# PI3Br2 is a nonpolar molecule, how would you determine the bond angle of I-P-I bond , Br-P-Br bond and I-P-Br bond angles?

Dec 14, 2015

$\text{∠I-P-I = 120°}$; $\text{∠Br-P-Br = 180°}$; $\text{∠I-P-Br = 90°}$.

#### Explanation:

${\text{PI"_3"Br}}_{2}$ is an ${\text{AX}}_{5}$ molecule.

According to VSEPR theory, it must have a trigonal bipyramidal geometry.

There are three "equatorial" bonds in a trigonal planar arrangement and two "axial" bonds in a linear arrangement.

So, how are the five halogen atoms arranged in the molecule?

Since the molecule is nonpolar, the two $\text{P-Br}$ bond dipoles must cancel, and the three $\text{P-I}$ bond dipoles must also cancel.

The $\text{P-Br}$ dipoles can cancel only if the $\text{Br}$ atoms occupy the two axial positions (top and bottom in the drawing).

Also, the three $\text{P-I}$ bond dipoles can cancel only if they are pointing toward the corners of an equilateral triangle.

The $\text{I}$ atoms must occupy the equatorial (horizontal) positions.

∴ The bond angles must be:

$\text{∠ I-P-I = 120°}$; $\text{∠ Br-P-Br = 180°}$; $\text{∠ I-P-Br = 90°}$