Question

draw the lewis structure of xecl₂ and then determine the hybridization of the central atom.

Explanation:

Step1: Calculate the number of valence electrons

Xe has 8 valence electrons, and each Cl has 7 valence electrons. So the total number of valence electrons for \(XeCl_{2}\) is \(8 + 2\times7= 22\)

Step2: Draw the Lewis structure

• Place Xe in the center.
• Connect each Cl to Xe with a single bond (each single bond uses 2 electrons). So far, \(2\times2 = 4\) electrons are used.
• Then distribute the remaining \(22 - 4=18\) electrons as lone - pairs. Each Cl gets 3 lone - pairs (\(2\times3\times2 = 12\) electrons), and Xe gets 3 lone - pairs (\(3\times2=6\) electrons)

Step3: Determine the electron - domain geometry

The central atom Xe has 2 bonding domains (Xe - Cl bonds) and 3 non - bonding domains (lone - pairs). The total number of electron - domains is \(2 + 3=5\)

Step4: Determine the hybridization

According to the VSEPR (Valence - Shell Electron - Pair Repulsion) theory, when the number of electron - domains \(n = 5\), the hybridization is \(sp^{3}d\)

Answer:

The Lewis structure of \(XeCl_{2}\) has Xe in the center with two Cl atoms bonded to it via single bonds and Xe having 3 lone - pairs. The hybridization of the central Xe atom is \(sp^{3}d\) (Option C)