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# In vinyl acetylene, $CH\equiv C-CH=C{{H}_{2}}$, the type of overlapping in $({{C}_{2}}\sigma {{C}_{3}})$ bond is:A. $s{{p}^{2}}-sp$B. $sp-s{{p}^{2}}$C. $s{{p}^{3}}-s{{p}^{3}}$D. $s{{p}^{3}}-s{{p}^{2}}$

Last updated date: 13th Jul 2024
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Hint: Take into consideration the number of bonds each carbon is forming and whether they are $\sigma$-bonds or$\pi$-bonds. This will help you determine the hybridization of each carbon atom.

First, let us identify and name the carbon atoms that are asked in the question:
${{C}^{1}}H\equiv {{C}^{2}}-{{C}^{3}}H={{C}^{4}}{{H}_{2}}$
We have to find the hybridization of the two carbons present in the middle.
Consider carbon number 2.
We can see that it is forming 1 $\sigma$-bond and 2 $\pi$-bonds with carbon number 1, and 1 $\sigma$-bond with carbon number 3. It does not form any bonds with a hydrogen atom.
Thus, it has 2 $\sigma$-bonds whose bond pairs we need to consider while calculating the hybridization of the atom. $\pi$-bonds are not considered while calculating the hybridization. As there are only 2 bond pairs involved, 2 hybridized orbitals are required to accommodate them. Thus, the hybridization of this carbon is $sp$.
Consider carbon number 3.
We can see that it is forming 1 $\sigma$-bond with carbon number 2, and 1 $\sigma$-bond and 1 $\pi$-bond with carbon number 4. It also forms 1 $\sigma$-bond with a hydrogen atom.
Now, to calculate the hybridization of this carbon, consider the number of $\sigma$-bonds present. We see that 3 $\sigma$-bonds are present and 3 hybridized orbitals are required to accommodate the 3 bond pairs. Thus, the hybridization of this carbon is $s{{p}^{2}}$.

So, the correct answer is “Option B”.

Note: Please be careful while calculating the number of $\sigma$-bonds that each of the carbon atoms forms with not only other carbon atoms but also the hydrogen atoms since they are not explicitly shown and you might miss them.