In the organic compound CH2=CH–CH2–CH2–C≡CH, the pair of hydridised orbitals involved in the formation
In the organic compound $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C}=\mathrm{CH}$, the pair of hydridised orbitals involved in the formation of: $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond is:
(a) $s p-s p^{2}$ (b) $s p-s p^{3}$ (c) $s p^{2}-s p^{3}$ (d) $s p^{3}-s p^{3}$
$\mathrm{C} \stackrel{6}{\mathrm{H}}_{2}=\stackrel{5}{\mathrm{C}} \mathrm{H}-\mathrm{C} \stackrel{4}{\mathrm{H}}_{2}-\mathrm{C} \stackrel{3}{\mathrm{H}}_{2}-\stackrel{2}{\mathrm{C}} \equiv \stackrel{1}{\mathrm{C}} \mathrm{H}$
In the given organic compound, the carbon atoms numbered as $1,2,3,4,5$, and 6 are $s p, s p, s p^{3}, s p^{3}, s p^{2}$, and $s p^{2}$ hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond is $s p-s p^{3}$.