Chemical Bonding
Atoms bond to achieve lower energy states. The type of bond — ionic, covalent, or metallic — depends on the electronegativity difference and metallic character of the atoms involved. Once bonds form, the three-dimensional arrangement of atoms determines molecular shape, polarity, and physical properties. This skill covers bond formation, Lewis structures, VSEPR geometry prediction, hybridization, and the intermolecular forces that govern bulk behavior.
Agent affinity: pauling (bonding/molecular chemistry, primary)
Concept IDs: chem-ionic-bonding, chem-covalent-bonding, chem-molecular-geometry, chem-intermolecular-forces
Bond Type Classification
| Bond type | Electronegativity difference | Electron behavior | Example |
|---|---|---|---|
| Nonpolar covalent | < 0.4 | Shared equally | H-H, Cl-Cl |
| Polar covalent | 0.4 - 1.7 | Shared unequally | H-Cl, O-H |
| Ionic | > 1.7 | Transferred | NaCl, MgO |
| Metallic | Between metals | Delocalized "sea" | Fe, Cu, Al |
These boundaries are guidelines, not sharp cutoffs. Bond character exists on a continuum.
Ionic Bonding
Mechanism. Metal atoms lose electrons to form cations; nonmetal atoms gain electrons to form anions. The electrostatic attraction between oppositely charged ions forms the ionic bond.
Lattice energy. The energy released when gaseous ions assemble into a crystal lattice. Higher lattice energy means a more stable compound. Lattice energy increases with higher ion charges and smaller ion radii (Coulomb's law: E proportional to q1*q2/r).
Worked example. Predict the formula of the compound formed by aluminum and oxygen.
Aluminum (Group 3) loses 3 electrons: Al^3+. Oxygen (Group 16) gains 2 electrons: O^2-. To balance charges: 2(Al^3+) + 3(O^2-) gives total charge = 2(+3) + 3(-2) = 0. Formula: Al2O3.
Properties of ionic compounds. High melting points, brittle, conduct electricity when molten or dissolved (ions free to move), do not conduct as solids (ions locked in lattice).
Covalent Bonding
Mechanism. Nonmetal atoms share electron pairs to achieve stable electron configurations. A single bond shares 2 electrons, a double bond shares 4, a triple bond shares 6.
Bond order, length, and energy relationship:
| Bond | Bond order | Approximate length (pm) | Approximate energy (kJ/mol) |
|---|---|---|---|
| C-C | 1 | 154 | 347 |
| C=C | 2 | 134 | 614 |
| C-triple-C | 3 | 120 | 839 |
Higher bond order means shorter, stronger bonds. This pattern holds across all elements.
Lewis Structures
Lewis structures show valence electrons as dots or lines (bonding pairs). The systematic procedure:
Step 1. Count total valence electrons. Add electrons for negative charges; subtract for positive charges.
Step 2. Connect atoms with single bonds. The least electronegative atom is usually central (H is always terminal).
Step 3. Distribute remaining electrons as lone pairs, completing octets on terminal atoms first, then the central atom.
Step 4. If the central atom lacks an octet, convert lone pairs on adjacent atoms to multiple bonds.
Step 5. Calculate formal charges: FC = (valence electrons) - (lone pair electrons) - (1/2 bonding electrons). Minimize formal charges; negative FC should be on more electronegative atoms.
Worked Example: Lewis Structure of CO2
Step 1. C has 4, each O has 6. Total: 4 + 6 + 6 = 16 valence electrons.
Step 2. O-C-O uses 4 electrons for two single bonds. Remaining: 12.
Step 3. Place 6 electrons (3 lone pairs) on each O: 12 used. Carbon has only 4 electrons around it — incomplete octet.
Step 4. Convert one lone pair from each O into a bonding pair: O=C=O. Carbon now has 8 electrons (two double bonds). Each O has 4 lone pair electrons + 4 bonding electrons = 8. All octets satisfied.
Step 5. Formal charges: C = 4 - 0 - 4 = 0. Each O = 6 - 4 - 2 = 0. All zero — optimal.
Worked Example: Lewis Structure of NO3- (Nitrate)
Step 1. N has 5, each O has 6, plus 1 for the negative charge. Total: 5 + 18 + 1 = 24.
Step 2. Three N-O single bonds use 6 electrons. Remaining: 18.
Step 3. Place 6 electrons on each O (18 total). N has only 6 electrons — needs 2 more.
Step 4. Convert one lone pair from one O to a double bond. N=O with two N-O. N now has 8 electrons.
Step 5. Formal charges: N = 5 - 0 - 4 = +1. Double-bonded O = 6 - 4 - 2 = 0. Each single-bonded O = 6 - 6 - 1 = -1. Total: +1 + 0 + (-1) + (-1) = -1. Correct.
Resonance. The double bond could be on any of the three O atoms. Three equivalent resonance structures exist. The true structure is a hybrid — each N-O bond has bond order 4/3.
Octet Rule Exceptions
| Exception type | Example | Explanation |
|---|---|---|
| Incomplete octet | BF3 (B has 6 e-) | Boron is electron-deficient; stable with 6 |
| Expanded octet | SF6 (S has 12 e-) | Period 3+ elements use d orbitals |
| Odd electron | NO (11 e- total) | Free radical — unpaired electron on N |
Critical rule. Only elements in period 3 or below can exceed the octet. Never draw expanded octets for C, N, O, or F.
VSEPR Theory
Valence Shell Electron Pair Repulsion: electron domains (bonding pairs and lone pairs) around a central atom arrange themselves to maximize separation, determining molecular geometry.
Electron Domain Count to Geometry
| Electron domains | Electron geometry | Bond angle(s) |
|---|---|---|
| 2 | Linear | 180 deg |
| 3 | Trigonal planar | 120 deg |
| 4 | Tetrahedral | 109.5 deg |
| 5 | Trigonal bipyramidal | 90 deg, 120 deg |
| 6 | Octahedral | 90 deg |
Key distinction. Electron geometry counts ALL domains (bonding + lone pairs). Molecular geometry describes only the atom positions. Lone pairs are "invisible" in molecular geometry but still affect shape.
VSEPR Decision Table
| Electron domains | Bonding pairs | Lone pairs | Molecular geometry | Example |
|---|---|---|---|---|
| 2 | 2 | 0 | Linear | CO2, BeCl2 |
| 3 | 3 | 0 | Trigonal planar | BF3 |
| 3 | 2 | 1 | Bent | SO2, O3 |
| 4 | 4 | 0 | Tetrahedral | CH4 |
| 4 | 3 | 1 | Trigonal pyramidal | NH3 |
| 4 | 2 | 2 | Bent | H2O |
| 5 | 5 | 0 | Trigonal bipyramidal | PCl5 |
| 5 | 4 | 1 | Seesaw | SF4 |
| 5 | 3 | 2 | T-shaped | ClF3 |
| 5 | 2 | 3 | Linear | XeF2 |
| 6 | 6 | 0 | Octahedral | SF6 |
| 6 | 5 | 1 | Square pyramidal | BrF5 |
| 6 | 4 | 2 | Square planar | XeF4 |
Worked Example: Molecular Geometry of Water (H2O)
Step 1. Lewis structure: O has 2 bonding pairs (to H) and 2 lone pairs. Total electron domains = 4.
Step 2. Electron geometry: tetrahedral (4 domains).
Step 3. Molecular geometry: bent (2 bonding pairs visible, 2 lone pairs hidden).
Step 4. Bond angle: approximately 104.5 deg (compressed from ideal 109.5 deg because lone pairs occupy more space than bonding pairs).
Worked Example: Molecular Geometry of XeF4
Step 1. Xe has 8 valence electrons, 4 F atoms contribute 4 bonds (8 electrons), leaving 2 lone pairs on Xe. Total electron domains = 6.
Step 2. Electron geometry: octahedral.
Step 3. Molecular geometry: square planar (lone pairs occupy opposing axial positions to minimize repulsion).
Step 4. Bond angles: 90 deg. The molecule is flat and symmetric.
Hybridization
Orbital hybridization explains observed geometries by mixing atomic orbitals:
| Hybrid | Orbitals mixed | Geometry | Angle | Example |
|---|---|---|---|---|
| sp | 1s + 1p | Linear | 180 deg | BeCl2, CO2 (C) |
| sp2 | 1s + 2p | Trigonal planar | 120 deg | BF3, C2H4 (each C) |
| sp3 | 1s + 3p | Tetrahedral | 109.5 deg | CH4, NH3, H2O |
| sp3d | 1s + 3p + 1d | Trigonal bipyramidal | 90/120 deg | PCl5 |
| sp3d2 | 1s + 3p + 2d | Octahedral | 90 deg | SF6 |
Shortcut. The number of hybrid orbitals equals the number of electron domains. 4 domains = sp3. 3 domains = sp2. 2 domains = sp.
Sigma and pi bonds. A single bond = 1 sigma bond. A double bond = 1 sigma + 1 pi. A triple bond = 1 sigma + 2 pi. Sigma bon