Question

According to the VSEPR theory, what are the most stable shapes of \(\text{XeF}_4\) and \(\text{SF}_4\), respectively?

• A. Square planar and see-saw
• B. Both see-saw
• C. See-saw and square planar
• D. Both square planar

Hint:

For a quick recall of steric configurations:
- \(\text{AB}_4\text{E}_2\) systems (like \(\text{XeF}_4\)) are always square planar.
- \(\text{AB}_4\text{E}\) systems (like \(\text{SF}_4\)) are always see-saw.
This simple classification covers most of the shape questions in exams.

Answer 1

The Correct Option is A

To determine the most stable shapes of \(\text{XeF}_4\) (Xenon Tetrafluoride) and \(\text{SF}_4\) (Sulfur Tetrafluoride) according to the VSEPR (Valence Shell Electron Pair Repulsion) theory, we need to analyze the molecular geometries based on the number of bonding pairs and lone pairs of electrons around the central atom.

Step-by-Step Analysis:

1. XeF₄ (Xenon Tetrafluoride):
   • Xenon (Xe) is the central atom and forms 4 bonds with fluorine (F).
   • Xe has 8 valence electrons, and after bonding with four fluorines, 4 electrons are left as 2 lone pairs.
   • According to the VSEPR theory, the arrangement of 4 bonding pairs and 2 lone pairs is octahedral.
   • However, the presence of lone pairs affects the shape but not the arrangement.
   • The lone pairs occupy positions opposite to each other to minimize repulsion, resulting in a square planar shape.
2. SF₄ (Sulfur Tetrafluoride):
   • Sulfur (S) is the central atom and forms 4 bonds with fluorine (F).
   • Sulfur has 6 valence electrons. After forming 4 bonds, it has 2 electrons left as 1 lone pair.
   • The arrangement of 4 bonding pairs and 1 lone pair is trigonal bipyramidal.
   • The lone pair occupies an equatorial position to minimize repulsion.
   • This results in a see-saw molecular shape.

Conclusion: Based on the VSEPR theory, \(\text{XeF}_4\) most stably forms a square planar shape, while \(\text{SF}_4\) takes on a see-saw shape. Therefore, the correct answer is Square planar and see-saw.