Vsepr Theory
According to VSEPR theory, the shape of XeF₄ is square planar. The number of lone pairs present on the central atom in XeF₄ is:
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About This Question
- Subject
- chemistry
- Chapter
- chemical bonding and molecular structure
- Topic
- vsepr theory
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
2
- Determine Total Valence Electrons: Xe has 8 valence electrons and each F contributes 7, giving a total of 8+4×7=36 valence electrons.
- Assign Bonding Electrons: Four Xe–F bonds use 4×2=8 electrons. Each F atom gets 3 lone pairs (6 electrons each), consuming 4×6=24 more electrons.
- Remaining Electrons on Xe: 36−8−24=4 electrons remain on Xe, forming 2 lone pairs.
- Determine Geometry: With 4 bond pairs + 2 lone pairs = 6 electron domains, the electron-pair geometry is octahedral (sp³d² hybridisation). The lone pairs occupy opposite axial positions to minimise lp–lp repulsion.
- Resulting Molecular Shape: Removing the lone pairs from the octahedral arrangement leaves the four F atoms in a plane — hence square planar.
- Conclusion: XeF₄ has 2 lone pairs on the central Xe atom.
This easy difficulty chemistry question is from the chapter chemical bonding and molecular structure, covering the topic of vsepr theory. It appeared in the 2025 exam.
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