MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Hybridisation of a carbon atom depends on the number of sigma bonds and pi bonds it forms.Explanation:The given compound is $${}^{1}\mathrm{CH}_{2}={}^{2}\mathrm{CH}-{}^{3}\mathrm{CH}_{2}-{}^{4}\mathrm{CH}_{2}-{}^{5}\mathrm{C}\equiv{}^{6}\mathrm{CH}$$Carbon $C_2$ forms a double bond, so it has one $\pi$ bond and three $\sigma$ bonds. Hence, $C_2$ is $sp^2$ hybridised.Carbon $C_3$ forms only single bonds with $C_2$, $C_4$, and two hydrogen atoms. Thus, $C_3$ has four $\sigma$ bonds and is $sp^3$ hybridised.Therefore, the $C_2-C_3$ bond is formed between an $sp^2$ carbon and an $sp^3$ carbon.Answer:The $C_2-C_3$ bond is of type $sp^2-sp^3$, which is Option D.