Question

Determine the type of hybrid orbitals formed by the boron atom in a molecule of boron fluoride, \(\mathrm{BF}_{3}\).

Short answer

The boron atom in \(\text{BF}_3\) forms \(\text{sp}^2\) hybrid orbitals.

Step-by-step solution

Determine the Electron Domain Geometry

Identify the regions of electron density around the central boron atom. In \(\text{BF}_3\), boron has three bonds with three fluorine atoms and no lone pairs on boron. Therefore, boron has three regions of electron density.

Identify the Molecular Geometry

Given the three regions of electron density, the electron domain geometry for boron in \(\text{BF}_3\) is trigonal planar. According to VSEPR theory, this arrangement corresponds to a molecular geometry that's also trigonal planar.

Determine the Hybridization

For a trigonal planar arrangement, boron needs to form three equivalent \(\text{sp}^2\) hybrid orbitals. This is achieved by mixing one \(\text{s}\) orbital and two \(\text{p}\) orbitals of boron.

Key concepts

electron domain geometry

The concept of electron domain geometry helps us understand the spatial arrangement of electron pairs around a central atom. In \(\text{BF}_3\), the boron atom is central and surrounded by three fluorine atoms. To determine the electron domain geometry, we need to count the number of regions with electron density around the boron. Each bond counts as one region, so boron in \(\text{BF}_3\) has three bonds and no lone pairs.
Thus, it has three regions of electron density.
Such an arrangement leads to a specific geometry known as 'trigonal planar,' where boron's electron domains form a plane with 120-degree angles between them each.

VSEPR theory

VSEPR theory, which stands for Valence Shell Electron Pair Repulsion, helps us predict the shape of molecules. It is based on the idea that electron pairs around a central atom will arrange themselves to minimize repulsion, thus determining the molecule's geometry.
In the case of \(\text{BF}_3\), boron has three electron domains from its three bonds with fluorine atoms. According to VSEPR theory, these regions will position themselves as far apart as possible. This results in a 'trigonal planar' geometry.
Both the electron domain geometry and the resulting molecular shape are the same in this molecule.

sp2 hybrid orbitals

To understand how boron forms bonds in \(\text{BF}_3\), we need to consider hybridization. Hybridization is a process where atomic orbitals mix to form new hybrid orbitals.
For trigonal planar geometry, boron must form three equivalent \(\text{sp}^2\) hybrid orbitals. This is achieved by mixing one \(\text{s}\) orbital and two \(\text{p}\) orbitals. The result is three \(\text{sp}^2\) hybrid orbitals lying 120 degrees apart, providing the framework for strong sigma bonds with each fluorine atom. By using \(\text{sp}^2\) hybrid orbitals, the molecule attains a stable, symmetrical structure.