---
title: "Both oxygen and sulfur belong to Group 16 of the periodic table. When reacting with excess fluorine, sulfur readily forms the stable molecular compounds \\(\\text{SF}_4\\) and \\(\\text{SF}_6\\), whereas oxygen forms \\(\\text{OF}_2\\) but cannot form \\(\\text{OF}_4\\) or \\(\\text{OF}_6\\). Which of the following statements best justifies why sulfur can form \\(\\text{SF}_6\\) while oxygen is unable to form \\(\\text{OF}_6\\)?"
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url: "https://nerd-notes.com/ubq/123482/"
date_modified: "2026-09-28T11:59:53+00:00"
---

# Both oxygen and sulfur belong to Group 16 of the periodic table. When reacting with excess fluorine, sulfur readily forms the stable molecular compounds \(\text{SF}_4\) and \(\text{SF}_6\), whereas oxygen forms \(\text{OF}_2\) but cannot form \(\text{OF}_4\) or \(\text{OF}_6\). Which of the following statements best justifies why sulfur can form \(\text{SF}_6\) while oxygen is unable to form \(\text{OF}_6\)?

Both oxygen and sulfur belong to Group 16 of the periodic table. When reacting with excess fluorine, sulfur readily forms the stable molecular compounds \(\text{SF}_4\) and \(\text{SF}_6\), whereas oxygen forms \(\text{OF}_2\) but cannot form \(\text{OF}_4\) or \(\text{OF}_6\). Which of the following statements best justifies why sulfur can form \(\text{SF}_6\) while oxygen is unable to form \(\text{OF}_6\)?

- **A.** Sulfur can form \(\text{SF}_6\) whereas oxygen cannot because sulfur has a lower electronegativity than oxygen, which minimizes repulsive electrostatic interactions between adjacent fluorine ligands.
- **B.** Sulfur can form \(\text{SF}_6\) whereas oxygen cannot because sulfur possesses a greater effective nuclear charge than oxygen, enabling its nucleus to attract and stabilize twelve shared valence electrons.
- **C.** Sulfur can form \(\text{SF}_6\) whereas oxygen cannot because the valence electrons of sulfur occupy the \(n = 3\) shell where accessible \(d\) orbitals allow an expanded valence shell, whereas oxygen is limited to the \(n = 2\) shell containing only \(s\) and \(p\) subshells.
- **D.** Sulfur can form \(\text{SF}_6\) whereas oxygen cannot because the \(\text{S}-\text{F}\) covalent bonds have a higher bond enthalpy than hypothetical \(\text{O}-\text{F}\) bonds, providing the necessary thermodynamic driving force for hypervalent coordination.

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