Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the shape of benzene, and what hybridization do you expect for each carbon?

Short Answer

Expert verified
Benzene is a planar hexagonal ring with each carbon atom \(sp^2\) hybridized.

Step by step solution

01

Understand Benzene's Structure

Benzene is a well-known aromatic hydrocarbon with the molecular formula \(C_6H_6\). It consists of six carbon atoms arranged in a planar hexagonal ring. Each carbon atom is bonded to two other carbons and one hydrogen.
02

Identify Benzene’s Shape

The structure of benzene is a perfect planar hexagon where all carbon-carbon bond angles are 120 degrees. This arrangement results in a symmetrical, flat, ring structure that is characteristic of aromatic compounds.
03

Determine the Hybridization of Carbon Atoms

In benzene, each carbon atom is part of a double bond and is bonded to three atoms (two carbon and one hydrogen). This trigonal planar arrangement requires \(sp^2\) hybridization. The \(sp^2\) hybridized orbitals form the sigma bonds, while the unhybridized p orbitals overlap to form the delocalized pi bonds that contribute to benzene's aromaticity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Aromatic Hydrocarbon
Aromatic hydrocarbons are a special class of organic compounds known for their stability and unique chemical properties. Benzene is the most famous example of this group.
  • They contain a notable ring structure with alternating single and double bonds, known as conjugation.
  • This conjugation is responsible for their remarkable stability and is a key feature of aromatic compounds.
  • Benzene's formula, \(C_6H_6\), showcases this stability, as it has equal bond lengths and energies, unlike typical alkanes or alkenes.
Aromatic compounds are typically very stable due to the delocalization of electrons. These electrons are free to move across the entire ring, which maintains a lower energy state. This contributes to Benzene's exceptional stability and its distinct characteristics as an aromatic hydrocarbon.
Planar Hexagon
The structure of benzene forms a planar hexagon, which is a flat, two-dimensional shape. This configuration is essential to its stability and aromatic nature.
  • Each of the six carbon atoms in benzene is arranged at a 120-degree angle, forming a perfect hexagon.
  • This symmetry makes benzene highly stable, as all bond angles and bond distances are equal.
  • This planar structure allows the overlap of p orbitals (perpendicular to the plane of the ring) to form continuous pi systems.
Because the hexagon is planar, the electrons in the pi system can evenly distribute across the entire structure. This even electron distribution is an essential factor in benzene's stability, which is unusual compared to other non-aromatic hydrocarbons.
Carbon Hybridization
In the benzene molecule, each of the six carbon atoms undergoes \(sp^2\) hybridization. Hybridization is a critical concept for understanding the bond formation in organic molecules.
  • Each carbon atom combines one s orbital and two p orbitals to form three \(sp^2\) hybrid orbitals.
  • These \(sp^2\) orbitals arrange themselves in a trigonal planar geometry around each carbon atom, with an angle of 120 degrees between them.
  • The remaining unhybridized p orbital is oriented perpendicular to this plane and is involved in forming pi bonds with its neighbors.
This hybridization results in strong sigma bonds between the carbon atoms and a delocalized pi system, which is crucial for benzene's aromaticity. It explains why all C-C and C-H bonds in benzene are of similar length and energy, showcasing its uniformity.
Trigonal Planar Arrangement
The trigonal planar arrangement is a geometric configuration that offers insight into the structure and behavior of benzene's carbon atoms.
  • Each carbon atom in benzene is bonded to three other atoms: two carbon atoms and one hydrogen atom, resulting in three groups arranged in a planar fashion.
  • This arrangement ensures that the bond angles are all 120 degrees, contributing to benzene's overall planar structure.
  • It facilitates the formation of the delocalized pi bonds as the p orbitals from each carbon overlap efficiently.
The trigonal planar configuration around each carbon atom is integral to achieving benzene's stability and aromatic characteristics. This geometric property explains the uniform distribution of electrons in the pi system that supports the overall structure and aromatic nature of benzene.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Draw structures for the following molecules, showing lone pairs: (a) Acrylonitrile, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N},\) which contains a carbon-carbon double bond and a carbon-nitrogen triple bond (b) Ethyl methyl ether, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O},\) which contains an oxygen atom bonded to two carbons (c) Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), which contains a chain of four carbon atoms (d) Cyclohexene, \(\mathrm{C}_{6} \mathrm{H}_{10}\), which contains a ring of six carbon atoms and one carbon-carbon double bond

Divalent carbon species called carbenes are capable of fleeting existence. For example, methylene, : \(\mathrm{CH}_{2}\), is the simplest carbene. The two unshared electrons in methylene can be either paired in a single orbital or unpaired in different orbitals. Predict the type of hybridization you expect carbon to adopt in singlet (spin-paired) methylene and triplet (spin-unpaired) methylene. Draw a picture of each, and identify the valence orbitals on carbon.

Draw a line-bond structure for 1,3 -butadiene, \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} .\) Indicate the hybridization of the orbitals on each carbon, and predict the value of each bond angle.

The following model is a representation of aspartame, \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{5}\), known commercially under many names, including NutraSweet. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure for aspartame, and indicate the positions of multiple bonds (gray \(=\mathrm{C},\) red \(=\mathrm{O},\) blue \(=\mathrm{N},\) ivory \(=\mathrm{H})\)

How many valence electrons does each of the following dietary trace elements have? (a) Zinc (b) Iodine (c) Silicon (d) Iron

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free