To draw the Lewis structure for xenon tetroxide (XeO4), we start by determining the total number of valence electrons. Xenon (Xe) has 8 valence electrons, and each oxygen (O) atom has 6 valence electrons. Since there are 4 oxygen atoms, we have:
- Valence electrons from Xe: 8
- Valence electrons from 4 O: 4 × 6 = 24
- Total = 8 + 24 = 32 valence electrons
Next, we place the central atom, xenon, in the center and connect it to the four oxygen atoms. Each Xe-O bond uses 2 electrons, which means:
- 4 bonds × 2 electrons per bond = 8 electrons used
- Remaining electrons = 32 – 8 = 24 electrons
Now, we distribute the remaining electrons to the oxygen atoms. Each oxygen atom needs 8 electrons to be stable, and since we have 4 oxygen atoms:
- 4 O × 6 electrons = 24 electrons
We can place 6 electrons (3 lone pairs) on each oxygen atom. After that, each oxygen will be satisfied (having 8 electrons: 2 from the bond with xenon and 6 from their lone pairs).
The structure will resemble a tetrahedral geometry since the four oxygen atoms surround the central xenon atom. Since the bonding in XeO4 is symmetrical, the dipoles cancel out.
Now, regarding polarity: the molecule is nonpolar because its symmetrical shape leads to an even distribution of electrical charge. Even though the bonds between Xe and O are polar due to the difference in electronegativity, the molecular geometry ensures that any dipoles cancel each other out, rendering the overall molecule nonpolar.