Question

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?

Short answer

Both PF3 and BF3 have a trigonal planar geometry. However, the difference in dipole moments can be attributed to the electronegativity differences between their constituent atoms and the resulting effect on the electric dipoles. In PF3, the dipoles do not cancel out due to the geometry and significant electronegativity differences between P and F, leading to a dipole moment of \(1.03\,\mathrm{D}\). In contrast, the dipoles in BF3 cancel out because of the trigonal planar arrangement of the B-F dipoles, resulting in an overall dipole moment of zero.

Step-by-step solution

Determine the molecular geometry of PF3 and BF3

First, we need to understand the structures of both PF3 and BF3 molecules. Both molecules have a central atom (P in PF3 and B in BF3) surrounded by three F atoms. In both cases, the central atoms obey the octet rule: - P (Phosphorus) has 5 electrons in its valence shell and can form three covalent bonds with three F atoms. - B (Boron) has 3 electrons in its valence shell and forms three covalent bonds with three F atoms. Based on the VSEPR (valence shell electron pair repulsion) theory, the electron pairs surrounding the central atom will try to minimize their repulsion, positioning themselves as far apart as possible. In the case of PF3 and BF3, both have a trigonal planar geometry with bond angles of approximately 120 degrees.

Understand the concept of electronegativity and dipole moment

Electronegativity is the ability of an atom to attract bonding electrons towards itself. In a covalent bond between two atoms with different electronegativities, the electrons are not shared equally. As a result, an electric dipole is created, with a partial positive charge on the less electronegative atom and a partial negative charge on the more electronegative atom. A molecule's dipole moment is the vector sum of the individual dipole moments of its constituent atoms, and it depends on the geometry of the molecule. In general, the higher the difference in electronegativity between the bonded atoms, the stronger the dipole moment.

Analyze the electronegativity in PF3 and BF3

In the case of PF3, the phosphorus atom (P) has an electronegativity value of 2.19, while fluorine atoms (F) have an electronegativity value of 3.98. There is a significant electronegativity difference between P and F atoms, which leads to the formation of electric dipoles along the P-F bonds. In the case of BF3, the boron atom (B) has an electronegativity value of 2.04, while the electronegativity of fluorine atoms (F) is still 3.98. The electronegativity difference between B and F in BF3 is even larger than in PF3, also resulting in electric dipoles along the B-F bonds.

Explain the difference in dipole moments of PF3 and BF3

Despite the electronegativity differences between the central and surrounding atoms in both molecules, their overall dipole moments are different because of their molecular geometry. In PF3, the trigonal planar geometry and the difference in electronegativity between P and F atoms cause the electric dipoles along the P-F bonds to not cancel each other out, resulting in a nonzero dipole moment for the entire molecule (\(1.03\,\mathrm{D}\)). However, in the case of BF3, the trigonal planar geometry and the symmetric arrangement of the three B-F dipoles cause them to cancel each other out, resulting in an overall dipole moment of zero for the molecule. In conclusion, the difference in dipole moments of PF3 and BF3 can be explained by their molecular geometry and the electronegativity differences between the constituents atoms. The electric dipoles in PF3 do not cancel out due to the geometry and electronegativity differences, while in BF3, the dipoles cancel out because of the trigonal planar arrangement of the B-F dipoles.