chemical-bonding

# 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 | Tr