Chapter 17: Problem 25

The geometry about each carbon in a particular hydrocarbon molecule is tetrahedral. What does this say about the possibility of there being \(C=C\) or \(C \equiv C\) bonds in the molecule? Explain.

Short Answer

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The given hydrocarbon molecule has a tetrahedral geometry around each carbon atom, indicating that all carbon atoms are sp3 hybridized and form only single bonds. As a result, there can be no \(C=C\) (double) or \(C \equiv C\) (triple) bonds in the molecule, since they would require different hybridization states (sp2 for double bonds and sp for triple bonds) and result in different geometries (trigonal planar and linear, respectively).

Step by step solution

Review Tetrahedral Geometry and Hybridization

In a molecule, the carbon atom can form different geometries based on its hybridization state. For carbon atoms to form a tetrahedral geometry, they need to be sp3 hybridized, which means the carbon atom has a combination of one 2s orbital and three 2p orbitals, resulting in a total of four orbitals (sp3). In this hybridization state, the carbon atom forms four sigma bonds with other atoms.

Analyze Carbon-Carbon Double Bonds

A double bond between two carbon atoms (\(C=C\)) consists of one sigma bond and one pi bond. Carbon atoms that form a double bond are sp2 hybridized since they have one 2s orbital and two 2p orbitals to form a total of three orbitals (sp2). In this hybridization state, the carbon atom forms three sigma bonds. As a result, the geometry around carbon atoms with double bonds is typically trigonal planar, not tetrahedral.

Analyze Carbon-Carbon Triple Bonds

A triple bond between two carbon atoms (\(C \equiv C\)) consists of one sigma bond and two pi bonds. Carbon atoms that form a triple bond are sp hybridized since they have one 2s orbital and one 2p orbital to form a total of two orbitals (sp). In this hybridization state, the carbon atom forms two sigma bonds. The geometry around carbon atoms with triple bonds is linear, not tetrahedral.

Conclusion

Since we are given that the geometry around each carbon atom in the hydrocarbon molecule is tetrahedral, it implies that all the carbon atoms are sp3 hybridized and form only single bonds. Therefore, there cannot be any \(C=C\) (double bonds) or \(C \equiv C\) (triple bonds) in the molecule.

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The geometry about each carbon in a particular hydrocarbon molecule is tetrahedral. What does this say about the possibility of there being \(C=C\) or \(C \equiv C\) bonds in the molecule? Explain.

Short Answer

Step by step solution

Review Tetrahedral Geometry and Hybridization

Analyze Carbon-Carbon Double Bonds

Analyze Carbon-Carbon Triple Bonds

Conclusion

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