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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, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\) A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2}\) (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(\mathrm{C}\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

Short Answer

Expert verified
In azobenzene, N atoms are sp^2 hybridized which allows for greater π electron delocalization and planar molecular geometry, making it a good candidate for solar energy conversion applications. In hydrazobenzene, N atoms are sp^3 hybridized with non-planar geometry, resulting in less π electron delocalization. Azo dyes like azobenzene have strong absorption in visible light, while hydrazobenzene is nearly colorless, making it less effective for solar energy conversion.

Step by step solution

01

(a) Determining Hybridization of Nitrogen Atoms

To determine the hybridization at the Nitrogen (N) atoms in both azobenzene and hydrazobenzene, first consider the number of sigma bonds and lone pairs around the N atom. For azobenzene, the N atom is connected to two carbon atoms through double bonds (amide bond) and a single bond with another N atom. This results in three sigma/molecule bonds, and no lone pairs. Therefore, the hybridization of the N atom in azobenzene is sp^2. For hydrazobenzene, the N atom is connected to one carbon atom through a single bond, another N atom through a single bond, and has a lone pair. This means that there are three sigma/molecule bonds, and one lone pair. Therefore, the hybridization of the N atom in hydrazobenzene is sp^3.
02

(b) Unhybridized Atomic Orbitals in Azo and Hydrazobenzenes

In azobenzene, the Nitrogen atoms are sp^2 hybridized, meaning they have one unhybridized p atomic orbital. Thus, the π electron density can be delocalized over the entire π system via resonance. In addition, azobenzene’s carbon atoms are sp^2 hybridized and also have one unhybridized p atomic orbital each. In hydrazobenzene, the Nitrogen atoms are sp^3 hybridized, meaning they don't have any unhybridized p orbitals. Thus, the π electron density cannot be delocalized via resonance between the N atoms. The carbon atoms in the benzene ring are sp^2 hybridized and have one unhybridized p atomic orbital each.
03

(c) Predicting N-N-C Angles

To predict the N-N-C angles in each of the substances, identify the hybridization of the nitrogen atoms. For azobenzene, the Nitrogen is sp^2 hybridized, so the ideal bond angle for the N-N-C is approximately 120°. For hydrazobenzene, the Nitrogen is sp^3 hybridized, so the ideal bond angle for the N-N-C is approximately 109.5°.
04

(d) Delocalization Comparison

Azobenzene has unhybridized p orbitals on both the Nitrogen and Carbon atoms. This allows for greater π electron delocalization, which leads to the extended conjugation of double bonds in the molecule. This extended conjugation explains the statement that azobenzene has greater delocalization of its π electrons than hydrazobenzene. In hydrazobenzene, the Nitrogen atoms are sp^3 hybridized and lack unhybridized p-orbitals, preventing π electron delocalization between the N atoms and resulting in less extended conjugation of double bonds.
05

(e) Planarity and Delocalization Consistency

The observation that all of the atoms in azobenzene lie in one plane, while those in hydrazobenzene do not, is consistent with the greater delocalization of π electrons in azobenzene. The sp^2 hybridized N and C atoms in azobenzene have planar molecular geometries, allowing the p orbitals to overlap, resulting in efficient delocalization of π electrons. Conversely, the sp^3 hybridized N atoms in hydrazobenzene result in a non-planar geometry, impeding π electron delocalization.
06

(f) Solar Energy Conversion Application

In a solar energy conversion device, we are generally looking for a molecule with strong light absorption in the visible region of the spectrum. Azo dyes, like azobenzene with its intense red-orange color, are more suitable for capturing a broader range of the spectrum, resulting in efficient solar energy conversion. On the other hand, hydrazobenzene, being nearly colorless, does not absorb light effectively in the visible range and is therefore less effective for solar energy conversion applications.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hybridization
Hybridization is a key concept in understanding the structure and bonding of molecules. It describes the mixing of atomic orbitals to create new hybrid orbitals for bonding.
In the case of azo dyes, particularly azobenzene (\(\mathrm{C}_{12}\mathrm{H}_{10}\mathrm{N}_{2}\)), the nitrogen atoms are involved in a specific type of hybridization. Azobenzene's nitrogen atoms form three sigma bonds without lone pairs, leading to an \(sp^2\) hybridization. This hybridization of \(sp^2\) includes participation of one s orbital and two p orbitals.
Contrast this with hydrazobenzene (\(\mathrm{C}_{12}\mathrm{H}_{12}\mathrm{N}_{2}\)), where the nitrogen shows \(sp^3\) hybridization due to its involvement in forming three sigma bonds and hosting a lone pair. Here, the atom uses one s orbital and three p orbitals. Understanding how each nitrogen fits into these hybridizations helps predict molecule shape, angle between atoms, and the overall electronic properties of the substances.
  • \(sp^2\) hybridization results in a planar structure and a bond angle similar to 120°.
  • \(sp^3\) hybridization leads to a tetrahedral structure with angles close to 109.5°.
Electron Delocalization
Electron delocalization is a phenomenon where \(\pi\) electrons extend over multiple atoms, enhancing molecule stability.
In azobenzene, the \(\pi\) electron delocalization is pronounced due to the presence of unhybridized p orbitals at both nitrogen and carbon atoms. This extensive delocalization across the conjugated system largely contrasts with hydrazobenzene, which lacks unhybridized orbitals on its nitrogen atoms.
Because azobenzene has sp2 hybridized nitrogen, it allows for \(\pi\) electrons to spread across the molecule, enabling resonance and contributing to its vivid color. Meanwhile, hydrazobenzene, with sp3 hybridization, doesn’t permit efficient delocalization of \(\pi\) electrons.
The extended conjugation in azobenzene underlies the term 'greater delocalization,' reflecting the aesthetic and chemical impacts of its molecular structure. Enhanced delocalization in azobenzene means it absorbs visible light, attributing to its color.
  • Azobenzene features strong \(\pi\) conjugation.
  • Hydrazobenzene’s \(\pi\) system is more localized.
Solar Energy Conversion
Solar energy conversion involves transforming sunlight into usable energy, often requiring materials with efficient light absorption.
Azo dyes like azobenzene are exceptional for this purpose due to their intense colors, resulting from extensive \(\pi\) electron delocalization. Azobenzene's red-orange color covers a broad part of the visible spectrum, making it advantageous in capturing solar energy.
Hydrazobenzene, however, is nearly colorless due to its inefficient electron delocalization, thus it doesn't absorb visible light effectively, rendering it less suitable for such applications. Azobenzene's planar structure contributes to its utility in solar devices by allowing optimal orbital overlap for electron movement, necessary for efficient energy conversion.
In conclusion, the molecular properties of azobenzene highlight its superiority over hydrazobenzene in solar applications—its coloration and structure both optimize its function for harnessing solar energy.
  • Azobenzene captures a wide wavelength spectrum.
  • Hydrazobenzene is not effective in light absorption.

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Most popular questions from this chapter

In which of the following AF \(_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{PF}_{6}^{-}, \mathrm{SbF}_{\mathrm{s}}, \mathrm{SF}_{4} ?\)

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3},(\mathbf{b}) \mathrm{CO},(\mathbf{c}) \mathrm{BF}_{3},\) (d) \(\mathrm{SO}_{2} ?\)

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(96 \mathrm{pm}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? \(\mathrm{In}\) what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxvgen?

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

(a) An \(\mathrm{AB}_{2}\), molecule is linear. How many nonbonding electron pairs are around the A atom from this information? (b) How many nonbonding electrons surround the Xe in \(\mathrm{XeF}_{2} ?(\mathbf{c})\) Is \(\mathrm{XeF}_{2}\) linear?

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