Final answer:
The solution to the "sp" hybridization dilemma is understanding that carbon atoms involved in a triple bond, like in ethyne, are sp hybridized with linear geometry, while a double bond results in sp² hybridization with trigonal planar geometry.
Step-by-step explanation:
The solution to the dilemma of equivalency in "sp" hybridization involves understanding how carbon atoms can have different hybrid states depending on the types of bonds they form. In the context of organic molecules, when a carbon atom forms a triple bond, as in ethyne (HC=CH), the hybridization is described as sp. This process involves the mixing of one s orbital with one p orbital to create two sp hybrid orbitals. These orbitals align at 180° from each other, resulting in a linear geometry.
The remaining two p orbitals, which do not participate in the hybridization, are available for forming pi bonds perpendicular to the axis of the sp hybrid orbitals, contributing to the multiple bonds with adjacent carbon atoms. Notably, in compounds like ethene or other alkenes, carbon atoms connected by a double bond exhibit sp² hybridization, where one s orbital and two p orbitals mix to form three sp² hybrid orbitals in a trigonal planar arrangement at 120° angles.