The \(\mathrm{PF}_{3}\) molecule has a dipole moment of \(1.03 \mathrm{D}\), but \(\mathrm{BF}_{3}\) has a dipole moment of zero. How can you explain the difference?

To describe the geometries of the molecules, we’ll use the principles of VSEPR (Valence Shell Electron Pair Repulsion) theory. Both \(\mathrm{PF}_{3}\) and \(\mathrm{BF}_{3}\) have a central atom with three bonding pairs and no lone pairs around it, so they both have a trigonal planar molecular geometry.

The dipole moment arises due to the difference in electronegativity between the central atom and the surrounding atoms in a polar covalent bond. The electronegativity values for the elements involved are approximately: - Phosphorus (P): 2.19 - Boron (B): 2.04 - Fluorine (F): 3.98 In both molecules, the electronegativity difference exists between the central atom and fluorine atoms. In \(\mathrm{PF}_{3}\), the electronegativity difference between P and F is 1.79, which indicates polar covalent bonds. However, in \(\mathrm{BF}_{3}\), the electronegativity difference between B and F is 1.94, which also leads to polar covalent bonds.

In a trigonal planar geometry, the bond dipoles can be arranged in such a way that they either cancel out each other, resulting in a net dipole moment of zero, or they don't, resulting in a nonzero net dipole moment. In the case of \(\mathrm{PF}_{3}\), the three P-F polar bonds do not completely cancel each other out because of the difference in electronegativity between P and F, thus leading to an overall nonzero dipole moment of \(1.03\,\mathrm{D}\). On the other hand, in the case of \(\mathrm{BF}_{3}\), though there are three B-F polar bonds due to electronegativity difference, they are symmetrically distributed around the central boron atom. Hence, the bond dipoles are evenly distributed and cancel each other out, resulting in an overall dipole moment of zero.

The difference in dipole moments between \(\mathrm{PF}_{3}\) and \(\mathrm{BF}_{3}\) arises from the combination of the molecular geometry (trigonal planar) and the electronegativity differences between the atoms involved. In \(\mathrm{PF}_{3}\), the bond dipoles do not cancel each other, giving it a nonzero dipole moment, while in \(\mathrm{BF}_{3}\), the bond dipoles cancel each other out resulting in a dipole moment of zero.