JAMB Chemistry 2019 Paper

Concept:A carbon atom's hybridization is determined by the number of sigma bonds and lone pairs around it.Four sigma bonds give $sp^3$, while two sigma bonds indicate $sp$ hybridization.Explanation:The given compound is a terminal alkyne represented as $\mathrm{CH_3-CH_2-C\equiv CH}$.The $\mathrm{CH_3}$ carbon has four single bonds, so it is $sp^3$ hybridized.The $\mathrm{CH_2}$ carbon also forms four single bonds, so it is $sp^3$ hybridized.The two carbons involved in the triple bond each form only two sigma bonds: one to the neighbouring carbon and one to the other atom on the other side.Therefore, each of these triple-bonded carbons has two sigma bonds and is $sp$ hybridized.Hence, the molecule contains both $sp^3$ and $sp$ hybridized carbons.Answer:A. $sp^3$ and $sp$