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

The molecular geometry of \( \text{XeF}_4 \) involves the xenon atom bonded to four fluorine atoms. Xenon has 8 valence electrons, and the bonding arrangement involves four single bonds with fluorine atoms, leaving two lone pairs on the xenon atom. The hybridization of the xenon atom in \( \text{XeF}_4 \) can be determined by considering the number of electron pairs around the central atom. In this case, there are 4 bonding pairs and 2 lone pairs, making a total of 6 electron pairs. To accommodate 6 electron pairs, the xenon atom undergoes \( sp^3d^2 \) hybridization, which involves the mixing of one s orbital, three p orbitals, and two d orbitals to form six hybrid orbitals. These hybrid orbitals form bonds with the fluorine atoms and also accommodate the lone pairs. Thus, the correct hybridization of \( \text{XeF}_4 \) is \( sp^3d^2 \).