Final answer:
XeF2 has a total of three lone pairs on the Xe atom, which are represented as six electrons not involved in bonding with the fluorine atoms.
Step-by-step explanation:
In XeF2, or xenon difluoride, each fluorine (F) atom contributes seven valence electrons, for a total of 14 electrons from the two F atoms. Xenon (Xe) has eight valence electrons. When we calculate the number of valence electrons in XeF2 (8 + 2 × 7), we get 22 electrons. After forming two single bonds with the fluorine atoms, we use up four electrons, leaving 18 electrons. These 18 electrons are then distributed as lone pairs.
Given that a lone pair consists of two electrons, the Xe atom ends up having three lone pairs or six electrons that are not participating in bonding. The Lewis structure of XeF2 is represented as :F-Xe-F:, where the dots around Xe represent the lone pairs.