# Why is the "FPF" bond angle larger than the "BrPBr" angle in "PF"_3 and "PBr"_3 respectively?

Jun 8, 2017

It's not... The $\text{FPF}$ bond angle of ${\text{PF}}_{3}$ is ${97.7}^{\circ}$, while the $\text{BrPBr}$ bond angle of ${\text{PBr}}_{3}$ is ${101.00}^{\circ}$. Hence, the bond angle on ${\text{PF}}_{3}$ is SMALLER.

http://cccbdb.nist.gov/expangle1.asp

In fact, if you look closely here at the electron density maps, you can kind of see the increase in bond angle from ${\text{PF}}_{3}$ to ${\text{PBr}}_{3}$:

We actually have two factors here:

• size of the terminal atoms (and thus the $\text{XPX}$ bond lengths)
• electronegativity of the terminal atoms

SIZE FACTOR

Since $\text{F}$ is smaller than $\text{Br}$, the optimal bond distance is shorter (the distance that optimizes the balance between nucleus-A-nucleus-B repulsion and nucleus-A-electron-B / nucleus-B-electron-A attraction).

The atomic radius difference also allows each fluorine atom to be closer together than the bromine atoms can be, without experiencing as much electron repulsions from each terminal atom's lone pairs.

(Or, we could even say that the repulsive torque from the bonding-electron pairs is less on the $\text{P"-"F}$ bonds due to perpendicular electron repulsion acting on a shorter lever arm length.)

ELECTRONEGATIVITY FACTOR

Furthermore, the electron density is more localized onto the fluorine atoms due to their higher electronegativity than that of bromine atoms. Electron density localized in the bonds is on average farther away when we move away from the peak of the molecule (the lone pair).

These together facilitate the smaller bond angle of ${\text{PF}}_{3}$ than that of ${\text{PBr}}_{3}$.