In the following compound :CH12=CH2−CH23−C4≡CH5 the C3−C4 bond is of the type :

In the following compound : ${\overset{1}{C H}}_{2} = \overset{2}{CH} - \overset{3}{{CH}_{2}} - \overset{4}{C} \equiv \overset{5}{CH}$ the $C_{3} - C_{4}$ bond is of the type :

A
$s p - s p^{2}$
B
$s p^{3} - s p^{3}$
C
$s p^{3} - s p$
D
$s p^{2} - s p^{3}$

Solution:

The hybridisation of the given compound can be represent as,

${\overset{1}{C H}}_{2} = \overset{2}{CH} - \overset{3}{{CH}_{2}} - \overset{4}{C} \equiv \overset{5}{CH}$

$s p^{2} s p^{2} s p^{3} s p s p$

Thus, The $C_{3} - C_{4}$ bond is of the type $s p^{3} - s p$ .

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