After drawing the Lewis structure of XeCl2, what is the hybridization of the central xenon atom?

• A. sp2
• B. sp
• C. sp3d
• D. sp3

Answer: (C)

Verified step by step guidance

1. First, draw the Lewis structure of XeCl\_2. Xenon (Xe) is the central atom bonded to two chlorine (Cl) atoms. Count the total valence electrons: Xe has 8 valence electrons, each Cl has 7, so total electrons = 8 + 2 \times 7 = 22 electrons.
2. Distribute the electrons to form bonds and complete octets. Place single bonds between Xe and each Cl (using 4 electrons), then assign lone pairs to complete the octets of Cl atoms. The remaining electrons will be placed as lone pairs on Xe.
3. Count the regions of electron density (bonding and lone pairs) around the central Xe atom. Each bond counts as one region, and each lone pair counts as one region. For XeCl\_2, Xe has 2 bonding pairs and 3 lone pairs, totaling 5 regions of electron density.
4. Determine the electron geometry based on the number of electron density regions. Five regions correspond to a trigonal bipyramidal electron geometry, which requires dsp\^3 (or sp\^3d) hybridization.
5. Conclude that the hybridization of the central Xe atom in XeCl\_2 is sp\^3d, consistent with the trigonal bipyramidal electron geometry and the presence of three lone pairs.