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Chemical Bonding and Structure - Covalent bonding

Grade 12IBChemistry

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

🔑Concepts

A covalent bond is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the atoms involved.

The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration of eight valence electrons, similar to a noble gas. Exceptions include incomplete octets (e.g., BeCl2BeCl_2, BF3BF_3) and expanded octets for elements in Period 3 and below (e.g., PCl5PCl_5, SF6SF_6).

Bond polarity is determined by the difference in electronegativity (Δχ\Delta\chi) between atoms. A polar covalent bond occurs when 0.5<Δχ<1.70.5 < \Delta\chi < 1.7, leading to a dipole moment represented by δ+\delta+ and δ\delta-.

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry based on minimizing repulsion between electron domains. Total domains include both bonding pairs and lone pairs.

Resonance occurs when more than one valid Lewis structure can be drawn for a molecule (e.g., O3O_3, C6H6C_6H_6, CO32CO_3^{2-}). The actual structure is a resonance hybrid with delocalized electrons.

Sigma (σ\sigma) bonds are formed by the head-on overlap of atomic orbitals along the internuclear axis. Pi (π\pi) bonds are formed by the sideways overlap of pp-orbitals above and below the internuclear axis.

Hybridization involves the mixing of atomic orbitals (such as ss and pp) to form new hybrid orbitals (sp,sp2,sp3sp, sp^2, sp^3) that are degenerate (equal in energy).

Giant covalent structures like diamond, graphite, graphene, and SiO2SiO_2 have high melting points and varying electrical conductivities based on electron delocalization.

📐Formulae

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

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

Δχ=χAχB\Delta\chi = |\chi_A - \chi_B| (Electronegativity difference)

💡Examples

Problem 1:

Predict the molecular geometry, electron domain geometry, and hybridization of the central atom in SF4SF_4.

Solution:

Electron domain geometry: Trigonal bipyramidal. Molecular geometry: Seesaw. Hybridization: sp3dsp^3d.

Explanation:

SS has 6 valence electrons. In SF4SF_4, it forms 4 single bonds with FF and has 1 lone pair. This results in 5 electron domains. According to VSEPR, 5 domains lead to a trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimize repulsion, resulting in a 'seesaw' molecular shape.

Problem 2:

Calculate the formal charge of the central oxygen atom in the ozone (O3O_3) molecule for the resonance structure O=O+OO=O^+-O^-.

Solution:

FC=6(2+12(6))=+1FC = 6 - (2 + \frac{1}{2}(6)) = +1

Explanation:

The central oxygen atom (V=6V=6) has one lone pair (2 electrons) and forms one double bond and one single bond (total 3 bonds, 6 bonding electrons). Using the formula FC=VL12BFC = V - L - \frac{1}{2}B, we get 623=+16 - 2 - 3 = +1.

Problem 3:

Identify the number of σ\sigma and π\pi bonds in ethyne (C2H2C_2H_2).

Solution:

3 σ\sigma bonds and 2 π\pi bonds.

Explanation:

The structure of ethyne is HCCHH-C\equiv C-H. Each single bond (CHC-H) is a σ\sigma bond (2 total). The triple bond between carbons consists of 1 σ\sigma bond and 2 π\pi bonds. Summing these gives 3 σ\sigma and 2 π\pi bonds.

Covalent bonding - Revision Notes & Key Formulas | IB Grade 12 Chemistry