Ch.9 - Molecular Geometry and Bonding Theories

Problem 107b2

Chapter 9, Problem 107b2

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C12H10N2. A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are

(Recall the shorthand notation used for benzene.) (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? How many unhybridized atomic orbitals are there on the N and the C atoms in hydrazobenzene?

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Identify the hybridization of each carbon (C) and nitrogen (N) atom in both azobenzene and hydrazobenzene.

For azobenzene, note that each carbon in the benzene rings is sp2 hybridized, and the nitrogen atoms are also sp2 hybridized.

For hydrazobenzene, observe that the carbon atoms in the benzene rings remain sp2 hybridized, while the nitrogen atoms are sp3 hybridized due to the presence of single bonds and lone pairs.

Count the number of unhybridized p orbitals for each sp2 hybridized atom (each sp2 hybridized atom has one unhybridized p orbital).

Sum the unhybridized p orbitals for all the sp2 hybridized atoms in both azobenzene and hydrazobenzene to determine the total number of unhybridized atomic orbitals.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They help visualize the arrangement of electrons and the connectivity of atoms, which is crucial for understanding molecular geometry and reactivity. In the context of azo dyes and hydrazobenzene, Lewis structures illustrate how atoms are bonded and the distribution of electrons, aiding in the analysis of hybridization and orbital characteristics.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding requirements of atoms in a molecule. For example, carbon atoms in organic compounds often undergo sp3, sp2, or sp hybridization, depending on their bonding environment. Understanding hybridization is essential for determining the geometry of molecules and the number of unhybridized orbitals, which is a key aspect of the question regarding the nitrogen and carbon atoms in hydrazobenzene.

Unhybridized Atomic Orbitals

Unhybridized atomic orbitals are the orbitals that remain in their original state after hybridization has occurred. These orbitals are important for understanding the bonding and lone pairs in a molecule. In the case of hydrazobenzene, identifying the number of unhybridized orbitals on the nitrogen and carbon atoms is crucial for predicting the molecule's reactivity and interaction with other substances, particularly in the context of azo dye applications.

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Atomic Orbitals Example

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Textbook Question

One of the molecular orbitals of the H2− ion can be sketched as follows:

a. Is the molecular orbital a 𝜎 or 𝜋 MO? Is it bonding or antibonding?

Textbook Question

One of the molecular orbitals of the H2− ion can be sketched as follows:

d. Compared to the H—H bond in H2, the H—H bond in H2− is expected to be which of the following?

i. shorter and stronger

ii. longer and stronger

iii. shorter and weaker

iv. longer and weaker or

v. the same length and strength

Textbook Question

Place the following molecules and ions in order from smallest to largest bond order: H2+,B2,N2+,F2+, and Ne2.

Textbook Question

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C12H10N2. A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are

(Recall the shorthand notation used for benzene.) (c) Predict the N¬N¬C angles in each of the substances.

Textbook Question

Carbon monoxide, CO, is isoelectronic to N2. (d) Would you expect the p2p MOs of CO to have equal atomic orbital contributions from the C and O atoms? If not, which atom would have the greater contribution?

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Textbook Question

The energy-level diagram in Figure 9.40 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 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 to the molecular orbital. a. Assuming this electronic transition corresponds to the HOMO–LUMO transition, what is the HOMO in ethylene?