Final answer:
The empirical formula is XeF2, which has a linear molecular shape and sp3d hybridization for the xenon atom.
Step-by-step explanation:
To determine the empirical formula of a compound with 77.55% Xe and 22.45% F by mass, we'll assume that we have 100 grams of the compound, meaning we have 77.55 grams of Xe and 22.45 grams of F. Next, we'll convert these masses to moles by using the atomic masses: Xe (131.29 g/mol) and F (19.00 g/mol).
Moles of Xe = 77.55 g / 131.29 g/mol = 0.5906 mol Moles of F = 22.45 g / 19.00 g/mol = 1.181 mol
We then divide by the smallest number of moles to get the simplest ratio: Ratio of Xe:F = 0.5906 mol : 1.181 mol ≈ 1:2
Therefore, the empirical formula is XeF2.
To draw the Lewis structure for XeF2, we arrange two fluorine atoms around the xenon central atom with two bonds and include three lone pairs on the xenon to satisfy its octet, resulting in a linear molecular geometry.
Predicting the shape of the molecule, we know that XeF2 is linear because of the two bonded pairs and three lone pairs arranging in a way to minimize repulsion, which is described by the VSEPR theory.
The hybridization of Xe in XeF2 is sp3d because there are five areas of electron density (two bonding pairs and three lone pairs).