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 bonds form from head-on orbital overlap; pi bonds form from side-by-side p orbital overlap.

Worked example. Determine the hybridization of each carbon in CH3CHO (acetaldehyde).

The CH3 carbon has 4 electron domains (3 C-H bonds + 1 C-C bond) = sp3 hybridized, tetrahedral.

The CHO carbon has 3 electron domains (1 C-H bond + 1 C-C bond + 1 C=O double bond; the double bond counts as 1 domain) = sp2 hybridized, trigonal planar.

Molecular Polarity

A molecule is polar if (1) it contains polar bonds AND (2) the bond dipoles do not cancel by symmetry.

Worked example. Is CO2 polar?

Each C=O bond is polar (electronegativity difference = 1.0). But CO2 is linear with identical bonds pointing in opposite directions. The dipoles cancel. CO2 is nonpolar.

Worked example. Is H2O polar?

Each O-H bond is polar. H2O is bent (104.5 deg), so the dipoles do not cancel. The net dipole points from the H atoms toward the O atom. H2O is polar (dipole moment = 1.85 D).

Symmetry test. Molecules with identical surrounding atoms and no lone pairs on the central atom (CO2, BF3, CH4, SF6, XeF4) are typically nonpolar even with polar bonds. Lone pairs on the central atom usually break symmetry and create polarity (NH3, H2O, SO2).

Intermolecular Forces

Intermolecular forces (IMFs) act between molecules. They are much weaker than intramolecular bonds but determine physical properties: boiling point, melting point, viscosity, surface tension, and solubility.

The Four Types (Weakest to Strongest)

1. London Dispersion Forces (LDF). Present in ALL molecules. Arise from instantaneous dipoles created by electron motion. Strength increases with molar mass and surface area (more electrons = more polarizable). These are the ONLY forces in nonpolar molecules (He, N2, CH4).

2. Dipole-Dipole Forces. Between polar molecules. The positive end of one molecule attracts the negative end of another. Strength increases with dipole moment.

3. Hydrogen Bonding. Special strong case of dipole-dipole. Requires H bonded to F, O, or N (the three most electronegative small atoms) interacting with a lone pair on F, O, or N of a neighboring molecule. Approximately 5-30 kJ/mol — about 10x a typical dipole-dipole interaction.

4. Ion-Dipole Forces. Between ions and polar molecules. Strongest IMF type. Responsible for the dissolution of ionic compounds in water (hydration of ions).

Worked Example: Ranking Boiling Points

Problem. Rank the following by increasing boiling point: CH4, CH3OH, CH3CH3, CH3Cl.

Analysis.

• CH4 (16 g/mol): Nonpolar. Only LDF. Lowest BP.
• CH3CH3 (30 g/mol): Nonpolar. Only LDF. Slightly higher than CH4 (more electrons).
• CH3Cl (50.5 g/mol): Polar. LDF + dipole-dipole. Higher than CH3CH3.
• CH3OH (32 g/mol): Polar with O-H bond. LDF + dipole-dipole + hydrogen bonding. Highest BP despite lower molar mass than CH3Cl.

Order: CH4 (-161 C) < CH3CH3 (-89 C) < CH3Cl (-24 C) < CH3OH (65 C).

The hydrogen bonding in methanol overcomes its lower molar mass — this is the signature anomaly of H-bonding.

Worked Example: Why Ice Floats

Water's hydrogen bonding network forces molecules into an open hexagonal lattice when frozen. This lattice has more space between molecules than liquid water, making ice less dense than liquid water (0.917 g/mL vs. 1.000 g/mL). This is anomalous — most solids are denser than their liquids. The ecological consequence is enormous: ice floats, insulating bodies of water from above and allowing aquatic life to survive winter.

Metallic Bonding

Electron sea model. Metal atoms release valence electrons into a delocalized "sea" shared by all atoms in the lattice. The positive metal ion cores sit in this sea.

Properties explained: Electrical conductivity (electrons flow freely), thermal conductivity (electrons transfer kinetic energy), malleability/ductility (layers slide without breaking bonds — the electron sea reforms around shifted atoms), luster (free electrons absorb and re-emit photons).

Band theory. A more complete model — overlapping atomic orbitals form continuous energy bands. Metals have partially filled bands allowing electron mobility. Insulators have a large band gap. Semiconductors have a small band gap bridgeable by thermal energy or doping.

Common Mistakes

Mistake	Why it fails	Fix
Expanded octets for period 2	C, N, O, F have no accessible d orbitals	Only period 3+ elements can exceed 8
Confusing electron geometry with molecular geometry	Lone pairs affect shape but are "invisible"	Count ALL domains for electron geometry; show only atoms for molecular
Forgetting lone pairs in VSEPR count	Undercounting domains gives wrong geometry	Always draw the Lewis structure first
Calling CO2 polar	Symmetric cancellation of dipoles	Linear + identical bonds = nonpolar
Hydrogen bond to any H	H-bonding requires H-F, H-O, or H-N	H-C bonds do NOT form hydrogen bonds
Ignoring LDF for polar molecules	Polar molecules still have LDF too	LDF is always present; other forces add on top

Cross-References

• pauling agent: Bonding analysis, molecular geometry, hybridization. Primary agent for this skill.
• atomic-structure skill: Electron configurations determine bonding capacity and electronegativity.
• reactions-stoichiometry skill: Bond energies feed into enthalpy calculations (Hess's law via bond energies).
• organic-chemistry skill: Carbon bonding, hybridization, and functional groups build directly on covalent bonding concepts.
• materials-chemistry skill: Intermolecular forces determine phase behavior and material properties.

References

• Pauling, L. (1960). The Nature of the Chemical Bond. 3rd edition. Cornell University Press.
• Gillespie, R. J. & Hargittai, I. (2012). The VSEPR Model of Molecular Geometry. Dover Publications.
• Zumdahl, S. S. & Zumdahl, S. A. (2017). Chemistry. 10th edition. Cengage Learning.
• Atkins, P. & de Paula, J. (2014). Atkins' Physical Chemistry. 10th edition. Oxford University Press.
• Lewis, G. N. (1916). "The Atom and the Molecule." Journal of the American Chemical Society, 38(4), 762-785.
• Pauling, L. (1931). "The Nature of the Chemical Bond." Journal of the American Chemical Society, 53(4), 1367-1400.