Question

what orbitals are used to form each indicated bond in the molecule shown? part 1 of 4 a bond a c3 $sp^{2}$ - c4 $sp^{2}$ part: 1 / 4 part 2 of 4 π bond a c3 - c4

Explanation:

Step1: Recall pi - bond formation

Pi - bonds are formed by the overlap of unhybridized p - orbitals. In a double - bond, one bond is a sigma bond (formed by the overlap of hybridized orbitals) and the other is a pi - bond.

Step2: Identify unhybridized p - orbitals in C3 and C4

Carbon atoms C3 and C4 are sp² hybridized. Each sp² - hybridized carbon has one unhybridized p - orbital. The pi - bond between C3 and C4 is formed by the side - on overlap of these unhybridized p - orbitals.

Answer:

C3 $p$ - C4 $p$