Question

A: Tetracyanomethane
B: Carbon dioxide
C: Benzene
D: 1,3-buta-di-ene
Ratio of $\sigma $ and $\pi $:
(A) A and B are equal
(B) C and D are equal
(C) same in A, B, C and D
(D) B and D are equal

Answer 1

Hint: One single bond has only one sigma bond $\left( \sigma \right)$ and no pi bond $\left( \pi \right)$. One double bond contributes to one sigma bond $\left( \sigma \right)$ and one pi-bond $\left( \pi \right)$. One triple bond has one sigma bond $\left( \sigma \right)$ and two pi-bond $\left( \pi \right)$. Describe the structure and find the number of single, double and triple bonds and calculate the ratios.

Complete step by step solution:
Let us draw the structures of the four structures given in the question:

S. No.	Name of the compounds	Structure of the compounds	Explanation of the bonds and formation of the structure	Number of sigma bonds $\left( \sigma \right)$	Number of pi-bonds $\left( \pi \right)$	Ratio of $\sigma $ and $\pi $
1.	Tetracyanomethane		The compound has four cyano groups. A carbon atom is attached to the carbon atom of a cyano group by a single bond. There is a triple bond between nitrogen and carbon atoms in the cyano group. So, there are four triple bonds.	$\left( 1\times 4 \right)+\left( 1\times 4 \right)$, which is equal to 8.	$\left( 2\times 4 \right)$, which is equal to 8.	$\dfrac{8}{8}$ or 1.
2.	Carbon dioxide		The compound has a single carbon atom which forms a double bond with two oxygen atoms.	$\left( 1\times 2 \right)$ or 2	$\left( 1\times 2 \right)$ or 2	$\dfrac{2}{2}$ or 1
3.	Benzene		The compound has six carbon atoms which form an alternative double bond with each other. Every carbon atom is attached to a hydrogen atom with a single bond. The chemical formula is ${{\text{C}}_{6}}{{\text{H}}_{6}}$.	$\left( 1\times 6 \right)+\left( 1\times 6 \right)$ or 12	$\left( 1\times 3 \right)$ or 3	$\dfrac{12}{3}$ or 4
4.	1,3-buta-di-ene		The compound has an alternative double bond. The total number of hydrogen atoms is 6.	$\left( 1\times 6 \right)+\left( 1\times 3 \right)$or 9	$\left( 1\times 2 \right)$ or 2	$\dfrac{9}{2}$ or 4.5

The ratio of $\sigma $ and $\pi $ in A and B are equal, which is option (A).

Note: An element forms a sigma bond $\left( \sigma \right)$ with every different element only once. That is if an element forms multiple bonds with another element, then, it means that there is one sigma bond $\left( \sigma \right)$ and rest are pi-bonds. The compounds with single bonds are known as saturated and compounds with multiple bonds are unsaturated.