The state of hybridisation of carbons marked 2, 3, 5 and 6 in the following hydrocarbon are, respectively: $\overset{6}{\text{CH}_3}-\overset{5}{\text{CH}}=\overset{4}{\text{CH}}-\overset{3}{\text{CH}_2}-\overset{2}{\text{C}}\equiv\overset{1}{\text{CH}}$

In the given hydrocarbon $\overset{6}{\text{CH}_3}-\overset{5}{\text{CH}}=\overset{4}{\text{CH}}-\overset{3}{\text{CH}_2}-\overset{2}{\text{C}}\equiv\overset{1}{\text{CH}}$:

• Carbon 2 ($\text{-C}\equiv$): Forms 2 $\sigma$ bonds and 2 $\pi$ bonds, so it is $sp$ hybridised.
• Carbon 3 ($\text{-CH}_2\text{-}$): Forms 4 $\sigma$ bonds, so it is $sp^3$ hybridised.
• Carbon 5 ($\text{-CH=}$): Forms 3 $\sigma$ bonds and 1 $\pi$ bond, so it is $sp^2$ hybridised.
• Carbon 6 ($\text{-CH}_3$): Forms 4 $\sigma$ bonds, so it is $sp^3$ hybridised. Therefore, the states of hybridisation for carbons 2, 3, 5 and 6 are $sp$, $sp^3$, $sp^2$, and $sp^3$ respectively.