Question:
Identify the structure of \(\text{XeF}_{6}\) molecule from following.
Identify the structure of \(\text{XeF}_{6}\) molecule from following.
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Xenon in XeF6 has six bond pairs and one lone pair, giving a distorted octahedral shape.
Updated On: Nov 12, 2026
- Square planar
- Trigonal bipyramidal
- Distorted octahedral
- Square pyramidal
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The shape follows from VSEPR theory using the number of bond pairs and lone pairs on the central atom.
Step 2: Count the electron pairs
Xe has 8 valence electrons and forms 6 bonds with F, leaving 2 electrons, which is one lone pair.
Total pairs \(=6+1=7\), which is \(sp^{3}d^{3}\) hybridisation.
Step 3: Shape
Seven electron pairs give a pentagonal bipyramidal arrangement. The lone pair distorts the octahedron of fluorine atoms.
Step 4: Compare the options
The shape is distorted octahedral, option (C).
Final Answer:
XeF6 is distorted octahedral. \[ \boxed{\text{(C) }\text{Distorted octahedral}} \]
The shape follows from VSEPR theory using the number of bond pairs and lone pairs on the central atom.
Step 2: Count the electron pairs
Xe has 8 valence electrons and forms 6 bonds with F, leaving 2 electrons, which is one lone pair.
Total pairs \(=6+1=7\), which is \(sp^{3}d^{3}\) hybridisation.
Step 3: Shape
Seven electron pairs give a pentagonal bipyramidal arrangement. The lone pair distorts the octahedron of fluorine atoms.
Step 4: Compare the options
The shape is distorted octahedral, option (C).
Final Answer:
XeF6 is distorted octahedral. \[ \boxed{\text{(C) }\text{Distorted octahedral}} \]
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