Question

The number of sigma and $\pi$ bonds present in pent-2-en-4-yne are:
(a) $10 \sigma$ bonds and $3 \pi$ bonds
(b) $8 \sigma$ bonds and $5 \pi$ bonds
(c) $11 \sigma$ bonds and $2 \pi$ bonds
(d) $13 \sigma$ bonds and no $\pi$ bonds

Answer 1

First of all, write the structure of the pent-2-en-4-yne and after that count the number of single bonds i.e.-H bond and C-C bond which represents the sigma bond. In carbon double carbon, there is one sigma bond and one pi-bond but in carbon triple bond carbon there is one sigma bond and two pi-bonds. Now you can easily answer the given statement accordingly.

Complete step by step solution:
First of let’s discuss what are sigma and pi-bonds. The Sigma bond is formed by the end to end overlapping of the two half-filled atomic orbitals and in this, the orbitals overlap along the internuclear axis and the bond is symmetrical about the internuclear axis.
On the other hand, pi-bond is formed by the sidewise overlapping of the two atomic orbitals. The sidewise overlapping can take place when the orbitals have their lobes perpendicular to the molecular axis.
Now considering the statement as;
The structure of pent-2en-3yne is as;

From the structure, we can see that there are 10 $\sigma$ bonds ( I.e. six C-H bonds and four -C-C bonds)and 3$\pi$bonds (i.e. one pi- bond from carbon-carbon double bond and two pi-bonds from carbon-carbon triple bond).
So, The number of sigma and $\pi$ bonds present in pent-2-en-4-yne are:10 $\sigma$ bonds and 3$\pi$bonds.

Hence, option (a) is correct.

Note: Sigma bond is stronger than the pi-bond. It is so because in sigma bonds, the orbitals overlap end to end and the extent of overlapping is large but in case of pi-bond, the sidewise overlapping to form the pi-bond is not efficient.