According to VSEPR theory, XeF2 has what molecular shape?

Question

a) tetrahedral

b) trigonal bipyramidal

c) trigonal pyramidal

d) linear

e) bent

please help I know the answer is linear but not sure why, will give 5 pts to best answer

Answer 1

Check the hybridization. If we consider Xe to have eight valence electrons, and each fluorine to have seven, there are a total of twenty-two to distribute. Six can be placed on each fluorine atom (since we know these form single bonds), leaving fourteen. Each single bond accounts for two electrons, leaving six. Therefore, xenon has three lone pairs as well as three bonding pairs around it. Five electron pairs -----> sp3d hybridization. Now, lone pairs repel each other more strongly than bonding pairs. Therefore, it is only natural for them to take the three equatorial positions of the bipyramidial structure of sp3d hybridization, where they can be separated by 120 degrees... whereas if one of them were to occupy the axial position, there would be a 90-degree LP-LP interaction. Hence, the two bonding pairs occupy the axial positions, and the molecule is linear.

Answer 2

Linear

Answer 3

d) 3 lone pairs on Xe and  2 Xe-F bonds across from each other; electron pair geometry is trigonal bipyramidal