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Hybridization of l and 2 carbon atoms in
$\overset{1}{{CH}_{2}} = \overset{2}{C} = {CH}_{2}$

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${sp}^{3}, {sp}^{2}$

${sp}^{2}, sp$

${sp}^{2}, {sp}^{2}$

$sp, sp$

answer is C.

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Detailed Solution

$\overset{1}{\underset{{sp}^{2}}{{CH}_{2}}} = \overset{2}{\underset{sp}{C}} = {CH}_{2} .$

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