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Which type of hybridization is shown by carbon atom from left to right in the given compound:
$C{{H}_{2}}=CH-C\equiv N$?
(a)- $s{{p}^{2}},\text{ }s{{p}^{2}},\text{ }sp$
(b)- $s{{p}^{2}},\text{ }sp,\text{ }sp$
(c)- $sp,\text{ }s{{p}^{2}},\text{ }s{{p}^{3}}$
(d)- $s{{p}^{3}},\text{ }s{{p}^{2}},\text{ }sp$

Answer
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Hint: If the carbon atom has a single bond, the hybridization is $s{{p}^{3}}$. If the carbon atom has a double bond then the hybridization is $s{{p}^{2}}$. If the carbon atom has a triple bond then the hybridization is $sp$.

Complete answer:
All the orbitals in an atom are not of the same energy, so when the atoms attach with other atoms then these energies redistribute among themselves and become equal in energy, and this process is known as hybridization. With the help of hybridization, we can predict the number of orbitals in the atom of the molecule. For finding the hybridization of the carbon atom, the rules are as follows:
If the carbon atom has a single bond, the hybridization is $s{{p}^{3}}$.
If the carbon atom has a double bond then the hybridization is $s{{p}^{2}}$.
If the carbon atom has a triple bond then the hybridization is $sp$.
So, in the molecule:
$C{{H}_{2}}=CH-C\equiv N$
The first carbon atom from the left has a double bond, so its hybridization will be $s{{p}^{2}}$, the second carbon atom on the left side has a double bond and on the right side has a single bond, so a double bond will be considered, therefore, the hybridization will be $s{{p}^{2}}$. The last carbon atom has a triple bond, so its hybridization will be $sp$.

Therefore, the correct answer is an option (a)- $s{{p}^{2}},\text{ }s{{p}^{2}},\text{ }sp$.

Note:
The maximum hybridization of carbon can be $s{{p}^{3}}$, but some other atoms like transition metals have higher hybridization like $s{{p}^{3}}d,\ s{{p}^{3}}{{d}^{2}}$, etc. this is due to vacant d-orbital in transition metal. But the carbon has maximum covalency of 4 and cannot extend beyond it.