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Chemical Bonding and Structure - Further aspects of covalent bonding and structure (HL only)

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Sigma (σ\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 (π\pi) bonds are formed by the sideways overlap of parallel pp orbitals, resulting in electron density above and below the internuclear axis. A double bond consists of one σ\sigma and one π\pi bond, while a triple bond consists of one σ\sigma and two π\pi bonds.

Formal Charge (FCFC) 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 ss and pp) to form new hybrid orbitals (spsp, sp2sp^2, sp3sp^3) that are degenerate (equal in energy).

sp3sp^3 hybridization results in tetrahedral geometry (109.5109.5^{\circ}), sp2sp^2 results in trigonal planar geometry (120120^{\circ}), and spsp results in linear geometry (180180^{\circ}).

Delocalization occurs when π\pi 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 C6H6C_6H_6 or O3O_3).

The catalytic depletion of ozone (O3O_3) by NOxNO_x and CFCs involves the breaking of weaker bonds in ozone by UV radiation. Oxygen (O2O_2) requires higher energy UV-C (shorter wavelength) to break its double bond compared to the resonance-stabilized 1.51.5 bond order in ozone, which is broken by UV-B.

📐Formulae

FC=V(L+12B)FC = V - (L + \frac{1}{2}B) (where VV is valence electrons, LL is non-bonding electrons, and BB is bonding electrons)

Bond Order=Total number of bonding pairsTotal number of resonance positions\text{Bond Order} = \frac{\text{Total number of bonding pairs}}{\text{Total number of resonance positions}}

E=hcλE = \frac{hc}{\lambda} (Energy of photon required to break bonds)

💡Examples

Problem 1:

Calculate the formal charge for each atom in the ozone (O3O_3) molecule for the resonance structure: O(1)=O(2)O(3)O(1)=O(2)-O(3).

Solution:

For the central Oxygen (O2O2): V=6,L=2,B=6V=6, L=2, B=6. FC=6(2+12(6))=+1FC = 6 - (2 + \frac{1}{2}(6)) = +1. For the double-bonded terminal Oxygen (O1O1): V=6,L=4,B=4V=6, L=4, B=4. FC=6(4+12(4))=0FC = 6 - (4 + \frac{1}{2}(4)) = 0. For the single-bonded terminal Oxygen (O3O3): V=6,L=6,B=2V=6, L=6, B=2. FC=6(6+12(2))=1FC = 6 - (6 + \frac{1}{2}(2)) = -1.

Explanation:

The sum of formal charges is 0+11=00 + 1 - 1 = 0, 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 (C2H4C_2H_4).

Solution:

Each Carbon atom undergoes sp2sp^2 hybridization. This creates three sp2sp^2 hybrid orbitals and one unhybridized pp orbital. The CHC-H bonds are σ\sigma bonds formed by sp2ssp^2-s overlap. The CCC-C bond consists of one σ\sigma bond (sp2sp2sp^2-sp^2 overlap) and one π\pi bond (sideways ppp-p overlap).

Explanation:

The sp2sp^2 hybridization results in a trigonal planar geometry around each carbon atom with bond angles of approximately 120120^{\circ}.

Further aspects of covalent bonding and structure (HL only) Revision - Grade 12 Chemistry IB