Problem 49 - Chemistry The Central Science

Question

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated B atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B–F bonds in \(\mathrm{B} \mathrm{F}_{3}\)?(d) What valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3} ?\)

Step-by-Step Solution

Answer

The electron configuration of an isolated B atom is \(1s^{2}2s^{2}2p^{1}\), and for an isolated F atom is \(1s^{2}2s^{2}2p^{5}\). In BF3, the B atom forms 3 sp2 hybrid orbitals, \(sp^2_x + sp^2_y + sp^2_z\), to create bonds with the three F atoms. The unhybridized 2p orbital, \(2p_{z}\), on the B atom remains empty, giving BF3 its electron deficiency and causing it to act as a Lewis acid.

Step 1: : Boron has an atomic number of 5, indicating that it has 5 electrons. To determine the electron configuration, we fill the orbitals following the aufbau principle, according to which, electrons fill the lower energy orbitals first. The electron configuration of boron is: \[1s^{2}2s^{2}2p^{1}\] #b# Find the electron configuration of an isolated F atom

: Fluorine has an atomic number of 9, meaning that it has 9 electrons. Following the aufbau principle, the electron configuration of fluorine is: \[1s^{2}2s^{2}2p^{5}\] #c# Determine the hybrid orbitals constructed on the B atom to make the B–F bonds in \(\mathrm{BF}_{3}\)

Step 2: : Boron in BF3 forms one bond with each of its three neighboring fluorine atoms. To form these bonds, the B atom needs 3 half-filled orbitals. From its ground state electron configuration, it has only one half-filled 2p orbital. To form the required number of half-filled orbitals, atomic orbitals of B mix to form hybrid orbitals. In this case, the 2s and the two 2p orbitals of B hybridize to form 3 sp2 hybrid orbitals: \[2s + 2p_{x} + 2p_{y} \rightarrow sp^2_x + sp^2_y + sp^2_z\] These sp2 hybrid orbitals form 3 sigma bonds with the half-filled 2p orbitals of each fluorine atom for a total of 3 B-F bonds in BF3. #d# Determine which valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3}\)

: After the hybridization process in step c, one 2p orbital remains unused in the B atom: \[2p_{z}\] This unhybridized 2p orbital remains empty in BF3, contributing to the molecule's electron deficiency and its ability to act as a Lewis acid.

Key Concepts

Electron Configuration

Atoms are made up of protons, neutrons, and electrons. The electrons are organized into orbitals that surround the nucleus of an atom. To find an atom’s electron configuration, we use the aufbau principle, which states that electrons fill the lowest energy orbitals first. Each electron contributes to the atom's electron cloud.

• Boron (B), with an atomic number of 5, has the electron configuration: \(1s^{2}2s^{2}2p^{1}\). This shows boron has two electrons in the 1s orbital, two in the 2s orbital, and one in the 2p orbital. The single electron in the 2p orbital makes boron highly versatile in forming bonds.
• Fluorine (F), with an atomic number of 9, has the electron configuration: \(1s^{2}2s^{2}2p^{5}\). Here, fluorine almost completes the p orbital, making it eager to gain one more electron, contributing to its high electronegativity.

Understanding how electrons are arranged helps us predict how atoms will bond and interact with each other.

sp2 Hybrid Orbitals

Hybridization is a key concept in molecular chemistry that explains how atoms form bonds. In hybridization, atomic orbitals mix to form new hybrid orbitals. These new orbitals have different shapes and energies than the original orbitals.
Boron in \(\mathrm{BF}_{3}\) serves as a great example. The boron atom starts with its ground state electron configuration, but to form three bonds, it requires three half-filled orbitals. Initially, boron has one half-filled 2p orbital, not enough to make a \(\mathrm{BF}_{3}\) molecule with three bonds.
During hybridization, one 2s orbital merges with two 2p orbitals to create three equivalent \(sp^2\) hybrid orbitals:

• These \(sp^2\) hybrid orbitals are arranged in a plane, forming 120-degree angles with each other.
• The hybrid orbitals then overlap with the half-filled 2p orbitals of fluorine to form three sigma (σ) bonds.

This explains the planar triangular shape of the \(\mathrm{BF}_{3}\) molecule.

Lewis Acid

A Lewis acid is a substance that can accept an electron pair. In a chemical context, it behaves differently from traditional acids. Instead of donating protons, a Lewis acid accepts electron pairs from a base, making it integral in many chemical reactions.
\(\mathrm{BF}_{3}\) exhibits Lewis acidity. Even after hybridization, one of boron's p orbitals (\(2p_z\)) remains empty. This empty orbital can accept an electron pair from other atoms or molecules, which is what makes \(\mathrm{BF}_{3}\) act as a Lewis acid.

• This characteristic is often utilized in forming adducts, where \(\mathrm{BF}_{3}\) bonds with a molecule that can supply an electron pair.
• The electron deficiency of \(\mathrm{BF}_{3}\) can make it very reactive in seeking out atoms that can share electrons.

Understanding Lewis acids helps in predicting how certain chemical reactions will proceed and why certain compounds form.