Question
Predict the geometry, hybridisation, and bond angle of using VSEPR theory.
Solution — Step by Step
Xe has 8 valence electrons. Each F contributes 1 (it shares one). Total around Xe: electrons, or electron pairs.
Of the pairs around Xe: are bond pairs (with the four F atoms) and are lone pairs.
The arrangement of electron pairs is octahedral. Two lone pairs occupy axial positions to minimise repulsion (since axial-axial repulsion at is less than equatorial-equatorial). The remaining four bond pairs go to the four equatorial corners.
- Geometry (only counting atoms, not lone pairs): square planar
- Hybridisation:
- Bond angle (F-Xe-F):
Final answer: Square planar, , .
Why This Works
VSEPR ranks lone pairs as occupying more space than bond pairs (lone pairs are held by only one nucleus, so they spread out more). With six electron pairs, the parent geometry is octahedral. Lone pairs in trans positions cancel each other’s repulsion, leaving a flat square of fluorines.
The hybridisation accommodates six electron domains. Both Xe and the F atoms contribute to the bonding orbitals.
Alternative Method
Draw the Lewis structure first: Xe in centre, 4 F single bonds, 2 lone pairs on Xe. Count electron domains (4 + 2 = 6). Apply VSEPR. Same result, but Lewis-first is clearer for students new to VSEPR.
Memorise the lone-pair pattern for octahedral systems: ABE gives square pyramidal, ABE gives square planar, ABE gives T-shape. JEE asks about XeF and BrF almost every year.
Common Mistake
Students often say XeF is tetrahedral (confusing it with ). The presence of two lone pairs forces square planar geometry, NOT tetrahedral. The hybridisation involves six orbitals, including two from the d-shell.