In the carbon skeleton structure below, identify the carbon atom(s) with sp2 hybridization. 4th attempt Jd See Periodic Table See Hint TIP: Select each sp2 hybridized carbon: Selected atoms will be highlighted in green
The Correct Answer and Explanation is:
To determine which carbon atoms in a carbon skeleton are sp² hybridized, focus on those that form double bonds or are part of planar trigonal arrangements, such as carbonyl groups or aromatic rings. An sp² hybridized carbon forms three sigma bonds and one unhybridized p orbital that participates in pi bonding.
Correct Identification:
- Carbon atoms involved in a double bond (C=C).
- Carbon atoms in benzene rings (aromatic).
- Carbon atoms bonded to oxygen in carbonyl groups (C=O).
These carbon atoms have trigonal planar geometry, with bond angles close to 120 degrees. In contrast:
- sp³ hybridized carbon atoms form four sigma bonds and exhibit tetrahedral geometry.
- sp hybridized carbon atoms form two sigma bonds and two pi bonds, usually found in triple bonds or linear arrangements.
If the image provided a structure like ethene (C₂H₄), the two central carbon atoms are sp² hybridized due to the double bond. In benzene (C₆H₆), all six carbon atoms are sp² hybridized due to the delocalized pi system. In molecules with aldehyde or ketone groups, the carbonyl carbon is sp² hybridized.
Example:
For a compound such as propene (CH₃–CH=CH₂):
- The second carbon (C₂), which is double-bonded to another carbon, is sp² hybridized.
- The third carbon (C₃), which is also part of the double bond, is sp² hybridized.
- The first carbon (C₁), part of the methyl group (CH₃), is sp³ hybridized.
Understanding electron domain geometry and recognizing bond types allows accurate identification of hybridization states. The sp² hybridization enables the formation of planar structures, critical in determining molecular shape and reactivity.
