BF3 Hybridization
The hybridization of a molecule is the mixing of atomic orbitals to form new hybrid orbitals. These hybrid orbitals determine the geometry and bonding properties of the molecule.
In the case of BF3, the boron atom has a valence electron configuration of 2s2 2p1. In order to form three bonds with fluorine atoms, it needs to have three unpaired electrons. This can be accomplished by promoting one of the 2s electrons to the 2p orbital.
The resulting excited state electron configuration of boron is 2s1 2px1 2py1. The 2s and two 2p orbitals then hybridize to form three sp2 hybrid orbitals.
These sp2 hybrid orbitals are arranged in a trigonal planar geometry with bond angles of 120°. Each sp2 hybrid orbital overlaps with a 2p orbital from a fluorine atom to form a sigma bond.
The BF3 molecule is therefore sp2 hybridized with a trigonal planar geometry and no lone pairs of electrons.
Questions and Answers
Q: What is the hybridization of BF3?
A: The hybridization of BF3 is sp2.
Q: What is the shape of BF3?
A: The shape of BF3 is trigonal planar.
Q: What are the bond angles in BF3?
A: The bond angles in BF3 are 120°.
Q: How many sigma bonds are in BF3?
A: There are three sigma bonds in BF3.
Q: How many lone pairs of electrons are in BF3?
A: There are no lone pairs of electrons in BF3.
Additional Information
The sp2 hybridization of BF3 is also reflected in its molecular polarity. Boron is more electronegative than fluorine, so the fluorine atoms have a partial negative charge and the boron atom has a partial positive charge. This results in a polar molecule with a dipole moment of 0.61 D.
BF3 is a colorless, odorless gas that is used in a variety of applications, including the manufacture of plastics, pesticides, and pharmaceuticals. It is also a Lewis acid and can react with bases to form adducts.
