | \(8\) | \(10\) |
| \(\text{F}^-\)| \(9\) | \(10\) |
| \(\text{Na}^+\) | \(11\) | \(10\) |
| \(\text{Mg}^{2+}\) | \(12\) | \(10\) |

Which of the following correctly ranks the ions in order of decreasing ionic radius, and provides the best explanation for this trend?

A student investigating trends in ionic crystal lattices examines the data below for four isoelectronic ions with the ground-state electron configuration \(1s^2 2s^2 2p^6\).

| Ion | Number of protons | Number of electrons |
| :--- | :--- | :--- |
| \(\text{O}^{2-}\) | \(8\) | \(10\) |
| \(\text{F}^-\)| \(9\) | \(10\) |
| \(\text{Na}^+\) | \(11\) | \(10\) |
| \(\text{Mg}^{2+}\) | \(12\) | \(10\) |

Which of the following correctly ranks the ions in order of decreasing ionic radius, and provides the best explanation for this trend?

- **A.** \(\text{Mg}^{2+} > \text{Na}^+ > \text{F}^- > \text{O}^{2-}\), because \(\text{Mg}^{2+}\) has the highest atomic number and experiences the greatest electron-electron repulsion among its \(10\) electrons.
- **B.** \(\text{Mg}^{2+} > \text{Na}^+ > \text{F}^- > \text{O}^{2-}\), because \(\text{Mg}^{2+}\) has the greatest nuclear charge, which expands the electron cloud to accommodate the higher positive charge.
- **C.** \(\text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+}\), because \(\text{O}^{2-}\) has more occupied electron shells than \(\text{Mg}^{2+}\), resulting in greater shielding of the valence electrons.
- **D.** \(\text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+}\), because as the number of protons increases across the series, the effective nuclear charge increases and pulls the valence electrons closer to the nucleus.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121459/*