Chapter 3, Problem 3.55P

Interpretation Introduction

(a)

Interpretation:

MO energy diagram of twisted ethane is to be drawn and the HOMO and LUMO identified.

Concept introduction:

The molecular orbital energy diagram of a molecule is built by considering the interactions of all the valence shell orbitals from each atom. The total number of MOs produced is the same as the number of interacting AOs.

An end-on overlap of two orbitals produces a pair of MOs of $\text{σ}$ symmetry. One of these, the bonding MO, is considerably lower in energy compared to the contributing AOs. The other, an antibonding MO, is higher in energy compared to the contributing AOs by an almost equal amount.

Interactions between p orbitals produce MOs of two types, one pair with $\text{σ}$ symmetry and two pairs with $\text{π}$ symmetry. The $\text{σ}$ bonding MO is considerably lower in energy than the two $\text{π}$ bonding MO. Conversely, the ${\text{σ}}^{*}$ antibonding MO is considerably higher in energy than the two ${\text{π}}^{*}$ antibonding MOs.

The valence electrons of the contributing atoms are then filled in these MOs in increasing order of energy. The MO of highest energy that contains any electrons is called the highest occupied molecular orbital (HOMO), and the empty MO immediately above it is called the lowest unoccupied molecular orbital (LUMO).

Interpretation Introduction

(b)

Interpretation:

Why the all planar ethane molecule is more stable than the twisted one is to be explained by comparing the MO energy diagrams of the two.

Concept introduction:

Bonding MOs have lower energy than the contributing AOs; nonbonding MOs are essentially at the same level as the AOs, while the antibonding MOs have higher energy compared to the AOs. Therefore, the presence of electrons in bonding MOs stabilizes the molecule compared to the individual atoms. Electrons in nonbonding MOs have no effect. Any electrons in antibonding MOs destabilize the molecule.