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Chemical Bonding and Structure - Covalent bonding

Grade 11IBChemistry

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

🔑Concepts

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

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Bond polarity is determined by the difference in electronegativity (Δχ\Delta \chi) between atoms. A bond is considered polar covalent if 0.5<Δχ<1.70.5 < \Delta \chi < 1.7.

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Bond strength (bond enthalpy) increases and bond length decreases as the number of shared electron pairs increases (e.g., triple bonds are shorter and stronger than single bonds).

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The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry. Electron domains (bonding and non-bonding pairs) arrange themselves to minimize repulsion, following the order: lone pair−lone pair>lone pair−bonding pair>bonding pair−bonding pairlone\ pair-lone\ pair > lone\ pair-bonding\ pair > bonding\ pair-bonding\ pair.

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A coordinate covalent (dative) bond is a type of covalent bond where both electrons in the shared pair originate from the same atom, often represented by an arrow (e.g., in NH4+NH_4^+ or COCO).

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Resonance occurs when more than one valid Lewis structure can be drawn for a molecule. The actual structure is a resonance hybrid with delocalized electrons, such as in O3O_3, C6H6C_6H_6, and CO32−CO_3^{2-}.

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Giant covalent structures, such as diamond (CC), graphite (CC), and silicon dioxide (SiO2SiO_2), involve atoms linked by covalent bonds in a continuous network, resulting in high melting points.

📐Formulae

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

Δχ=∣χA−χB∣\Delta \chi = |\chi_A - \chi_B| (Electronegativity difference determines bond character)

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}}

💡Examples

Problem 1:

Predict the molecular geometry and the bond angle of the NH3NH_3 molecule using VSEPR theory.

Solution:

The central Nitrogen atom has 5 valence electrons. It forms 3 single bonds with Hydrogen and has 1 lone pair. This results in 4 electron domains (tetrahedral arrangement). Due to the lone pair, the molecular geometry is trigonal pyramidal. The bond angle is approximately 107∘107^\circ.

Explanation:

Nitrogen in NH3NH_3 has 3 bonding pairs and 1 lone pair. The repulsion from the lone pair reduces the ideal tetrahedral angle of 109.5∘109.5^\circ to 107∘107^\circ.

Problem 2:

Calculate the formal charge of the Carbon atom in the Carbon Monoxide (COCO) molecule, given the Lewis structure consists of a triple bond :C≡O::C \equiv O:.

Solution:

FC(C)=4−(2+12(6))=4−5=−1FC(C) = 4 - (2 + \frac{1}{2}(6)) = 4 - 5 = -1

Explanation:

Carbon has 4 valence electrons (V=4V=4). In :C≡O::C \equiv O:, it has 2 non-bonding electrons (N=2N=2) and 6 bonding electrons (B=6B=6) from the triple bond. The resulting formal charge is −1-1.

Problem 3:

Explain why the bond lengths in the Nitrate ion (NO3−NO_3^-) are all equal.

Solution:

The NO3−NO_3^- ion exhibits resonance. It can be represented by three equivalent Lewis structures where the double bond shifts between the three Oxygen atoms.

Explanation:

Because the electrons are delocalized across the N−ON-O bonds, each bond has an identical bond order of 1.331.33 (calculated as 4 bonds3 positions\frac{4\text{ bonds}}{3\text{ positions}}). This results in bond lengths that are intermediate between a single N−ON-O bond and a double N=ON=O bond.

Covalent bonding Grade 11 Notes & Examples | IB Chemistry