Question

ch 2
question 2 of 32 (1 point) | question attempt: 1 of 5
part 1 of 4
σ bond a
c3 $sp^{2}$ - c4 $sp^{2}$
part 2 of 4
π bond a
c3 $p$ - c4 $p$
part 3 of 4
bond b
c2 - c3

Explanation:

Step1: Identify hybridization of C - atoms

Carbon atoms in double - bonded systems like the one between C3 and C4 have $sp^{2}$ hybridization for $\sigma$ bond formation and unhybridized p - orbitals for $\pi$ bond formation. C2 and C3 in the given structure: C2 is $sp^{3}$ hybridized as it has 4 single bonds around it, and C3 (in the double - bond system) is $sp^{2}$ hybridized.

Step2: Determine bond types

For Bond b (C2 - C3), since C2 is $sp^{3}$ hybridized and C3 is $sp^{2}$ hybridized, the $\sigma$ bond between them is formed by the overlap of an $sp^{3}$ orbital of C2 and an $sp^{2}$ orbital of C3.

Answer:

C2 $sp^{3}$ - C3 $sp^{2}$