Question
In the organic compound $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2-\text{C}\equiv\text{CH}$, the pair of hybridised orbitals involved in the formation of $\mathrm{C}_2-\mathrm{C}_3$ bond is

Answer

When double and triple bonds are present at equivalent positions, then preference is given to double bond while numbering the carbon chain. Thus
$\stackrel{{1}}{\hbox{CH}_2}=\stackrel{{2}}{\hbox{CH}}-\stackrel{{3}}{\hbox{CH}_2}-\stackrel{{4}}{\hbox{CH}_2}-\stackrel{{5}}{\hbox{C}}\equiv\stackrel{{6}}{\hbox{CH}}\\ \text{sp}^2\ \ \ \ \ \ \text{sp}^2\ \ \ \ \ \ \ \text{sp}^3\ \ \ \ \ \ \ \text{sp}^3\ \ \ \ \ \ \ \text{sp}\ \ \ \text{sp}$
$\therefore \mathrm{C}_2-\mathrm{C}_3$ bond is formed by overlap of $\mathrm{sp}^2$ and $\mathrm{sp}^3$ orbitals.

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