---
title: "A sample of gaseous selenium atoms, \\(\\text{Se}\\), is subjected to conditions that produce both \\(\\text{Se}^{2+}\\) cations and \\(\\text{Se}^{2-}\\) anions. Which of the following correctly predicts the relative sizes of the species from smallest to largest radius, and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/120184/"
date_modified: "2026-08-21T16:02:51+00:00"
---

# A sample of gaseous selenium atoms, \(\text{Se}\), is subjected to conditions that produce both \(\text{Se}^{2+}\) cations and \(\text{Se}^{2-}\) anions. Which of the following correctly predicts the relative sizes of the species from smallest to largest radius, and provides the correct justification?

A sample of gaseous selenium atoms, \(\text{Se}\), is subjected to conditions that produce both \(\text{Se}^{2+}\) cations and \(\text{Se}^{2-}\) anions. Which of the following correctly predicts the relative sizes of the species from smallest to largest radius, and provides the correct justification?

- **A.** \(\text{Se}^{2-} < \text{Se} < \text{Se}^{2+}\), because \(\text{Se}^{2+}\) has fewer electrons, resulting in less core shielding and a weaker attraction to the nucleus, whereas \(\text{Se}^{2-}\) has greater shielding that pulls the valence electrons closer.
- **B.** \(\text{Se}^{2-} < \text{Se} < \text{Se}^{2+}\), because adding valence electrons increases the effective nuclear charge (\(Z_{\text{eff}}\)) experienced by the outer shell, whereas removing valence electrons reduces \(Z_{\text{eff}}\).
- **C.** \(\text{Se}^{2+} < \text{Se} < \text{Se}^{2-}\), because removing electrons to form \(\text{Se}^{2+}\) decreases electron-electron repulsions in the valence shell, whereas adding electrons to form \(\text{Se}^{2-}\) increases electron-electron repulsions with no change in nuclear charge.
- **D.** \(\text{Se}^{2+} < \text{Se} < \text{Se}^{2-}\), because \(\text{Se}^{2+}\) has lost its entire \(n = 4\) principal energy level, whereas \(\text{Se}^{2-}\) has added electrons into a higher \(n = 5\) principal energy level.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120184/*