JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Chemical Bonding and Molecular Structure

Hybridization of a carbon atom follows from the number of sigma-bond-forming regions around it: two regions (a triple bond, or two cumulated double bonds) give $sp$, three regions (one double bond) give $s{p}^2$, and four single bonds give $s{p}^3$.

In the given chain, the ketonic carbonyl carbon carries one $\pi$ bond to oxygen and is $s{p}^2$. Further along, an isolated $\mathrm{C}\equiv \mathrm{C}$ triple bond contributes two $sp$ carbons, and the terminal carbon bonded to nitrogen in the nitrile ($\mathrm{C}\equiv \mathrm{N}$) contributes a third $sp$ carbon. Two conjugated $\mathrm{CH}=\mathrm{CH}$ units between the alkyne and the nitrile contribute four more $s{p}^2$ carbons. The remaining $\mathrm{C}{\mathrm{H}}_2$ and $\mathrm{C}{\mathrm{H}}_3$ carbons, having only single bonds, stay $s{p}^3$ and are not counted.

Adding these up: $sp$ carbons $=2+1=3$, and $s{p}^2$ carbons $=1+4=5$. The molecule has 3 $sp$ and 5 $s{p}^2$ hybridized carbon atoms.