Concept explainers
(a)
Interpretation:
The number of nodal planes perpendicular to the bonding axes of each of the three
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.
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
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
When atomic orbitals with different phases overlap, a node (zero electron density) is formed at the center, increasing the energy of the MO.
(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
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.
The p orbitals contributing to each MO are determined on the basis of the phases and the presence of an adjacent nodal plane.
(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 (
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.
The type of interaction is determined on the basis of the phases of interacting AOs.
(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.
The overall character of a multicenter MO is determined by the numbers of bonding and antibonding interactions.
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Chapter 14 Solutions
Organic Chemistry: Principles And Mechanisms: Study Guide/solutions Manual (second)
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- On one-electron oxidation of a five valence electron AH₂ compound (C₂, point group) to a closed shell (no unpaired electrons) four valence electron C₂, cationic species, what is the symmetry of the highest energy occupied molecular orbital, and what happens to the energy of this orbital when the four valence electron AH2 C2y cation changes to a Doch geometry? O 2a₁ and stabilised (decreases in energy) O 2a₁ and destabilised (increases in energy) O 1b2 and stabilised (decreases in energy) O b₁ and no change in energyarrow_forward2. Prepare a molecular orbital energy-level diagram for the cyanide ion c) Which molecular orbital of CN¯ would you predict to interact most strongly with a hydrogen 1s orbital to form an H - C bond in the reaction CN +H* → HCN? Explain.arrow_forwardWhich of the following arrangements of p atomic orbitals gives the highest energy p-bonding molecular orbital of 1,3- butadiene?arrow_forward
- or each of the following molecules, construct the MOS from the 2p, atomic orbitals perpendicular to the plane of the carbon atoms. (a) Cyclobutadiene HC=CH НС—СН HC=CH НС —СH (b) Allyl radical H H H H C=C •C-C, H C-H H C-H H H Indicate which, if any, of these orbitals have identical ener- gies from symmetry considerations. Show the number of electrons occupying each w MO in the ground state, and indicate whether either or both of the molecules are para- magnetic. Assume that the C atoms in the allyl radical are all sp? hybridized. Activate io to Sarrow_forward. The pyridine molecule (C3H;N) is obtained by replacing one C-H group in benzene with a nitrogen atom. Because nitrogen is more electronegative than the C-H group, orbitals with electron density on nitrogen are lower in energy. How do you expect the MOs and energy levels of pyridine to differ from those of benzene?arrow_forward4) How many nodes, other than the node coincident with the molecular plane, are found in the highest energy л MO of 1,3-butadiene?arrow_forward
- 3 Convert the following infrared wavelengths to cm-1. 3.38 mm, typical for a saturated C¬H bondarrow_forwardIn the highest energy TT MO of 1,3-butadiene nodes are found.arrow_forwardExplain why 1og is the ground state for Hi. By combining your answer with the answer to Problem 5, what conclu- sions can you draw about the molecular orbital descrip- tion of the bond in H¿?arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning