Question

The number of $\pi$ and $\sigma$ bonds present in benzonitrile respectively are:

• A. 5, 10
• B. 5, 13
• C. 4, 10
• D. 4, 13

Hint:

Count π bonds from double and triple bonds; count σ bonds from all single bonds including those in triple bonds.

Answer 1

The Correct Option is B

Step 1: Draw benzonitrile as a benzene ring with one hydrogen replaced by a $-C\equiv N$ group.
Benzonitrile is $C_6H_5-C\equiv N$. Starting from unsubstituted benzene, one ring hydrogen is swapped out for the nitrile carbon, so the ring keeps its six carbons but now carries only five hydrogens.

Step 2: Count the sigma bonds inside the ring itself.
The six ring carbons are joined by 6 C-C sigma bonds around the ring, and the five remaining ring hydrogens each add one C-H sigma bond:
\[ 6 \, (C-C) + 5 \, (C-H) = 11 \, \text{sigma bonds from the ring} \]

Step 3: Count the sigma bond linking the ring to the nitrile group.
The bond joining the substituted ring carbon to the nitrile carbon is a single bond, contributing 1 more sigma bond.

Step 4: Count the bonds inside the nitrile group.
$C \equiv N$ is a triple bond, made of 1 sigma bond and 2 pi bonds.

Step 5: Add up the sigma bonds.
\[ 11 \, (\text{ring}) + 1 \, (\text{ring-to-CN}) + 1 \, (C\equiv N \text{ sigma}) = 13 \]

Step 6: Add up the pi bonds.
The aromatic ring contributes 3 pi bonds (its three formal double bonds), and the nitrile group contributes 2 more:
\[ 3 + 2 = 5 \]

Final Answer:
Benzonitrile has 5 pi bonds and 13 sigma bonds.
\[ \boxed{5, \; 13} \]