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Structure 2. Models of bonding and structure - The covalent model

Grade 12IBChemistry

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 bonded atoms.

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The Octet Rule states that atoms tend to gain, lose, or share electrons to achieve a stable configuration of eight valence electrons, similar to a noble gas. Exceptions include hydrogen (H2H_{2}), which requires only 2 electrons, and elements in Period 3 or below that can expand their octet.

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Bond polarity is determined by the difference in electronegativity (Δχ\Delta \chi) between atoms. If 0.0<Δχ<0.50.0 < \Delta \chi < 0.5, the bond is non-polar covalent; if 0.5≤Δχ<1.80.5 \le \Delta \chi < 1.8, the bond is polar covalent.

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Bond length and bond strength: As the number of shared electron pairs increases (single to double to triple), bond length decreases and bond enthalpy (strength) increases.

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Coordinate (Dative) Covalent Bonding occurs when one atom provides both electrons for the shared pair. This is represented by an arrow (→\rightarrow) in Lewis structures.

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VSEPR Theory (Valence Shell Electron Pair Repulsion) predicts molecular geometry based on minimizing repulsion between electron domains. Total domains determine the electron domain geometry: 2 = Linear (180∘180^{\circ}), 3 = Trigonal Planar (120∘120^{\circ}), 4 = Tetrahedral (109.5∘109.5^{\circ}).

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Resonance occurs when more than one valid Lewis structure can be drawn for a molecule (e.g., O3O_{3}, CO32−CO_{3}^{2-}, C6H6C_{6}H_{6}). The actual structure is a resonance hybrid with delocalized electrons.

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Giant covalent structures (Network solids) like Diamond, Graphite, and Silicon Dioxide (SiO2SiO_{2}) have high melting points and hardness because atoms are linked by a continuous network of strong covalent bonds throughout the entire crystal.

📐Formulae

FC=V−L−12BFC = 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 bonding positions\text{Bond Order} = \frac{\text{Total number of bonding pairs}}{\text{Total number of bonding positions}}

Δχ=∣χA−χB∣\Delta \chi = |\chi_{A} - \chi_{B}|

💡Examples

Problem 1:

Predict the molecular geometry and bond angle for the Hydronium ion, H3O+H_{3}O^{+}.

Solution:

  1. Count valence electrons: Oxygen (6) + 3 Hydrogen (3×13 \times 1) - 1 (positive charge) = 88 electrons.
  2. Draw Lewis structure: Oxygen is the central atom with 3 single bonds to Hydrogen and 1 lone pair.
  3. Identify electron domains: 3 bonding pairs + 1 lone pair = 4 total domains.
  4. Electron domain geometry: Tetrahedral.
  5. Molecular geometry: Trigonal Pyramidal.
  6. Predicted bond angle: Approximately 107∘107^{\circ} (less than 109.5∘109.5^{\circ} due to lone pair-bonding pair repulsion).

Explanation:

The VSEPR theory states that lone pairs exert more repulsion than bonding pairs, compressing the bond angles between the O−HO-H bonds.

Problem 2:

Calculate the formal charge for each atom in the Carbon Dioxide molecule (O=C=OO=C=O).

Solution:

For Carbon: V=4,L=0,B=8V=4, L=0, B=8. FCC=4−0−12(8)=0FC_{C} = 4 - 0 - \frac{1}{2}(8) = 0 For each Oxygen: V=6,L=4,B=4V=6, L=4, B=4. FCO=6−4−12(4)=0FC_{O} = 6 - 4 - \frac{1}{2}(4) = 0

Explanation:

Since the formal charge on all atoms is 0, this represents the most stable Lewis structure for CO2CO_{2}.