---
title: "The polyatomic ion \\(\\text{IO}_2\\text{F}_2^-\\), which has the most favorable formal-charge arrangement with an expanded octet on iodine, contains a central iodine atom bonded to two oxygen atoms and two fluorine atoms. Which of the following correctly identifies the molecular geometry of \\(\\text{IO}_2\\text{F}_2^-\\) and provides the best justification using VSEPR theory?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/119546/"
date_modified: "2026-08-21T03:26:48+00:00"
---

# The polyatomic ion \(\text{IO}_2\text{F}_2^-\), which has the most favorable formal-charge arrangement with an expanded octet on iodine, contains a central iodine atom bonded to two oxygen atoms and two fluorine atoms. Which of the following correctly identifies the molecular geometry of \(\text{IO}_2\text{F}_2^-\) and provides the best justification using VSEPR theory?

The polyatomic ion \(\text{IO}_2\text{F}_2^-\), which has the most favorable formal-charge arrangement with an expanded octet on iodine, contains a central iodine atom bonded to two oxygen atoms and two fluorine atoms. Which of the following correctly identifies the molecular geometry of \(\text{IO}_2\text{F}_2^-\) and provides the best justification using VSEPR theory?

- **A.** Tetrahedral, because the central \(\text{I}\) atom forms four bonds and has an octet of valence electrons without any nonbonding electron pairs.
- **B.** Square planar, because the central \(\text{I}\) atom has five electron domains that orient the four bonding pairs into a single plane to minimize repulsion with a lone pair in the axial position.
- **C.** See-saw, because the central \(\text{I}\) atom is surrounded by five electron domains (four bonding domains and one lone pair), and placing the lone pair in an equatorial position minimizes \(90^\circ\) electron-pair repulsions.
- **D.** See-saw, because the central \(\text{I}\) atom is surrounded by four electron domains, and resonance delocalization between the two \(\text{I}=\text{O}\) bonds creates unequal electron repulsions that distort the ideal tetrahedral shape.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119546/*