Chemical Bonding and Structure - Further aspects of covalent bonding and structure (HL only)
Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Sigma () bonds are formed by the head-on (axial) overlap of atomic orbitals, resulting in electron density concentrated between the nuclei of the bonding atoms.
Pi () bonds are formed by the sideways overlap of parallel orbitals, resulting in electron density above and below the internuclear axis. A double bond consists of one and one bond, while a triple bond consists of one and two bonds.
Formal Charge () is a tool used to determine the most stable Lewis structure. The preferred structure is the one where the formal charges on atoms are closest to zero.
Hybridization is the mixing of atomic orbitals (such as and ) to form new hybrid orbitals (, , ) that are degenerate (equal in energy).
hybridization results in tetrahedral geometry (), results in trigonal planar geometry (), and results in linear geometry ().
Delocalization occurs when electrons are shared by more than two nuclei, often represented by resonance structures. This leads to intermediate bond lengths and increased stability (e.g., in or ).
The catalytic depletion of ozone () by and CFCs involves the breaking of weaker bonds in ozone by UV radiation. Oxygen () requires higher energy UV-C (shorter wavelength) to break its double bond compared to the resonance-stabilized bond order in ozone, which is broken by UV-B.
📐Formulae
(where is valence electrons, is non-bonding electrons, and is bonding electrons)
(Energy of photon required to break bonds)
💡Examples
Problem 1:
Calculate the formal charge for each atom in the ozone () molecule for the resonance structure: .
Solution:
For the central Oxygen (): . . For the double-bonded terminal Oxygen (): . . For the single-bonded terminal Oxygen (): . .
Explanation:
The sum of formal charges is , which matches the neutral charge of the molecule. This distribution helps explain the reactivity of ozone.
Problem 2:
Describe the hybridization and bonding in ethene ().
Solution:
Each Carbon atom undergoes hybridization. This creates three hybrid orbitals and one unhybridized orbital. The bonds are bonds formed by overlap. The bond consists of one bond ( overlap) and one bond (sideways overlap).
Explanation:
The hybridization results in a trigonal planar geometry around each carbon atom with bond angles of approximately .