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In the compound
$\overset{}{C} H_{2} = \overset{}{C} H - \overset{}{C} H_{2} - \overset{}{C} H_{2} - \overset{}{C} \equiv \overset{}{C} H the C_{2} - C_{3} bond is of$

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$sp - {sp}^{2}$

${sp}^{2} - {sp}^{3}$

$sp - {sp}^{3}$

${sp}^{3} - {sp}^{3}$

answer is B.

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Detailed Solution

The correct numbering of parent chain is

$\overset{1}{C} H_{2} = \overset{2}{C} H - \overset{3}{C} H_{2} - \overset{4}{C} H_{2} - \overset{5}{C} \equiv \overset{6}{C} H$

Double and triple bond are equidistant from terminals, numbering is done in alphabetical order, giving lower number to ‘ene.’

Hence, $C_{2} - C_{3} are {sp}^{2} - {sp}^{3} hybridised .$

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