The correct answer is sp hybridization.

To determine the hybridization of carbon atoms in acetylene (C₂H₂), we need to consider its Lewis structure and apply the concept of valence bond theory. Each carbon atom in acetylene forms two sigma bonds (one with the other carbon atom and one with a hydrogen atom) and two pi bonds (with the other carbon atom).

Hybridization is determined by the number of sigma bonds and lone pairs around an atom. In this case, each carbon atom in acetylene has two sigma bonds and no lone pairs. According to the hybridization rules, two sigma bonds correspond to sp hybridization.

Therefore, each carbon atom in acetylene is sp hybridized, meaning it uses one s and one p orbital to form two hybrid orbitals. These sp hybrid orbitals participate in the formation of the two sigma bonds. The remaining two unhybridized p orbitals on each carbon atom overlap side-by-side to form two pi bonds, resulting in the triple bond between the two carbon atoms. This linear structure of acetylene directly reflects the sp hybridization.

The other options are incorrect because:

• sp²d hybridization requires more than 4 bonds to be formed around the central atom.
• sp² hybridization forms three sigma bonds.
• sp³ hybridization forms four sigma bonds.

Understanding hybridization is crucial for predicting molecular geometry and properties. The linear shape of acetylene is a direct consequence of the sp hybridization of its carbon atoms.