Question

The structural formula of a compound is ${CH}_3-CH=C={CH}_2$. The type of hybridization at the four carbon atoms from left to right is:

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

Answer 1

Answer: (C)

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

The first carbon is bonded to second $C$ and three $H$ atoms all by single bonds. It has tetrahedral geometry and thus, has ${sp}^3$ hybridisation.

The second and fourth $C$ are bonded by two single bonds and one double bond and thus, have trigonal planar geometry resulting in ${sp}^2$ hybridisation.

The third $C$ is constraint to move in any direction due to double bonds on either side and hence, has a linear geometry. Thus, third $C$ is $sp$ hybridised.

Option C is correct.