Boron trifluoride (BF3) has no dipole moment (μ = 0D). Explain how this observation confirms the geometry of BF3 predicted by VSEPR theory.

The observation that boron trifluoride (BF3) has no dipole moment (μ=0D) is consistent with the geometry predicted by the Valence Shell Electron Pair Repulsion (VSEPR) theory. VSEPR theory is a model used to predict the shapes of molecules based on the repulsion between electron pairs in the valence shell of an atom.

According to VSEPR theory, the central atom (boron, B) in BF3 is surrounded by three bonding pairs of electrons, each coming from the three fluorine atoms. These electron pairs are arranged in a trigonal planar geometry around the boron atom. In this arrangement, the bond angles between the B-F bonds are expected to be approximately 120 degrees.

The absence of a dipole moment in BF3 indicates that the vector sum of the bond dipoles cancels out, resulting in a net dipole moment of zero. A dipole moment arises when there is an uneven distribution of electron density in a molecule, leading to a separation of positive and negative charges. In a molecule like BF3, where the central atom and the surrounding atoms are all the same (fluorine), the individual bond dipoles are equal in magnitude and opposite in direction, effectively canceling each other out. As a result, the molecule as a whole has no net dipole moment.

This observation confirms the VSEPR-predicted trigonal planar geometry for BF3 because this arrangement of atoms and electron pairs allows for the cancellation of bond dipoles, resulting in a nonpolar molecule with no overall dipole moment.