Question

The correct geometry and hybridization for ${XeF_4}$ are

• A. octahedral, $sp^3d^2$
• B. trigonal bipyramidal, $sp^3d$
• C. planar triangle, $sp^3d^3$
• D. square planar, $sp^3d^2$

Answer 1

The Correct Option is D

Hybridization and Geometry of \( XeF_4 \) 

\(XeF_4\) is a compound where xenon is bonded to four fluorine atoms through four sigma bonds, and xenon has two lone pairs. The formula for the compound is \( AB_4 L_2 \), where \( A \) represents the central atom (xenon), \( B \) represents the bonding atoms (fluorine), and \( L \) represents the lone pairs on the central atom (xenon).

Explanation:

• Bonds: Xenon forms four sigma bonds with four fluorine atoms, and it has two lone pairs of electrons.
• Hybridization: The hybridization of xenon is \(sp^3 d^2\), which accounts for six regions of electron density (four bonds and two lone pairs).
• Geometry: The molecular geometry of \( XeF_4 \) is octahedral because it involves six regions of electron density (including the lone pairs).
• Shape: The shape of the molecule is square planar because the two lone pairs occupy two opposite positions, leaving the fluorine atoms in a square arrangement around the xenon atom.

Conclusion:

In \( XeF_4 \), xenon undergoes \( sp^3 d^2 \) hybridization, resulting in an octahedral geometry with a square planar shape.