The predicted geometry and polarity of the inorganic molecule XeF4 xenon hexafluoride can be determined using the VSEPR Valence Shell Electron Pair Repulsion theory.First, let's determine the electron domain geometry. Xenon Xe has 8 valence electrons, and each fluorine F atom has 7 valence electrons. In XeF4, there are four fluorine atoms bonded to the central xenon atom, which contributes 4 electrons. Additionally, there are two lone pairs of electrons on the xenon atom, contributing another 4 electrons. In total, there are 6 electron domains around the central Xe atom 4 bonding domains and 2 non-bonding domains .According to VSEPR theory, the electron domain geometry of XeF4 is octahedral. However, we are interested in the molecular geometry, which only considers the positions of the atoms, not the lone pairs. With four bonding domains and two non-bonding domains, the molecular geometry of XeF4 is square planar.Now, let's determine the polarity of XeF4. Polarity depends on the electronegativity difference between the atoms and the molecular geometry. Fluorine is more electronegative than xenon, so the Xe-F bonds are polar. However, due to the square planar geometry, the dipoles of the polar Xe-F bonds cancel each other out, resulting in a nonpolar molecule.In conclusion, the predicted geometry of the inorganic molecule XeF4 is square planar, and its polarity is nonpolar.