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Ch.9 - Molecular Geometry and Bonding Theories
Brown - Chemistry: The Central Science 15th Edition
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook
All textbooksBrown 15th EditionCh.9 - Molecular Geometry and Bonding TheoriesProblem 109c
Chapter 9, Problem 109c

The energy-level diagram in Figure 9.40 shows that the sideways overlap of a pair of p orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the to the molecular orbital. c. Is the bond in ethylene stronger or weaker in the excited state than in the ground state? Why?

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Identify the type of bond in ethylene in its ground state. Ethylene, C2H4, has a double bond between the two carbon atoms, which consists of a sigma (σ) bond and a pi (π) bond. The π bond is formed by the sideways overlap of p orbitals.
Understand the molecular orbital theory as it applies to the π bond in ethylene. In the ground state, the π bond is formed by the overlap of p orbitals, creating a bonding (π) molecular orbital that is filled with two electrons.
Recognize the effect of photon absorption on the electron configuration. When a photon is absorbed, it provides energy that can promote one of the electrons from the bonding π orbital to the antibonding π* orbital.
Analyze the impact of electron promotion on the π bond. Moving an electron to the antibonding π* orbital reduces the electron density in the bonding region, which weakens the π bond.
Conclude the relative strength of the π bond in the excited state compared to the ground state. Since the π bond is partially disrupted by the promotion of an electron to the antibonding orbital, the π bond in the excited state is weaker than in the ground state.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molecular Orbitals

Molecular orbitals (MOs) are formed by the linear combination of atomic orbitals (LCAO) when atoms bond. In ethylene, the sideways overlap of p orbitals creates a bonding molecular orbital, which stabilizes the molecule by lowering its energy, and an antibonding molecular orbital, which destabilizes it by raising its energy. The presence of electrons in these orbitals influences the bond strength and stability of the molecule.
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Molecular Orbital Theory

Excited State vs. Ground State

The ground state of a molecule refers to its lowest energy configuration, where electrons occupy the lowest available molecular orbitals. In contrast, the excited state occurs when an electron is promoted to a higher energy molecular orbital, resulting in a temporary increase in energy. This transition can affect the bond strength, as the distribution of electrons in bonding and antibonding orbitals changes.
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Ground State Electron Configurations

Bond Strength and Electron Configuration

Bond strength is influenced by the electron configuration of the molecular orbitals. In the ground state of ethylene, the bonding orbital is filled, contributing to a strong bond between the carbon atoms. When an electron is excited to an antibonding orbital, the effective bond order decreases, leading to a weaker bond in the excited state compared to the ground state, as the stabilizing effect of the bonding electrons is reduced.
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Textbook Question

The energy-level diagram in Figure 9.40 shows that the sideways overlap of a pair of p orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the to the molecular orbital. b. Assuming this electronic transition corresponds to the HOMO–LUMO transition, what is the LUMO in ethylene?

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Textbook Question

Sulfur tetrafluoride (SF4) reacts slowly with O2 to form sulfur tetrafluoride monoxide (OSF4) according to the following unbalanced reaction: SF4(g) + O2(g) → OSF4(g) The O atom and the four F atoms in OSF4 are bonded to a central S atom. (b) Write a Lewis structure of OSF4 in which the formal charges of all atoms are zero.

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Textbook Question

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C12H10N2. A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are

(Recall the shorthand notation used for benzene.) (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? How many unhybridized atomic orbitals are there on the N and the C atoms in hydrazobenzene?

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Textbook Question

Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur tetrafluoride monoxide 1OSF42 according to the following unbalanced reaction: SF41g2 + O21g2¡OSF41g2 The O atom and the four F atoms in OSF4 are bonded to a central S atom. (a) Balance the equation.

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Textbook Question

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C12H10N2. A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are

(Recall the shorthand notation used for benzene.) (c) Predict the N¬N¬C angles in each of the substances.

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