In which of the following molecules is the order of hybridisation $s{p}^2$, $s{p}^2$, $sp$, $sp$ for the carbon atoms from left to right?

Solution & Explanation

\textbf{Answer:} (b) \textbf{Solution:} The hybridisation of a carbon = number of sigma bonds + lone pairs around it. In $\mathrm{CH}{}_2=\mathrm{CH}-\mathrm{C}\equiv \mathrm{CH}$ the four carbons (left to right) are $s{p}^2$ ($=\mathrm{CH}{}_2$), $s{p}^2$ ($=\mathrm{CH}{}^{-}$), $sp$, $sp$ ($\equiv \mathrm{CH}$), matching the required sequence $s{p}^2$, $s{p}^2$, $sp$, $sp$. Option (a) is $s{p}^2$,$s{p}^2$,$s{p}^2$,$s{p}^2$; (c) is $sp$,$sp$,$sp$,$sp$; (d) is $s{p}^3$,$s{p}^2$,$s{p}^2$,$s{p}^3$. Hence (b).