Question

In the hydrocarbon $\underset{\left(1\right)}{C}{H}_2=\underset{\left(2\right)}{C}H-\underset{\left(3\right)}{C}\equiv CH$, the state of hybridisation of carbons 1, 2 and 3 respectively are:

• A. $s{p}^3,s{p}^2,sp$
• B. $s{p}^2,s{p}^2,sp$
• C. $s{p}^2,sp,s{p}^3$
• D. $s{p}^2,s{p}^3,sp$

Answer 1

The correct option is B $s{p}^2,s{p}^2,sp$

The hydrocarbon given is: $\underset{\left(1\right)}{C}{H}_2=\underset{\left(2\right)}{C}H-\underset{\left(3\right)}{C}\equiv CH$

The $C_1$ carbon is bonded to 2 $H$ atoms and 1 $C$ atom. Thus, a total of 3 sigma bonds and no lone pairs on carbon. Hence, it is $s{p}^2$ hybridised.

The $C_2$ carbon is bonded to 2 $C$ atoms and 1 $H$ atom. Thus, a total of 3 sigma bonds and no lone pairs on carbon. Hence, it is also $s{p}^2$ hybridised.

The $C_3$ carbon is bonded to 2 $C$ atoms only. Thus, a total of 2 sigma bonds and no lone pairs on carbon. Hence it is $sp$ hybridised.