Advanced Placement (AP) Chemistry Practice Exam

What hybridization is used to describe a molecule with four bonding pairs and no lone pairs?

1. sp hybridization

2. sp2 hybridization

3. sp3 hybridization

4. sp4 hybridization

The correct answer is: sp3 hybridization

The correct hybridization for a molecule with four bonding pairs and no lone pairs is sp3 hybridization. This type of hybridization involves the mixing of one s orbital and three p orbitals from the valence shell of the central atom, which results in four equivalent hybrid orbitals. These four sp3 hybrid orbitals are arranged in a tetrahedral geometry, which is ideal for maximizing the distance between the bonding pairs, thereby minimizing electron pair repulsion according to VSEPR (Valence Shell Electron Pair Repulsion) theory. This arrangement allows for bond angles of approximately 109.5°, characteristic of tetrahedral molecules. It's important to note that other hybridization states, such as sp and sp2, involve fewer hybrid orbitals and don’t accommodate four bonding pairs in their geometry. In sp hybridization, there are only two bonding pairs with a linear arrangement (180° angle), and sp2 hybridization involves three bonding pairs in a trigonal planar arrangement (120° angle). The sp4 hybridization, while not commonly referenced in standard hybridization terminology, does not apply here since it would imply the mixing of four orbitals to accommodate five bonding pairs, which exceeds the requirements of this specific scenario. Thus, the presence of four