Draw a Lewis Dot Structure for XeI4 and how do you find the electron domain around the central atom

Draw a Lewis Dot Structure for XeI4 and how do you find the electron domain around the central atom?

The correct answer and explanation is:

Lewis Dot Structure for XeI₄ (Xenon Tetraiodide)

Step-by-step Lewis Structure:

1. Count valence electrons:
   • Xenon (Xe): 8 valence electrons
   • Iodine (I): 7 valence electrons × 4 = 28
     Total: 8 + 28 = 36 valence electrons
2. Place Xe as the central atom.
   • Connect 4 Iodine atoms to Xe with single bonds.
   • Each Xe–I bond uses 2 electrons × 4 = 8 electrons.
3. Distribute remaining electrons:
   • 36 − 8 = 28 electrons left.
   • Distribute these around the 4 iodine atoms to complete their octets (6 electrons each) → 6 × 4 = 24 electrons used.
   • 28 − 24 = 4 electrons remain. Place these as 2 lone pairs on Xe.
4. Final structure:
   • Xe has 4 bonding pairs (from the I atoms) and 2 lone pairs.
   • Each I atom has 3 lone pairs (to complete their octet).

      I
      |
I — Xe — I
      |
      I
(With two lone pairs on Xe in the equatorial positions of an octahedral geometry)

Electron Domain Geometry:

To find the electron domain around the central atom (Xe), count:

• Bonding domains: 4 (each I–Xe bond is one domain)
• Non-bonding (lone pair) domains: 2
  Total electron domains = 6

With 6 electron domains, the electron geometry is octahedral. However, due to the two lone pairs, the molecular shape is square planar.

Explanation (300 words):

The Lewis Dot Structure of XeI₄ provides insight into the molecule’s bonding and geometry. Xenon, a noble gas, can expand its octet due to available d-orbitals, allowing it to form stable compounds like XeI₄. In this compound, xenon serves as the central atom bonded to four iodine atoms. Each iodine contributes one electron to form a single bond with xenon, consuming 8 electrons in total. Since there are 36 valence electrons available, 28 remain after bonding. These are used to complete the octets of the iodine atoms (each requiring 6 more electrons), leaving 4 electrons (or 2 lone pairs) on xenon.

To determine the electron domain geometry, we count all regions of electron density around the central atom—both bonding and lone pairs. Xenon in XeI₄ has six electron domains: four bonding pairs and two lone pairs. According to VSEPR (Valence Shell Electron Pair Repulsion) theory, six electron domains arrange themselves in an octahedral geometry to minimize repulsion. The two lone pairs occupy positions opposite each other, minimizing their repulsive impact on the molecule’s geometry. This leads to a square planar molecular shape, where the four iodine atoms lie in the same plane around the xenon atom.

This analysis helps in predicting physical and chemical properties, such as polarity, reactivity, and intermolecular forces. XeI₄, despite having polar Xe–I bonds, is a nonpolar molecule overall because the dipole moments cancel due to its symmetrical square planar shape.