---
title: "Information about four isoelectronic ions is shown in the table below.  | Ion | Number of protons | Number of electrons | Electron configuration | | :— | :— | :— | :— | | \\(\\text{O}^{2-}\\) | \\(8\\) | \\(10\\) | \\(1s^2 2s^2 2p^6\\) | | \\(\\text{F}^-\\)| \\(9\\) | \\(10\\) | \\(1s^2 2s^2 2p^6\\) | | \\(\\text{Na}^+\\) | \\(11\\) | \\(10\\) | \\(1s^2 2s^2 2p^6\\) | | \\(\\text{Mg}^{2+}\\) | \\(12\\) | \\(10\\) | \\(1s^2 2s^2 2p^6\\) |  Which of the following lists the ions in order of decreasing ionic radius and provides the correct explanation for the trend?"
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url: "https://nerd-notes.com/ubq/119608/"
date_modified: "2026-08-21T08:11:45+00:00"
---

# Information about four isoelectronic ions is shown in the table below.

| Ion | Number of protons | Number of electrons | Electron configuration |
| :— | :— | :— | :— |
| \(\text{O}^{2-}\) | \(8\) | \(10\) | \(1s^2 2s^2 2p^6\) |
| \(\text{F}^-\)| \(9\) | \(10\) | \(1s^2 2s^2 2p^6\) |
| \(\text{Na}^+\) | \(11\) | \(10\) | \(1s^2 2s^2 2p^6\) |
| \(\text{Mg}^{2+}\) | \(12\) | \(10\) | \(1s^2 2s^2 2p^6\) |

Which of the following lists the ions in order of decreasing ionic radius and provides the correct explanation for the trend?

Information about four isoelectronic ions is shown in the table below.

| Ion | Number of protons | Number of electrons | Electron configuration |
| :--- | :--- | :--- | :--- |
| \(\text{O}^{2-}\) | \(8\) | \(10\) | \(1s^2 2s^2 2p^6\) |
| \(\text{F}^-\)| \(9\) | \(10\) | \(1s^2 2s^2 2p^6\) |
| \(\text{Na}^+\) | \(11\) | \(10\) | \(1s^2 2s^2 2p^6\) |
| \(\text{Mg}^{2+}\) | \(12\) | \(10\) | \(1s^2 2s^2 2p^6\) |

Which of the following lists the ions in order of decreasing ionic radius and provides the correct explanation for the trend?

- **A.** \(\text{Mg}^{2+} > \text{Na}^+ > \text{F}^- > \text{O}^{2-}\), because ions with a greater positive charge have more protons to expand the valence electron cloud.
- **B.** \(\text{Mg}^{2+} > \text{Na}^+ > \text{F}^- > \text{O}^{2-}\), because \(\text{Mg}^{2+}\) has the highest atomic number and therefore the largest number of occupied electron shells.
- **C.** \(\text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+}\), because the valence electrons in \(\text{O}^{2-}\) experience greater shielding from core electrons than those in \(\text{Mg}^{2+}\).
- **D.** \(\text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+}\), because the increasing nuclear charge across the series exerts a stronger electrostatic attraction on the same number of electrons.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119608/*