Question

The cylindrical shape of an alkyne is due to

• A. three sigma C-C bonds
• B. three $π$C-C bonds
• C. two sigma C-C and one $π$ C-C bonds
• D. one sigma C-C and two $π$ C-C bonds

Answer 1

The Correct Option is D

The question asks about the structure of alkynes and the hybridization of bonds that gives alkynes their characteristic shape.

Alkynes are hydrocarbons that contain a carbon-carbon triple bond. The structure of alkynes can be understood by analyzing the nature of bonds in the triple bond. A carbon-carbon triple bond consists of:

1. One sigma (σ) bond
2. Two pi (π) bonds

The sigma bond is formed by the head-on overlap of orbitals, while the pi bonds are formed by the side-on overlap of p orbitals.

The hybridization of the carbon atoms involved in the triple bond is sp. In sp hybridization, one s orbital and one p orbital mix to form two equivalent sp hybrid orbitals that are linearly arranged at 180 degrees. The remaining two p orbitals participate in forming the two pi bonds.

The linear arrangement of the sp hybrid orbitals and the presence of this type of orbitals lead to the linear or cylindrical geometry around the carbon atoms involved in the triple bond.

Let's analyze the options:

• three sigma C-C bonds
  : This is incorrect because an alkyne does not have three sigma bonds; it has one sigma bond and two pi bonds in a triple bond.
• three $π$C-C bonds
  : This is incorrect because no such molecular geometry with three pi bonds exists for a stable main group hydrocarbon such as alkyne.
• two sigma C-C and one $π$ C-C bonds
  : This is incorrect because a triple bond in an alkyne is comprised of one sigma and two pi bonds, not vice versa.
• one sigma C-C and two $π$ C-C bonds
  : This is correct because a carbon-carbon triple bond consists of one sigma bond and two pi bonds.

Thus, the cylindrical shape of an alkyne is due to one sigma C-C and two $π$ C-C bonds.