Solving for the Ratio of Pure Orbitals and Hybrid Orbitals in Organic Compounds: Acetylene and Benzene

Acetylene:
- Acetylene, with the chemical formula C2H2, consists of two carbon atoms and two hydrogen atoms.
- Each carbon atom in acetylene undergoes sp hybridization, forming two sp hybrid orbitals.
- The sp hybrid orbitals overlap with the 1s orbitals of two hydrogen atoms to form two sigma bonds.
- Additionally, each carbon atom also has two unhybridized p orbitals that overlap laterally to form two pi bonds.
- Therefore, in acetylene, the ratio of pure orbitals to hybrid orbitals is 2:2, representing the two sp hybrid orbitals and two unhybridized p orbitals on each carbon atom.

Benzene:
- Benzene, with the chemical formula C6H6, is a cyclic aromatic hydrocarbon consisting of six carbon atoms and six hydrogen atoms.
- Each carbon atom in benzene undergoes sp2 hybridization, forming three sp2 hybrid orbitals and one unhybridized p orbital.
- The six carbon atoms in benzene form a ring structure where each carbon atom is bonded to two other carbon atoms and one hydrogen atom.
- The overlapping of the sp2 hybrid orbitals results in the formation of six sigma bonds, while the lateral overlap of the unhybridized p orbitals forms three pi bonds within the benzene ring.
- Thus, in benzene, the ratio of pure orbitals to hybrid orbitals is 1:3, representing the single unhybridized p orbital and three sp2 hybrid orbitals on each carbon atom.

In conclusion, by understanding the hybridization of carbon atoms in organic compounds like acetylene and benzene, one can determine the ratio of pure orbitals to hybrid orbitals present in these molecules.