Question

Which of the following is the correct hybridization of \( \text{XeF}_4 \)?

• A. sp\(^3\)d
• B. sp\(^3\)
• C. sp\(^3\)d\(^2\)
• D. sp\(^3\)d\(^3\)

Hint:

For molecules with six electron pairs around the central atom, the hybridization is typically \( sp^3d^2 \), which results in an octahedral arrangement.

Answer 1

The Correct Option is C

Xenon tetrafluoride (\( \text{XeF}_4 \)) exhibits a molecular structure where the central xenon atom is covalently linked to four fluorine atoms. The xenon atom possesses 8 valence electrons. The bonding configuration comprises four single bonds with fluorine atoms, resulting in two non-bonding electron pairs on the xenon atom.The hybridization state of the xenon atom in \( \text{XeF}_4 \) is ascertainable by counting the total number of electron domains (bonding pairs and lone pairs) surrounding the central atom. In this molecule, there are 4 bonding pairs and 2 lone pairs, totaling 6 electron domains.To accommodate 6 electron domains, the xenon atom undergoes \( sp^3d^2 \) hybridization. This process involves the atomic orbital hybridization of one s orbital, three p orbitals, and two d orbitals to generate six equivalent hybrid orbitals. These hybrid orbitals are utilized for bond formation with the fluorine atoms and for housing the non-bonding electron pairs.Consequently, the determined hybridization of \( \text{XeF}_4 \) is \( sp^3d^2 \).