Problem 112 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\pi\) 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 \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

Step-by-Step Solution

Answer

The HOMO in ethylene corresponds to the \(\pi_{2 p}\) molecular orbital. The LUMO corresponds to the \(\pi_{2 p}^{*}\) molecular orbital. The C-C bond in ethylene is weaker in the excited state compared to the ground state due to the promotion of an electron from the bonding to the antibonding orbital. The C-C bond is easier to twist in the excited state than in the ground state, as the bonding is weakened.

Step 1: 1. Identifying the HOMO in Ethylene

In ethylene, there is a pair of electrons in the bonding π orbital between the two carbons, which means that the carbon-carbon bond is formed. The HOMO corresponds to the highest energy occupied molecular orbital. Since the pair of electrons in the bonding π orbital are involved in forming the C-C bond, the HOMO in ethylene is this bonding π orbital. So, the HOMO corresponds to the \(\pi_{2 p}\) molecular orbital.

Step 2: 2. Identifying the LUMO in Ethylene

The LUMO is the lowest unoccupied molecular orbital. When a photon is absorbed, one of the electrons in the \(\pi_{2 p}\) bonding molecular orbital (HOMO) can be promoted to the higher energy \(\pi_{2 p}^{*}\) molecular orbital (LUMO). Hence, the LUMO corresponds to the \(\pi_{2 p}^{*}\) molecular orbital.

Step 3: 3. Comparing C-C bond strength in the ground state and the excited state

In the ground state, both the bonding electrons are in the \(\pi_{2 p}\) molecular orbital. Upon absorbing a photon and exciting an electron to the \(\pi_{2 p}^{*}\) molecular orbital, there is now only one electron left in the \(\pi_{2 p}\) bonding orbital, which weakens the C-C bond. Therefore, the C-C bond in ethylene is weaker in the excited state compared to the ground state.

Step 4: 4. Comparing C-C bond twisting in the ground state and the excited state

In the ground state, with both electrons in the \(\pi_{2 p}\) bonding orbital, the overlap and bonding between the p orbitals are strong, which makes it difficult to twist the C-C bond. However, in the excited state, with one electron promoted to the \(\pi_{2 p}^{*}\) molecular orbital, the overlap and bonding are weakened, making it easier to twist the C-C bond. Thus, the C-C bond in ethylene is easier to twist in the excited state than in the ground state.

Key Concepts

HOMO-LUMO transition

The HOMO-LUMO transition in molecular orbital theory is an essential concept. It involves the excitation of an electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). In ethylene, a classic example, the HOMO is the bonding \( \pi_{2p} \)\ orbital. This orbital has electrons involved in forming the carbon-carbon double bond. When ethylene absorbs a photon with enough energy, one of its electrons from the HOMO can "jump" to the LUMO.

• The LUMO in ethylene corresponds to the antibonding \( \pi_{2p}^{*} \)\ orbital.
• This electron transition alters the electronic state and affects the molecule's properties.

Transitioning from HOMO to LUMO decreases stability by weakening the bond, resulting in altered reactivity and physical properties. This concept is foundational for understanding reactions involving photo-excitation.

Ethylene molecular structure

Ethylene, a simple alkene, consists of a pair of carbon atoms connected by a double bond and each bonded to two hydrogen atoms. The carbon atoms are sp² hybridized, meaning their orbitals mix to form three identical structures, allowing formation of the \( \sigma \)\ bonds with hydrogen and one another. The fourth available p orbital from each carbon forms the \( \pi \)\ bond through sideways overlap.

• The \( \pi \)\ bond involves electrons shared in the lateral space above and below the plane of the ethylene molecule.
• These \( \pi \)\ electrons provide unique properties, including increased reactivity compared to alkanes.

Understanding ethylene's structure provides insights into its chemical behavior, including its participation in reactions like polymerization. The molecular geometry affects how the electrons distribute in space, influencing both physical properties and chemical reactivity.

C-C bond strength and twisting

The strength and flexibility of carbon-carbon (C-C) bonds in ethylene change with electron transitions. In the ground state, the C-C bond is formed by both the \( \pi \) and \( \sigma \)\ bonds, making it rigid and strong. In contrast, upon excitation when an electron moves to the antibonding \( \pi_{2p}^{*} \)\ molecular orbital, the bond weakens.

• The original bonding electron remains in the \( \pi \)\ orbital, but the \( \pi \)\ bond now has only one electron.
• This reduction weakens bond overlap, decreasing bond strength.

This weakening is crucial because it increases the ease of twisting or rotating around the bond. In the excited state, the weakened bond requires less energy to manipulate, making twisting much easier than in the ground state. This knowledge is vital in varied chemical applications, from synthesis to kinetic studies.