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title: "An aqueous solution of iodine is prepared by dissolving solid \\(\\text{I}_2\\) in an aqueous solution containing iodide ions, \\(\\text{I}^-\\), producing the triiodide complex ion according to the following equation:  \\[ \\text{I}_2\\text{(aq)} + \\text{I}^-\\text{(aq)} \\rightarrow \\text{I}_3^-\\text{(aq)} \\]  Based on VSEPR theory and the distribution of valence electrons in the Lewis diagram of \\(\\text{I}_3^-\\), which of the following correctly predicts and explains the molecular geometry of the \\(\\text{I}_3^-\\cos\\) ion?"
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url: "https://nerd-notes.com/ubq/119659/"
date_modified: "2026-08-21T08:11:57+00:00"
---

# An aqueous solution of iodine is prepared by dissolving solid \(\text{I}_2\) in an aqueous solution containing iodide ions, \(\text{I}^-\), producing the triiodide complex ion according to the following equation:

\[ \text{I}_2\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{I}_3^-\text{(aq)} \]

Based on VSEPR theory and the distribution of valence electrons in the Lewis diagram of \(\text{I}_3^-\), which of the following correctly predicts and explains the molecular geometry of the \(\text{I}_3^-\cos\) ion?

An aqueous solution of iodine is prepared by dissolving solid \(\text{I}_2\) in an aqueous solution containing iodide ions, \(\text{I}^-\), producing the triiodide complex ion according to the following equation:

\[ \text{I}_2\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{I}_3^-\text{(aq)} \]

Based on VSEPR theory and the distribution of valence electrons in the Lewis diagram of \(\text{I}_3^-\), which of the following correctly predicts and explains the molecular geometry of the \(\text{I}_3^-\cos\) ion?

- **A.** Linear, because the central iodine atom has \(5\) electron domains with the \(3\) nonbonding electron pairs located in equatorial positions.
- **B.** Bent, because the central iodine atom has \(4\) electron domains consisting of \(2\) bonding pairs and \(2\) nonbonding electron pairs.
- **C.** T-shaped, because the central iodine atom has \(5\) electron domains with \(2\) nonbonding electron pairs in axial positions and \(1\) nonbonding pair in an equatorial position.
- **D.** Linear, because the central iodine atom forms \(2\) double bonds with \(0\) nonbonding electron pairs, resulting in \(2\) electron domains.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119659/*