Question

How many sigma and pi bonds are present in the molecule given below:

$C{H}_2=C=CH-C{H}_3$

• A. sigma $C-C$ bonds $=5$, sigma $C-H$ bonds $=5$, pi $C-C$ bonds $=4$
• B. sigma $C-C$ bonds $=3$, sigma $C-H$ bonds $=6$, pi $C-C$ bonds $=2$
• C. sigma $C-C$ bonds $=4$, sigma $C-H$ bonds $=5$, pi $C-C$ bonds $=2$
• D. sigma $C-C$ bonds $=4$, sigma $C-H$ bonds $=6$, pi $C-C$ bonds $=4$

Answer 1

Answer: (B)

sigma $C-C$ bonds $=3$, sigma $C-H$ bonds $=6$, pi $C-C$ bonds $=2$

Sigma bonds ($\sigma$) are formed by the coaxial overlapping of the two orbitals of equal energies. Generally, the single bonds are sigma bonds.

Pi bonds ($\pi$) are formed by the lateral overlapping of two orbitals with equal energies.

Double bond contains one sigma bond and one pi bond whereas triple bond contains one sigma bond and two pi bond.

Here,

sigma $C-C$ bonds $=3$, sigma $C-H$ bonds $=2+1+3=6$ , pi $C-C$ bonds $=1+1=2$

Hence, option $B$ is correct.