Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Hybridization is the conceptual process of mixing atomic orbitals (such as and orbitals) to form new hybrid orbitals that are degenerate (equal in energy) and suitable for the pairing of electrons to form chemical bonds.
Hybridization: Formed by mixing one and three orbitals. This results in four hybrid orbitals arranged in a tetrahedral geometry with bond angles of approximately (e.g., in or ).
Hybridization: Formed by mixing one and two orbitals. This results in three hybrid orbitals arranged in a trigonal planar geometry with bond angles of . One unhybridized orbital remains perpendicular to the plane (e.g., in or ).
Hybridization: Formed by mixing one and one orbital. This results in two hybrid orbitals arranged in a linear geometry with bond angles of . Two unhybridized orbitals remain (e.g., in or ).
Sigma () bonds: These are formed by the head-on overlap of atomic or hybrid orbitals along the internuclear axis. All single covalent bonds are bonds.
Pi () bonds: These are formed by the sideways overlap of unhybridized orbitals. A double bond consists of one and one bond; a triple bond consists of one and two bonds.
The hybridization of an atom can be predicted by the number of electron domains: domains , domains , domains .
📐Formulae
💡Examples
Problem 1:
Identify the hybridization of the carbon atoms in ethene () and describe the bonding involved.
Solution:
Each carbon atom in is bonded to two hydrogen atoms and one carbon atom (via a double bond). This gives each carbon electron domains. Therefore, the hybridization is .
Explanation:
In hybridization, each carbon forms three bonds using its hybrid orbitals. The remaining unhybridized orbital on each carbon atom overlaps sideways to form one bond. Thus, the double bond consists of one bond and one bond.
Problem 2:
Determine the hybridization of the central nitrogen atom in the nitrate ion ().
Solution:
In , the central nitrogen atom has resonance structures, but in each, it is surrounded by electron domains (bonding regions to three oxygen atoms, with no lone pairs on ). Thus, the hybridization is .
Explanation:
Since there are electron domains around the nitrogen, it requires three hybrid orbitals. Mixing one and two orbitals gives three orbitals, resulting in a trigonal planar electron domain geometry.
Problem 3:
Calculate the formal charge of Carbon in the molecule ().
Solution:
Explanation:
Carbon () has zero lone pair electrons () and is sharing 8 electrons in four bonds (, two double bonds). A formal charge of indicates a stable distribution of electrons.