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?

Question

Elias Aguila · Accepted Answer

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

#"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°"#

Ellie Adams · Answer

undefined Iodine monobromide, PI3Br2, has bond angles of approximately 90 degrees for the I-P-I bond, 120 degrees for the Br-P-Br bond, and around 109.5 degrees for the I-P-Br bond.