Answer
The correct answer is the two resonance structures shown above. The delocalization of the double bonds around the benzene ring is a key feature of the aromatic nature of benzene, which contributes to its stability and unique chemical properties.
Work Step by Step
To write the Lewis structures for benzene (C6H6), including the resonance structures, we need to follow these steps:
1. Draw the basic structure of benzene, which is a six-membered ring of carbon atoms with one hydrogen atom bonded to each carbon atom.
2. Determine the number of valence electrons in the benzene molecule:
- Carbon (C) has 4 valence electrons, and there are 6 carbon atoms.
- Hydrogen (H) has 1 valence electron, and there are 6 hydrogen atoms.
- Total valence electrons = (6 × 4) + (6 × 1) = 24 + 6 = 30 valence electrons.
3. Draw the Lewis structure for benzene, showing the single bonds between the carbon atoms and the hydrogen atoms.
```
H-C=C-H
| |
C-C-C
| |
H-C=C-H
```
4. Observe that the above structure does not accurately represent the bonding in benzene. Benzene exhibits resonance, where the double bonds are delocalized around the ring.
5. Draw the resonance structures for benzene. There are two equivalent resonance structures:
Resonance Structure 1:
```
H-C=C-H
| |
C-C-C
| |
H-C=C-H
```
Resonance Structure 2:
```
H-C-C-H
| |
C=C-C=
| |
H-C-C-H
```
