Question:medium

According to VSEPR theory, the molecular shapes of \(XeF_4\), \(XeO_4\), \(XeO_2F_2\) and \(XeOF_4\) respectively are

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Always count lone pairs on central atom — they decide shape distortion.
Updated On: Apr 22, 2026
  • square planar, square planar, see-saw, square pyramidal.
  • square planar, tetrahedral, trigonal bipyramidal, octahedral.
  • square planar, tetrahedral, see-saw, square pyramidal.
  • octahedral, tetrahedral, trigonal bipyramidal, octahedral.
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Molecular shape is determined by the number of bonding pairs and lone pairs on the central atom (Xenon has 8 valence electrons). Hybridization and geometry are determined using the steric number.
Step 2: Detailed Explanation:
1. \(XeF_4\): Central Xe has 8 valence electrons. 4 electrons used for bonds with F. 4 electrons left \(\implies\) 2 lone pairs. Steric number = 4 bp + 2 lp = 6 (\(sp^3d^2\)). Geometry is octahedral, but shape is square planar.
2. \(XeO_4\): Xe uses all 8 valence electrons to form 4 double bonds with O. 0 lone pairs. Steric number = 4 sigma bonds + 0 lp = 4 (\(sp^3\)). Shape is tetrahedral.
3. \(XeO_2 F_2\): Xe has 8 val. e-. 4 used for 2 double bonds with O, 2 used for 2 bonds with F. 2 electrons left \(\implies\) 1 lone pair. Steric number = 4 sigma + 1 lp = 5 (\(sp^3d\)). Geometry is TBP, but shape is see-saw.
4. \(XeOF_4\): Xe has 8 val. e-. 2 used for 1 double bond with O, 4 used for 4 bonds with F. 2 electrons left \(\implies\) 1 lone pair. Steric number = 5 sigma + 1 lp = 6 (\(sp^3d^2\)). Geometry is octahedral, but shape is square pyramidal.
Step 3: Final Answer:
The shapes are square planar, tetrahedral, see-saw, and square pyramidal.
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