# Bond Angles in Trigonal Bipyramidal Molecules

1. Feb 25, 2014

### Qube

1. The problem statement, all variables and given/known data

What are the number of different bond angles for the isomer of $F_{3}PCl_{2}$ with each chlorine equatorial? What about for the isomer of $BrPF_{4}$ with the bromine axial?

According to the key the the first molecule has 5 unique bond angles.

2. Relevant equations

Trigonal bipyramidal usually has 2 different bond angles (90 and 120) when all the attachments are identical.

3. The attempt at a solution

I'm having a hard time visualizing what bond angles there could be when the attachments are not all identical. I see three different bond angles. When looking at the trigonal planar "center" of the molecule, I see two different bond angles. I also know about the 90 degree angle between the axial fluorine and the central phosphorous.

At best I see these 4 unique bond angles:

F-P-P: 90 and 180 degrees
Cl-P-Cl and Cl-P-F.

Where's the fifth one?

For $BrPF_{4}$ with the Br axial, I have three unique bond angles.

Last edited: Feb 26, 2014
2. Jan 28, 2017

### TeethWhitener

For the first molecule (PF3Cl2 with chlorine in the equatorial position), the chlorines will push the fluorines out of their normal positions. The symmetry of the molecule will be C2v. The 120° equatorial angles will be widened between the two chlorines and narrowed between the chlorines and the equatorial fluorine. There's your first 2 angles. The chlorines will also push the axial fluorines out of their positions (away from 90° and toward the fluorine). This increases the angle between the chlorines and the axial fluorines and decreases the angle between the equatorial and axial flurines. There's another two angles. The fifth angle comes from the fact that, since the axial fluorines are pushed out of position by the chlorines, they no longer sit at 180° from each other, but in fact are a little less. (I can go run a quick calculation on this if you want hard numbers. We'll see if I have time this weekend.)

In the second molecule (BrPF4 with Br in the axial position) the bromine will push the equatorial fluorines away from it, giving a C3v symmetric compound. The angles between the equatorial fluorines will be equal, but you'll have different angles between the axial and equatorial fluorines versus the bromine and the equatorial fluorines. So the answer here is 3 unique angles.