Question

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?

Answer 1

`"∠I-P-I = 120°"`; `"∠Br-P-Br = 180°"`; `"∠I-P-Br = 90°"`.

Explanation:

`"PI"_3"Br"_2` is an `"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 `"P-Br"` bond dipoles must cancel, and the three `"P-I"` bond dipoles must also cancel.

The `"P-Br"` dipoles can cancel only if the `"Br"` atoms occupy the two axial positions (top and bottom in the drawing).

Also, the three `"P-I"` bond dipoles can cancel only if they are pointing toward the corners of an equilateral triangle.

The `"I"` atoms must occupy the equatorial (horizontal) positions.

∴ The bond angles must be:

`"∠ I-P-I = 120°"`; `"∠ Br-P-Br = 180°"`; `"∠ I-P-Br = 90°"`