The hybridization of carbon atoms in ${\mathrm{C}}_2-{\mathrm{C}}_3$ single bond of  $\stackrel{4}{\mathrm{HC}}\equiv \stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{CH}}=\stackrel{1}{{\mathrm{CH}}_2}$ is :-

The correct answer is option (C)

Therefore,

  The hybridization of carbon atoms in the  $\mathrm{C}-\mathrm{C}$ single bond of $\mathrm{HC}=\mathrm{C}-\mathrm{CH}={\mathrm{CH}}_2$ because the right side of carbon having another side triple bond so it is sp hybridized whereas left side carbon having double bond $sp$ it is $s{p}^2$ hybridized.

$\stackrel{4}{\underset{sp}{\mathrm{HC}}}=\stackrel{3}{\underset{sp}{\mathrm{C}}}-\stackrel{2}{\underset{s{p}^{\mathit{2}}}{\mathrm{CH}}}=\stackrel{1}{\underset{s{p}^{\mathit{2}}}{{\mathrm{CH}}_2}}$