Based on the VSEPR Valence Shell Electron Pair Repulsion theory, the predicted geometry and polarity of the XeF2 Xenon Difluoride molecule can be determined as follows:1. First, we need to determine the central atom, which is Xenon Xe in this case.2. Next, we need to determine the number of valence electrons for each atom. Xenon has 8 valence electrons, and Fluorine has 7 valence electrons. Since there are two Fluorine atoms, the total number of valence electrons in the molecule is 8 + 2 7 = 22.3. Now, we need to distribute the valence electrons in the molecule. Xenon forms two single bonds with the two Fluorine atoms, using 2 2 = 4 electrons. The remaining 18 electrons are placed as lone pairs on the central atom Xenon and the two Fluorine atoms. Xenon will have 3 lone pairs 6 electrons , and each Fluorine atom will have 3 lone pairs 6 electrons each .4. To determine the molecular geometry, we need to consider the electron groups around the central atom. In XeF2, there are a total of 5 electron groups around Xenon 2 single bonds and 3 lone pairs . According to VSEPR theory, these electron groups will arrange themselves to minimize repulsion, resulting in a linear geometry.5. To determine the polarity of the molecule, we need to consider the electronegativity difference between the atoms and the molecular geometry. Fluorine is more electronegative than Xenon, which creates a dipole moment along the Xe-F bond. However, due to the linear geometry, the dipole moments of the two Xe-F bonds cancel each other out, resulting in a nonpolar molecule.In conclusion, the predicted geometry of the XeF2 molecule based on the VSEPR theory is linear, and the molecule is nonpolar.