To draw the Lewis structure for xenon tetrachloride (XeCl4), we start by determining the total number of valence electrons. Xenon (Xe) has 8 valence electrons, and each chlorine (Cl) atom has 7 valence electrons. Therefore, for XeCl4, the total number of valence electrons is:
- 1 Xe: 8 electrons
- 4 Cl: 4 x 7 = 28 electrons
- Total: 8 + 28 = 36 valence electrons
Now, we can draw the Lewis structure:
- Place the Xe atom in the center and surround it with 4 Cl atoms.
- Each Cl atom forms a single bond with Xe, utilizing 8 electrons (2 per bond).
- This leaves us with 28 – 8 = 28 electrons. Each Cl atom will have 6 additional nonbonding electrons (making a total of 8 electrons around each Cl, achieving the octet rule).
- Xe, being in period 4, can have more than 8 electrons around it. It has 4 bonds (8 electrons) plus 4 additional lone electrons, totaling 12 electrons.
The Lewis structure looks like this:
Now, let’s discuss the following information:
A. Formal Charge
To calculate the formal charge for each atom, we use the formula:
Formal Charge = Valence Electrons – (Nonbonding Electrons + 0.5 x Bonding Electrons)
- For Xe: 8 – (4 nonbonding + 4 bonding) = 0
- For each Cl: 7 – (6 nonbonding + 1 bonding) = 0
Thus, the formal charges are 0 for Xe and 0 for each Cl.
B. Total Number of Electron Domains
There are 4 bonding pairs and 0 lone pairs around the central Xe atom. Hence, the total number of electron domains is 4.
C. Electron Geometry
Since there are 4 electron domains, the electron geometry is Tetrahedral.
D. Molecular Geometry
The molecular geometry of XeCl4, considering all domains are bonding, is also Tetrahedral.
E. Polarity
XeCl4 is a nonpolar molecule because the dipoles from the Cl-Xe bonds cancel each other out due to the symmetric tetrahedral shape.