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Chapter 8: Problem 89

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) consists of a six- membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis structures for benzene, including resonance structures.

Short Answer

Expert verified
The Lewis structures of benzene (C6H6) consist of six carbon atoms in a ring with alternating single and double bonds, and one hydrogen atom bonded to each carbon atom. There are two resonance structures for benzene: 1. Resonance structure 1: C1=C2-C3=C4-C5=C6 | | | H1 H2 H3 H4 H5 2. Resonance structure 2: C1-C2=C3-C4=C5-C6=C1 | | | H1 H2 H3 H4 H5 Each carbon-carbon bond in benzene has an intermediate bond order of 1.5, which is an average of the two resonance structures.

Step by step solution

01

Determine the total number of valence electrons

: Each carbon atom has 4 valence electrons, and each hydrogen atom has 1 valence electron. There are six carbon atoms and six hydrogen atoms in benzene. The total number of valence electrons in the molecule is: Total valence electrons = (6 脳 4) + (6 脳 1) = 24 + 6 = 30
02

Connect the atoms using single bonds

: Connect the six carbon atoms in a ring and add one hydrogen atom to each carbon atom in the ring using single bonds. C1 - C2 - C3 - C4 - C5 - C6 | | | | H1 H2 H3 H4 H5
03

Distribute the remaining electrons in the molecule

: Next, distribute the remaining valence electrons among the atoms to fulfill the octet rule for carbon atoms. We've used 12 electrons so far to form single bonds. There are 18 electrons remaining: 30 total valence electrons - 12 electrons for single bonds = 18 remaining electrons To fulfill the octet rule, we can use the remaining 18 electrons to create three double bonds between alternating carbon atoms.
04

Check the octet rule

: Check that the octet rule is followed for each atom: H1, H2, H3, H4, H5, and H6: 2 electrons (Completed because hydrogen can only have 2 electrons) C1, C2, C3, C4, C5, and C6: 8 electrons (Completed because each carbon atom has 4 single electron pairs and one double bond) The octet rule is followed for each atom in the structure.
05

Draw resonance structures if applicable

: Resonance structures are possible when there is more than one possible arrangement of electrons in a molecule. In benzene, three double bonds can be rotated between the carbon atoms in the ring. As a result, there are two resonance structures for benzene: Resonance structure 1: C1=C2-C3=C4-C5=C6 | | | H1 H2 H3 H4 H5 Resonance structure 2: C1-C2=C3-C4=C5-C6=C1 | | | H1 H2 H3 H4 H5 In reality, the electron distribution is an average of these two resonance structures. The actual structure of benzene is a hybrid of these two resonance structures, with each carbon-carbon bond having an intermediate bond order of 1.5.

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Most popular questions from this chapter

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}, \mathrm{O}-\mathrm{H}\), \(\mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H}\)

Compare and contrast the bonding found in the \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{HF}(\mathrm{g})\) molecules with that found in \(\operatorname{NaF}(s)\).

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