Midterm 2015 #4

(Polar molecules, Non-polar molecules, etc.)

Moderators: Chem_Mod, Chem_Admin

Julia Hwang 3G
Posts: 22
Joined: Wed Sep 21, 2016 3:00 pm

Midterm 2015 #4

Postby Julia Hwang 3G » Wed Nov 02, 2016 10:56 am

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?

Alex Chen 1B
Posts: 24
Joined: Wed Sep 21, 2016 2:57 pm
Been upvoted: 1 time

Re: Midterm 2015 #4

Postby Alex Chen 1B » Wed Nov 02, 2016 11:23 am

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.

Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 4 guests