Chapter 9: Problem 49

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 \(\mathrm{BF}_{3} ?\)

Short Answer

Expert verified

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 by step solution

: 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 \(\mathrm{BF}_{3}\)

: 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.