Q3 / 150 · MHT CET 2026 (13 April Shift 2) · Chemistry
In the compound ${}^1\text{CH}_2 = {}^2\text{CH} - {}^3\text{CH}_2 - {}^4\text{CH}_2 - {}^5\text{C} \equiv {}^6\text{CH}$, The $C_2 - C_3$ bond is of type
- A$sp - sp^2$
- B$sp^3 - sp^3$
- C$sp - sp^3$
- Correct answerD$sp^2 - sp^3$
Solution
The given compound is $^{1}\text{CH}_2 = ^{2}\text{CH} - ^{3}\text{CH}_2 - ^{4}\text{CH}_2 - ^{5}\text{C} \equiv ^{6}\text{CH}$.
The hybridization of a carbon atom is determined by the number of sigma bonds it forms.
Carbon atom $C_2$ is bonded to one carbon atom via a double bond, one carbon atom via a single bond, and one hydrogen atom via a single bond. It forms $3$ sigma bonds and $1$ pi bond. Thus, $C_2$ is $sp^2$ hybridized.
Carbon atom $C_3$ is bonded to two carbon atoms via single bonds and two hydrogen atoms via single bonds. It forms $4$ sigma bonds and $0$ pi bonds. Thus, $C_3$ is $sp^3$ hybridized.
Therefore, the $C_2 - C_3$ bond is formed by the overlap of $sp^2$ and $sp^3$ hybridized orbitals, making it an $sp^2 - sp^3$ type bond.
Answer: $sp^2 - sp^3$
The hybridization of a carbon atom is determined by the number of sigma bonds it forms.
Carbon atom $C_2$ is bonded to one carbon atom via a double bond, one carbon atom via a single bond, and one hydrogen atom via a single bond. It forms $3$ sigma bonds and $1$ pi bond. Thus, $C_2$ is $sp^2$ hybridized.
Carbon atom $C_3$ is bonded to two carbon atoms via single bonds and two hydrogen atoms via single bonds. It forms $4$ sigma bonds and $0$ pi bonds. Thus, $C_3$ is $sp^3$ hybridized.
Therefore, the $C_2 - C_3$ bond is formed by the overlap of $sp^2$ and $sp^3$ hybridized orbitals, making it an $sp^2 - sp^3$ type bond.
Answer: $sp^2 - sp^3$
Hence, the correct answer is option: 'D'.