The correct geometry and hybridization for XeF4 are
- octahedral, sp3d2
- trigonal bipyramidal, sp3d
- planar triangle, sp3d3
- square planar, sp3d2
Answer
(1) octahedral, sp3d2
Solution
Xe has 8 valence electrons; 4 are used in four Xe–F bonds ⇒ 4 bond pairs + 2 lone pairs = 6 electron pairs.
Number of hybrid orbitals = 12(8+4)=6 ⇒ hybridisation sp3d2.
⇒ The six electron pairs (hybrid orbitals) are arranged octahedrally ⇒ geometry: octahedral.
∵ The two lone pairs occupy positions trans to each other, the shape of the molecule (positions of atoms only) is square planar.
∴ Geometry octahedral, hybridisation sp3d2.
Answer: (1)