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

Grade 11IBChemistry

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

🔑Concepts

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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.

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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.

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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.

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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).

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sp3sp^3 hybridization results in tetrahedral geometry (109.5∘109.5^{\circ}), sp2sp^2 results in trigonal planar geometry (120∘120^{\circ}), and spsp results in linear geometry (180∘180^{\circ}).

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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).

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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+1−1=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 C−HC-H bonds are σ\sigma bonds formed by sp2−ssp^2-s overlap. The C−CC-C bond consists of one σ\sigma bond (sp2−sp2sp^2-sp^2 overlap) and one π\pi bond (sideways p−pp-p overlap).

Explanation:

The sp2sp^2 hybridization results in a trigonal planar geometry around each carbon atom with bond angles of approximately 120∘120^{\circ}.