How many different types of hybridization are present in the compound, $H C \equiv C - C H = {C H}_{2}$ ?

Open in App

Solution

$H C \equiv C - C H = {C H}_{2}$
$s p s p {s p}^{2} {s p}^{2}$
Hence, 2 types of hybridisation are present in the compound, $H C \equiv C - C H = C H_{2}$ .

Suggest Corrections

Similar questions

H_{3} C - C \equiv C - {C H}_{3}

How many $σ$ and $π$ bonds are presents in each of the following molecule?

(a) $H C \equiv C - C H = C H - C H_{3}$ (b) $C H_{2} = C = C H - C H_{3}$

σ

π

H C \equiv C - C H = C H - {C H}_{3}

C - C

H C \equiv C - C H = C H_{2}

In the compound $CH_2 = CH - CH_2 - CH_2 - C \equiv CH$, the hybridization of $C_3$ is:

Join BYJU'S Learning Program