Xenon difluoride (XeF2) exhibits sp3d hybridization for xenon. This linear molecule’s bonding arises from xenon’s expanded octet, involving five hybrid orbitals.

Molecule Overview

XeF2 features central xenon bonded to two fluorine atoms, with three lone pairs on xenon. Xenon has eight valence electrons but expands its octet using d-orbitals.
Steric number equals bond pairs (2) plus lone pairs (3), yielding 5, which dictates sp3d hybridization.

Correct Answer: sp3d

Xenon’s ground state is 5s²5p⁶; excitation promotes one electron to 5d, yielding s¹p³d¹ with five unpaired electrons for hybridization.
Two sp3d orbitals form Xe-F sigma bonds; three hold lone pairs, resulting in linear geometry (180° bond angle).

Option Analysis

• (A) sp: Forms two hybrid orbitals for diatomic linear molecules like BeCl₂ (steric number 2). XeF2’s steric number 5 requires more orbitals.

• (B) sp²: Produces three orbitals for trigonal planar shapes like BF₃ (steric number 3). Inadequate for XeF2’s five electron domains.

• (C) sp³: Yields four tetrahedral orbitals like CH₄ (steric number 4). Fails to accommodate XeF2’s extra lone pair.

• (D) sp³d: Matches steric number 5, enabling trigonal bipyramidal electron geometry with linear molecular shape after lone pair repulsion.

Exam Relevance

For JEE/NEET, calculate steric number via Lewis dot structure: valence electrons (Xe:8, F:14) total 28, forming two bonds (4 electrons) and three Xe lone pairs. VSEPR confirms linear shape, solidifying sp3d as standard answer.