Question
Predict the shape and bond angle of using VSEPR theory. Also state the hybridisation of the central atom.
Solution — Step by Step
Xenon has valence electrons. Each F contributes electron to the bond pair, so 4 bond pairs use electrons. Remaining: electrons = lone pairs.
Total electron domains around Xe: bond pairs + lone pairs = .
Steric number means octahedral electron geometry. With lone pairs, they occupy opposite (axial) positions to minimise repulsion. The 4 F atoms sit in the equatorial plane.
Molecular shape: square planar. Bond angle: (F-Xe-F).
Hybridisation: (matches steric number 6).
Final answer: Square planar, , .
Why This Works
VSEPR theory says lone pairs and bond pairs around a central atom arrange themselves to minimise repulsion. For 6 domains, octahedral arrangement is optimal. Lone pairs go opposite to each other (rather than adjacent) because that maximises lone-pair-lone-pair distance.
The hybridisation labels the orbital scheme that matches the geometry. Six domains require six equivalent hybrid orbitals, which only provides.
Alternative Method
Use molecular orbital theory or the Bent’s rule analysis. Both give the same shape but require more setup. For NEET/JEE Main, VSEPR is the fastest route.
NEET asks about , , and shapes almost every year. Memorise: linear (), square planar (), distorted octahedral ().
Common Mistake
Saying is tetrahedral. That would be true if Xe had no lone pairs. With 2 lone pairs in axial positions, the four F atoms form a square plane — not a tetrahedron.