Problem 112 - Chemistry: The Central Science

Question

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of porbitals 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 $\pi_{2 p}^{\star}$ 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 C-Cbond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(C-C\) bond in ethylene easier to twist in the ground state or in the excited state?

Step-by-Step Solution

Answer

In ethylene, the HOMO is \(\pi_{2 p}\) and the LUMO is \(\pi_{2 p}^{\star}\). The C-C bond is weaker in the excited state than in the ground state due to an electron being promoted to an antibonding orbital. The C=C bond is easier to twist in the excited state as compared to the ground state.

Step 1: (a) Find the HOMO in ethylene

HOMO stands for the Highest Occupied Molecular Orbital, which means it is the highest energy orbital that has electrons. In the ground state of ethylene, the \(\pi_{2 p}\) orbital is filled with a pair of electrons, making it the HOMO. So, the HOMO in ethylene is \(\pi_{2 p}\).

Step 2: (b) Find the LUMO in ethylene

LUMO stands for the Lowest Unoccupied Molecular Orbital, which means it is the lowest energy orbital which does not have electrons currently. In the ground state of ethylene, the \(\pi_{2 p}^{\star}\) orbital is empty, making it the LUMO. So, the LUMO in ethylene is \(\pi_{2 p}^{\star}\).

Step 3: (c) Compare C-C bond strength in the excited state and ground state

When a photon is absorbed, an electron from the HOMO (\(\pi_{2 p}\)) gets promoted to the LUMO (\(\pi_{2 p}^{\star}\)), this creates an excited state of ethylene. In this excited state, the electrons are in an antibonding orbital instead of a bonding orbital. This in turn reduces the overall stability and increases the energy. Due to this reason, the C-C bond in ethylene is weaker in the excited state as compared to the ground state.

Step 4: (d) Compare the difficulty of twisting the C=C bond in ground state and excited state

In the ground state, the C=C bond has a pair of electrons in the bonding \(\pi_{2 p}\) orbital. This double bond creates a stronger bond between the two carbons which makes it harder to twist the C=C bond. However, in the excited state, an electron from the bonding \(\pi_{2 p}\) orbital is promoted to the antibonding \(\pi_{2 p}^{\star}\) orbital, weakening the bond between the two carbons. As a result, the C=C bond becomes easier to twist in the excited state as compared to the ground state.