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Midterm 2015 #4

Posted: Wed Nov 02, 2016 10:56 am
by Julia Hwang 3G
On practice midterm 2015 #4, we are supposed to draw the molecule XeF2 and name its shape and angle. Xe ends up having 3 lone pairs while bonded to the 2 flourines, and the shape of this is linear with 180 degree angles. I don't quite understand how a molecule with 3 lone pairs and 2 atoms could still end up being linear, because wouldn't the electron pairs cause repulsion so the shape would change?

Re: Midterm 2015 #4

Posted: Wed Nov 02, 2016 11:23 am
by Alex Chen 1B
The VSEPR formula for XeF2 is AX2E3. Since there are five regions of electron density around Xe, the geometry of the molecule is trigonal bipyramidal. The three lone pairs have to be the three regions of electron density that are in one plane, because this minimizes the lone pair-lone pair repulsion. Since the three regions in the plane are lone pairs, the actual shape of XeF2 is linear, because only the atoms above the plane and below the plane are left. The three lone pairs exert the same repulsion on the two bonding pairs because of the lone pairs are evenly spaced out at 120 degrees apart and this allows the two bonding pairs to remain in a linear shape.