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(a) Use the concept of resonance to explain why all six \(\mathrm{C}-\mathrm{C}\) bonds in benzene are equal in length. (b) Are the \(\mathrm{C}-\mathrm{C}\) bond lengths in benzene shorter than \(\mathrm{C}-\mathrm{C} \mathrm{sin}-\) gle bonds? Are they shorter than \(\mathrm{C}=\mathrm{C}\) double bonds?

Short Answer

Expert verified
In benzene, all six C-C bonds have equal lengths due to the molecule being a resonance hybrid of two different structures. This causes the electrons to be shared equally among the carbon atoms, resulting in a bond length intermediate between a single bond and a double bond. As a result, benzene's C-C bond lengths are shorter than C-C single bonds but longer than C=C double bonds.

Step by step solution

01

(Step 1: Understanding Resonance Structures in Benzene)

Resonance structures are used to describe molecular structures where a single Lewis structure does not fully describe the bonding in a molecule. Benzene, with a chemical formula \(\text{C}_6\text{H}_6\), has two resonance structures. In one structure, benzene has \(\text{C-C}\) single bonds adjacent to \(\text{C} = \text{C}\) double bonds, while the other structure has double bonds adjacent to single bonds. Since neither structure fully describes benzene, the actual molecule is a hybrid of these two resonance structures.
02

(Step 2: Explaining Equal Bond Lengths in Benzene)

In benzene, all \(\text{C-C}\) bond lengths are equal because the molecule is a resonance hybrid of two different structures. This means that the electrons are shared equally between all the carbon atoms, creating a bond length between a single bond and double bond. The actual structure of benzene is neither of the resonance structures but an intermediate form. This intermediate structure causes all \(\text{C-C}\) bonds to have equal bond lengths.
03

(Step 3: Comparing Bond Lengths in Benzene to C-C Single Bonds and C=C Double Bonds)

In general, a single bond has a longer bond length than a double bond. This is because double bonds involve more electron sharing between the bonded atoms, pulling them closer together. In benzene, the actual bond length lies between the length of a single bond and a double bond. To answer part (b) of the question, the C-C bond lengths in benzene are shorter than \(\text{C-C}\) single bonds but longer than \(\text{C}=\text{C}\) double bonds. The actual bond length of benzene is an intermediate length between single and double bonds due to the resonance hybrid structure.

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Key Concepts

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

Equal Bond Lengths in Benzene
In benzene, all carbon-carbon (C-C) bonds are of equal length, which is quite unique. Unlike typical molecules where bonds vary depending on single or double bond configurations, benzene is special. This uniform bond length occurs due to resonance, a phenomenon in chemistry where a molecule can be represented by two or more valid Lewis structures. In the case of benzene, these structures involve alternating single and double bonds.

However, benzene doesn't constantly switch between these structures. Instead, it exists as a resonance hybrid, where the actual structure is a blend of all possible resonance forms. This hybridization leads to the equal distribution of electron density over the C-C bonds. As a result, each bond has a length that is not purely single or double but rather a consistent intermediate length, making them all identical.
C-C Bond Comparison in Benzene
Benzene's C-C bonds are best understood when compared with standard single and double bonds. In general, single bonds are longer than double bonds because they involve fewer shared electrons. More electrons in a double bond create greater overlap between atomic orbitals, pulling the bonded atoms closer together.

In benzene, the situation is somewhat in between. The C-C bond lengths here are shorter than typical single bonds, which would normally be longer due to lesser electron sharing. However, they remain longer than standard double bonds found in alkenes. This unique bond length in benzene is due to its resonance hybrid form, where no bond is entirely single or double, but an average of both.
The Resonance Hybrid Structure of Benzene
The resonance hybrid structure is a key factor in explaining the unique properties of benzene. While individual resonance structures show clear single and double bonds, the real benzene molecule does not flip between these forms. Instead, it is a resonance hybrid. This hybrid is not an average of two structures but a true continuous state where electrons are delocalized. This delocalization allows the bonds to be shared equally among the carbons.

In this hybrid structure, electrons move freely across the molecule, providing stability and reducing the potential energy. This results in equal bond lengths, setting benzene apart from other compounds exhibiting isolated double or single bonds. The resonance hybrid contributes to benzene's remarkable stability and unique electrical and optical properties.

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Most popular questions from this chapter

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \%\) N. Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} \mathrm{~mol}^{-1}\). The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(\mathrm{S}-\mathrm{S}\) distance in the \(\mathrm{S}_{8}\) ring is \(2.05 \AA\).) (d) The enthalpy of formation of the compound is estimated to be \(480 \mathrm{~kJ} \mathrm{~mol}^{-1}\). \(\Delta H_{f}^{\circ}\) of \(S(g)\) is \(222.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Estimate the average bond enthalpy in the compound.

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) \(\mathrm{NO}^{+}\), (b) \(\mathrm{POCl}_{3}\) (P is bonded to the three \(\mathrm{Cl}\) atoms and to the \(\mathrm{O}\), (c) \(\mathrm{ClO}_{4}^{-}\), (d) \(\mathrm{HClO}_{3}(\mathrm{H}\) is bonded to \(\mathrm{O})\).

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) \(\mathrm{Al}\) and \(\mathrm{F}\), (b) \(\mathrm{K}\) and \(\mathrm{S}\), (c) \(\mathrm{Y}\) and \(\mathrm{O}\), (d) \(\mathrm{Mg}\) and \(\mathrm{N}\).

Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\) (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P}\) (c) \(\mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

How many elements in the periodic table are represented by a Lewis symbol with a single dot? Are all these elements in the same group? Explain.

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