Final answer:
To find the number of valence electrons in XeF₂, combine 8 from xenon with 14 from two fluorine atoms for a total of 22. The Lewis structure has two bonding pairs and three lone pairs around the central xenon atom, leading to a linear electron pair geometry and molecular structure.
Step-by-step explanation:
Determining Valence Electrons and Drawing Lewis Structure for XeF₂
To determine the number of valence electrons in XeF₂, we add the valence electrons of xenon (8) to those of two fluorine atoms (2 x 7). This gives us a total of 22 valence electrons for XeF₂. As for the Lewis structure, xenon (Xe) will be the central atom with fluorine (F) atoms bonded to it. Each fluorine atom gets three lone pairs to complete its octet, leaving us with three lone pairs to place on the xenon atom. Unlike most other elements, xenon can have more than eight electrons in its valence shell due to the availability of d orbitals.
The Lewis structure is:
:F-Xe-F:
with two bonding pairs between xenon and the fluorine atoms, and three lone pairs on xenon.
Check Your Learning: The electron pair geometry for XeF₂ is linear, and the molecular structure is also linear because the lone pairs on xenon are placed symmetrically around the atom to minimize electron pair repulsion.