1. The problem statement, all variables and given/known data Consider the geometric isomer of PBr2F3 with a Br-P-Br bond angle somewhat larger than 120 degrees. It is also true that for this isomer ... There are two sets of identical bond angles. 2. Relevant equations PBr2F3 is trigonal bipyramidal (although distorted). 3. The attempt at a solution Er, two identical sets of bond angles? I can come up with more than that. First, we consider the equatorial trigonal plane. Since the Br-P-Br bond angle is larger than 120 degrees, the Br-P-F bond angles should each be 1) less than 120 degrees and 2) still identical. If you think about it, we started with an equatorial triangle and we simply changed the length of one of the sides (the side with the two bromine atoms). This in turn changes only one of the angles of the triangle and should leave two identical bond angles in its wake. So that's one set of identical bond angles. Now we consider more steric effects and interactions between the axial and equatorial atoms. We expect the bromine to push down on the fluorines because bromine is huge compared to fluorine. This distorts the perfect 180 degree bond angle in a perfect trigonal bipyramidal molecule and similarly the 90 degree bond angles in an ideal trigonal bipyramidal molecule. This in turn creates two sets of identical bond angles for a total of three. We have: 1) F-axial, P, and Br-equatorial times 2 (there are two bromine atoms and two axial fluorines). This is somewhat greater than 90 degrees. 2) The somewhat smaller than 90 degrees F-axial, P, and F-equatorial bond. Diagram attached. So is my teacher wrong? Picture on left is an ideal representation with the perfect 90 degree bond angles explicitly shown; picture to the right below shows the repulsions.