Chapter 14, Problem 14.27P

Interpretation Introduction

(a)

Interpretation:

The number of nodal planes perpendicular to the bonding axes of each of the three $\text{π}$ MOs for the $\text{octa-1, 3, 5, 7-tetraene}$ is to be determined, and they are to be ranked in order of increasing energy.

Concept introduction:

Overlap of atomic orbitals (AOs) can be thought of as wave interference. It can be constructive or destructive. Constructive interference increases the electron density between the two nuclei (an antinode) and results in a molecular orbital (MO) of lower energy. The phases of the overlapping orbitals are the same in this case. Destructive interference decreases the electron density between the two nuclei and results in an MO of higher energy. The phases of such AOs are opposite. Since the electron density between the two nuclei decreases, there is a node (or a nodal plane) between the two atoms.

$\text{π}$ MOs, sometimes extend over three or more atoms. In this case, different combinations of constructive and destructive interference are possible. The overall bonding character and the energy depends on the number of nodes and antinodes. If there are more antinodes than nodes, the overall interaction is bonding and energy is lowered. If there are more nodes than antinodes, the overall interaction is antibonding and energy increases.

Answer to Problem 14.27P

The number of nodal planes is one for MO A, six for MO B, and none for MO C.

The order of increasing energy of the three MOs is $\text{F < D < E}$.

Explanation of Solution

The phases of the wave function of the two AOs on either side of a nodal plane are opposite each other. This results in destructive interference reducing the electron density to zero at the nodal plane.

Therefore, the nodal planes in the three MOs are:

The energy of the MO increases with the number of nodes. Therefore, the order of increasing energy of the three orbitals is $\text{F < D < E}$.

Conclusion

When atomic orbitals with different phases overlap, a node (zero electron density) is formed at the center, increasing the energy of the MO.

Interpretation Introduction

(b)

Interpretation:

The p orbital AO contributions on each carbon atom that would give rise to each MO are to be drawn.

Concept introduction:

The p orbitals that contribute to $\text{π}$ MOs are the ones perpendicular to the carbon-carbon bond exis. Only a pair of p AOs on adjacent carbon atoms overlap to produce MOs. When the overlapping orbitals have the same phase, constructive interference occurs increasing the electron density in the region between two nuclei, forming a lower energy bonding MO. If the phases of the two AOs are different, destructive interference occurs decreasing the electron density to zero at the center, forming a higher energy antibonding MO.

Answer to Problem 14.27P

The p orbitals that contribute to each of the three MOs are

Explanation of Solution

In case of D, there are two four-center MOs. Therefore, p orbitals that contribute to it must have the same phase. The second MO has the opposite phase. The p AOs that contribute to this MO will all have the same phase, but it will be opposite to that of the first group.

Therefore, the p orbitals that contribute to MOs shown in A are

E shows a total of five MOs. The first three from left are present in individual atoms. The fourth one is a two-center MO, followed again by three individual MOs. Only the AOs of C4 and C5 must have matching phases. All other adjacent AOs will have mismatched phases.

Therefore, the p orbitals contributing to the MOs as shown in D are

In case of F, a single MO covers all eight carbon atoms. Therefore, the contributing AOs must all have the same phase.

Conclusion

The p orbitals contributing to each MO are determined on the basis of the phases and the presence of an adjacent nodal plane.

Interpretation Introduction

(c)

Interpretation:

Each internuclear region is to be identified as having a bonding or an antibonding type of interaction.

Concept introduction:

A bonding interaction arises when the phases of the interacting AOs are the same. This increases the electron density between the two nuclei and lowers the energy of the MO. An antibonding interaction arises when the phases of the interacting AOs are different. This decreases the electron density between the two nuclei to zero at the center and increases the energy of the MO.

Answer to Problem 14.27P

The bonding ($\text{π}$) or antibonding (${\text{π}}^{*}$) interactions in each internuclear region in the three $\text{π}$ MOs are

Explanation of Solution

A bonding interaction requires p orbitals of the same phase on adjacent atoms. An antibonding interaction requires the interacting p orbitals to be of opposite phases. An antibonding interaction leads to a nodal plane between the two atoms.

Therefore, in D, there are six bonding interactions. There is only one antibonding interaction between C4 and C5 orbitals.

In case of E, there is only one bonding interaction between C4 and C5 orbitals. All other interactions are antibonding interactions.

In case of F, all are bonding interactions.

Conclusion

The type of interaction is determined on the basis of the phases of interacting AOs.

Interpretation Introduction

(d)

Interpretation:

Whether each MO is overall bonding, nonbonding, or antibonding is to be determined on the basis of the answer to part (c).

Concept introduction:

If the number of bonding interactions are more than the number of antibonding interactions, the MO is overall bonding. If the number of antibonding interactions is more than that of bonding interactions, the MO is overall antibonding. If the numbers are equal or there are no interactions, the MO is overall nonbonding.

Answer to Problem 14.27P

The MO shown in D is overall bonding. MO E is overall antibonding. MO F is overall antibonding. MO C is overall bonding.

Explanation of Solution

There are six bonding and only one antibonding interaction in this case. Therefore, the MO shown in D is overall bonding.

In the case of E, there is only one bonding interaction and six antibonding interactions. Therefore, MO E is overall antibonding.

All interactions in MO C are bonding interactions. Therefore, this MO is overall bonding.

Conclusion

The overall character of a multicenter MO is determined by the numbers of bonding and antibonding interactions.