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Ch.9 - Molecular Geometry and Bonding Theories
Chapter 9, Problem 112c

The energy-level diagram in Figure 9.36 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 π 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 p2p to the p*2p molecular orbital. (c) Is the C¬C bond in ethylene stronger or weaker in the excited state than in the ground state? Why?

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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 bonding (π) and antibonding (π*) molecular orbitals. The bonding orbital stabilizes the molecule, while the antibonding orbital destabilizes it. Understanding the nature of these orbitals is crucial for analyzing the strength of the C-C bond in different electronic states.
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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 orbital, such as from a bonding π orbital to an antibonding π* orbital. This transition can affect the bond strength, as the promotion of an electron can weaken the bond due to the presence of an antibonding orbital.
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Bond Strength and Electron Configuration

Bond strength is influenced by the electron configuration of the involved atoms. In ethylene, the C-C bond strength is determined by the occupancy of the bonding and antibonding orbitals. When an electron is excited from a bonding orbital to an antibonding orbital, the effective bond order decreases, leading to a weaker bond in the excited state compared to the ground state. This relationship is essential for understanding the stability of the molecule under different conditions.
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Related Practice
Textbook Question

Carbon monoxide, CO, is isoelectronic to N2. (d) Would you expect the p2p MOs of CO to have equal atomic orbital contributions from the C and O atoms? If not, which atom would have the greater contribution?

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Textbook Question

The energy-level diagram in Figure 9.36 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 π 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 p2p to the p*2p molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene?

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Textbook Question

The energy-level diagram in Figure 9.36 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 π 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 p2p to the p*2p 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 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

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

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. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic?

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