Final answer:
In C3H4, the double bond between the first and second carbon atoms is formed by the end-to-end overlap of sp² hybrid orbitals to create a sigma bond and the side-to-side overlap of unhybridized 2p orbitals to create a pi bond.
Step-by-step explanation:
Formation of a Double Bond in C3H4
In the molecule C3H4, a double bond between the first and second carbon atoms is formed by the interaction of hybridized and unhybridized orbitals. Each carbon atom involved in the double bond uses a set of sp² hybrid orbitals resulting from the hybridization of two of their 2p orbitals and the 2s orbital.
One sp² hybrid orbital from each carbon overlaps end-to-end to form a sigma bond, the first bond of the double bond. Additionally, the unhybridized 2p orbital, which is perpendicular to the plane of the sp² hybrid orbitals, overlaps side-to-side with the corresponding unhybridized 2p orbital on the adjacent carbon atom, forming a pi bond (π bond), the second bond of the double bond.
The formation of a sigma bond (σ bond) typically involves head-on overlap of orbitals, where the electron density is concentrated between the nuclei of the bonding atoms. In contrast, the formation of a pi bond involves the side-by-side overlap of p orbitals with electron density concentrated above and below the plane of the nuclei. Together, these two bonds constitute the double bond present in the molecule C3H4, granting it its unique structural and chemical properties.