To draw the Lewis dot structure for xenon tetraiodide (XeI4), we start by identifying the valence electrons. Xenon (Xe) has 8 valence electrons, and each iodine (I) has 7 valence electrons. Thus, the total valence electrons for XeI4 are:
8 (from Xe) + 4 × 7 (from 4 I) = 8 + 28 = 36 valence electrons.
1. **How many electron domains are around the central atom?**
In XeI4, the central atom (xenon) is surrounded by four iodine atoms. Each bond with iodine counts as one electron domain. Therefore, there are a total of 4 electron domains around the central atom (Xe).
2. **What is the electron domain geometry?**
The electron domain geometry is determined by the number of electron domains. With 4 electron domains, the electron domain geometry is tetrahedral.
3. **How many lone pairs are around the central atom?**
In XeI4, all four of the valence electrons of xenon are used to form bonds with the iodine atoms. Therefore, there are no lone pairs around the central atom. The number of lone pairs around the central atom is 0.
4. **What is the molecular geometry?**
Since there are no lone pairs on the central atom and all positions are occupied by iodine atoms, the molecular geometry is also tetrahedral. Thus, XeI4 has a tetrahedral molecular geometry.