Question

# 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)- $s{{p}^{3}},\text{ }s{{p}^{2}},\text{ }sp$

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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 $sp$.

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 $sp$.
$C{{H}_{2}}=CH-C\equiv N$
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)- $s{{p}^{2}},\text{ }s{{p}^{2}},\text{ }sp$.
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.