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Chapter 9: Problem 56

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

Short Answer

Expert verified
In the hydrogen fluoride molecule, the hydrogen 1s atomic orbital overlaps linearly with the fluorine 2p_z atomic orbital, forming both bonding and antibonding σ molecular orbitals. The bonding σ orbital has electron density concentrated between the two nuclei, strengthening the bond, while the antibonding σ orbital has electron density outside the internuclear region, creating a repulsive force between the nuclei.

Step by step solution

01

Understand hydrogen and fluorine atomic orbitals

We need to understand the hydrogen 1s atomic orbital and fluorine 2p atomic orbital. In hydrogen, there is only one electron in the 1s atomic orbital, while in fluorine, there are six valence electrons, two each in the 2s and three in 2p orbitals. The 2p orbitals consist of three orbitals, 2p_x, 2p_y, and 2p_z, having lobes along the x, y, and z-axis respectively.
02

Describe the overlap process

In the hydrogen fluoride molecule, the hydrogen 1s atomic orbital overlaps end-to-end with one of the fluorine 2p atomic orbitals, specifically the 2p_z orbital, because the z-axis is typically chosen as the inter-nuclear axis. This linear overlap forms a molecular orbital, where the electron density is concentrated between the two nuclei, leading to a bonding interaction. In addition to the bonding molecular orbital, an antibonding molecular orbital is formed as well. In this orbital, the electron density is outside the internuclear region, leading to a repulsive force between the nuclei.
03

Identify the type of molecular orbitals formed

Molecular orbitals can be classified as σ (sigma) or π (pi) orbitals, depending on their symmetry around the molecular axis. When the overlap is along the molecular axis, such as in hydrogen fluoride, the orbitals formed are σ orbitals. In the case of hydrogen fluoride, both the bonding and the antibonding molecular orbitals are σ orbitals since the overlap occurs between 1s and 2p_z atomic orbitals.
04

Summarize your findings

In summary, the hydrogen 1s and fluorine 2p_z atomic orbitals overlap linearly in hydrogen fluoride to form both bonding and antibonding σ molecular orbitals. The bonding σ orbital has electron density concentrated between the two nuclei, making the bond stronger, while the antibonding σ orbital has electron density outside the internuclear region, leading to a repulsive force between the nuclei.

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Most popular questions from this chapter

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity. a. \(\mathrm{CF}_{4}\) e. \(\mathrm{BeH}_{2}\) i. \(\mathrm{KrF}_{4}\) b. \(\mathrm{NF}_{3}\) f. \(\mathrm{TeF}_{4}\) j. \(\mathrm{SeF}_{6}\) c. \(\mathrm{OF}_{2}\) g. AsF \(_{5}\) k. IFs d. \(\mathrm{BF}_{3}\) h. \(\mathrm{KrF}_{2}\) L. \(\mathrm{IF}_{3}\)

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Draw the Lewis structures for \(\mathrm{SO}_{2}, \mathrm{PCl}_{3}, \mathrm{NNO}, \mathrm{COS}\), and \(\mathrm{PF}_{3}\). Which of the compounds are polar? Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit \(s p^{3}\) hybridization by the central atom? Which of the compounds have a linear molecular structure?

The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?

What are the relationships among bond order, bond energy, and bond length? Which of these quantities can be measured?
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