Question

PI3Br₂ is a nonpolar molecule. Based on this information, determine the I−P−I bond angle, the Br−P−Br bond angle, and the I−P−Br bond angle. Enter the number of degrees of the I−P−I, Br−P−Br, and I−P−Br bond angles, separated by commas (e.g., 30,45,90)

A) 90, 120, 180
B) 120, 180, 90
C) 180, 90, 120
D) 90, 180, 120

Answer 1

The bond angles in PI3Br2 with trigonal bipyramidal geometry are I-P-I: 180°, Br-P-Br: 120°, I-P-Br: 90°, making the correct answer D) 90, 180, 120.

Step-by-step explanation:

Considering that PI3Br2 is nonpolar and assuming it has a trigonal bipyramidal geometry, we would expect the phosphorus atom to be surrounded by five regions of electron density. This geometry arises because there are five bonding pairs of electrons with no lone pairs on the central phosphorus atom (AX5 molecular geometry). Since the molecule is nonpolar, this suggests that the more electronegative iodine atoms are in the axial positions to minimize repulsion, resulting in a linear arrangement for the I-P-I bond angle (180°).

To maintain the molecule's nonpolarity, the two bromine atoms would be situated in two of the equatorial positions, putting them 120° apart from each other. Hence, the Br-P-Br bond angle is 120°. The I-P-Br bond angle would then be 90°, as the axially-positioned iodine atoms are perpendicular to the plane of the bromine atoms.

In conclusion, the bond angles in PI3Br2 are I-P-I: 180°, Br-P-Br: 120°, I-P-Br: 90°. Therefore, the correct answer is D) 90, 180, 120.