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# How many $\sigma$ and $\pi$ bonds are there in the compound:(A) 14$\sigma$, 8$\pi$(B) 18$\sigma$,8$\pi$(C) 19$\sigma$,4$\pi$(D) 14$\sigma$,2$\pi$

Last updated date: 18th Jun 2024
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Hint: If there is a single bond between any atom, then that bond can be considered as a $\sigma$ bond. The multiplicity of a bond shows presence of a $\pi$-bond in addition to a $\sigma$ bond.

Complete step by step solution:
A $\pi$ bond is formed by overlap of orbitals in a side-by-side fashion and has zero electron density at its nodal planes. In other ways we can say that in a double bond, there is one $\pi$-bond and one $\sigma$-bond is present.
Let’s calculate the number of $\pi$-bonds present in the given compound.
You can see that there are 4 double bonds present in the compound and hence the number of $\pi$-bonds present in the given compound will be equal to 4.
Let’s find the number of $\sigma$-bonds present in the compound.
We can see that there are a total 19 $\sigma$-bonds present in the compound. Also draw C-H bonds and consider them in the counting of bonds as well.
So, correct answer is (C) 19$\sigma$,4$\pi$

A double bond is made up of one $\sigma$-bond and one $\pi$-bond. While a triple bond can be said to be made up of one $\sigma$-bond and two $\pi$-bonds.
$\sigma$-bond is more stronger than a $\pi$-bond.

Note:
Do not forget to include C-H $\sigma$-bonds in the counting process of total $\sigma$-bonds as they are often not shown in the structure and may lead to mistakes. Do not forget to include a $\sigma$-bond from a C-C double bond.