What is state of hybridisation of C in CO3^2
The Correct Answer and Explanation is:
The carbon atom in the carbonate ion (CO₃²⁻) is sp² hybridized.
Here’s an explanation:
In CO₃²⁻, the carbon atom forms bonds with three oxygen atoms. These bonds are arranged in a trigonal planar geometry, where the bond angles are approximately 120°. This suggests that the carbon atom is using three hybrid orbitals to form bonds with the oxygen atoms, indicating an sp² hybridization.
Step-by-step reasoning:
- Electron Domains:
In CO₃²⁻, carbon has three bonding regions of electron density (one for each C-O bond). It also has one lone pair of electrons. However, this lone pair is delocalized, so we don’t count it as a separate electron domain in the traditional sense. - Bonding:
Each of the three C-O bonds is a combination of sigma bonds and pi bonds, with resonance occurring between the bonds, meaning that the double bond character is spread across the three bonds. - Geometry:
The ideal geometry for sp² hybridized atoms is trigonal planar, which is exactly what we observe in CO₃²⁻, with bond angles close to 120° between the oxygen atoms. - Delocalization:
The resonance structures of CO₃²⁻ show that the electron density is spread evenly across the three C-O bonds. This delocalization of electrons further supports the idea of sp² hybridization, where one of the orbitals overlaps with the p orbital of oxygen to form a pi bond, and the other two form sigma bonds.
Therefore, the carbon atom in CO₃²⁻ is sp² hybridized, reflecting its trigonal planar structure and delocalized bonding.
