Chemical Bonding & Molecular Structure
Why atoms join and what shape the result takes — from Lewis dots and VSEPR to hybridization and molecular orbitals
- Why atoms bond — the octet rule, Lewis structures and formal charge.
- The difference between the ionic and covalent bond, and where Fajans' rules blur the line.
- Bond parameters — length, angle, enthalpy and order — and how resonance averages them.
- Predicting molecular shape with VSEPR, and explaining it with hybridization.
- Molecular orbital theory: bonding and antibonding levels, bond order and magnetism.
- Dipole moment, polarity, and the special role of the hydrogen bond.
Why Atoms Bond
Atoms bond because the joined arrangement is lower in energy than the separated atoms. The Kössel–Lewis view captured this as the octet rule: atoms gain, lose or share electrons to reach the stable eight-electron configuration of a noble gas (a duplet for hydrogen).
A Lewis (electron-dot) structure shows valence electrons as dots, a shared pair as a line. Formal charge tells us which of several valid structures is most plausible:
The best Lewis structure has formal charges closest to zero and any negative charge on the most electronegative atom. The octet rule has exceptions — incomplete octets \((\ce{BF3})\), expanded octets \((\ce{SF6})\) and odd-electron species \((\ce{NO})\).
The Ionic Bond
When a metal of low ionization energy meets a non-metal of high electron affinity, an electron transfers outright. The resulting cation and anion are held by electrostatic attraction in a giant crystal lattice. The energy released as gaseous ions assemble into one mole of solid is the lattice energy — the true measure of ionic-bond strength.
The Covalent Bond
When two atoms of similar electronegativity meet, neither wins the tug-of-war, so they share electron pairs. One shared pair is a single bond, two a double, three a triple. Where a single Lewis structure cannot capture the real molecule, two or more contributing structures are drawn and the molecule is a resonance hybrid of them.
The three \(\ce{C-O}\) bonds of \(\ce{CO3^2-}\) are identical, intermediate between single and double. No one Lewis structure shows this; the true ion is the average — the resonance hybrid — and is more stable than any single contributor (resonance energy).
Bond Parameters
Every covalent bond carries measurable quantities. Bond length is the equilibrium internuclear distance; bond angle is the angle between two bonds at an atom; bond enthalpy is the energy to break one mole of bonds in the gas phase; bond order is the number of bonds between two atoms.
| Bond | Order | Length (pm) | Enthalpy (kJ mol⁻¹) |
|---|---|---|---|
| \(\ce{C-C}\) | 1 | 154 | 348 |
| \(\ce{C=C}\) | 2 | 134 | 614 |
| \(\ce{C#C}\) | 3 | 120 | 839 |
VSEPR & the Shape of Molecules
Valence-shell electron-pair repulsion theory predicts geometry from one idea: electron pairs around the central atom arrange themselves to be as far apart as possible. Lone pairs repel more strongly than bonding pairs, so they compress the remaining bond angles.
| Bond + lone pairs | Electron geometry | Example shape |
|---|---|---|
| 2 + 0 | Linear | \(\ce{CO2}\) — linear |
| 3 + 0 | Trigonal planar | \(\ce{BF3}\) — trigonal planar |
| 4 + 0 | Tetrahedral | \(\ce{CH4}\) — tetrahedral |
| 3 + 1 | Tetrahedral | \(\ce{NH3}\) — pyramidal (107°) |
| 2 + 2 | Tetrahedral | \(\ce{H2O}\) — bent (104.5°) |
| 5 + 0 | Trigonal bipyramidal | \(\ce{PCl5}\) |
| 6 + 0 | Octahedral | \(\ce{SF6}\) |
Valence Bond Theory
VSEPR predicts shape but says nothing about how bonds form. Valence bond theory says a covalent bond is the overlap of two half-filled atomic orbitals; the greater the overlap, the stronger the bond.
Hybridization
Pure atomic orbitals cannot explain the equal bonds and observed angles of methane. The fix is hybridization: atomic orbitals of similar energy mix to form an equal number of identical hybrid orbitals pointing toward the bonded atoms. The number of hybrids equals the steric number (σ bonds + lone pairs).
| Hybridization | Steric number | Geometry | Example |
|---|---|---|---|
| \(sp\) | 2 | Linear | \(\ce{BeCl2},\ \ce{C2H2}\) |
| \(sp^2\) | 3 | Trigonal planar | \(\ce{BF3},\ \ce{C2H4}\) |
| \(sp^3\) | 4 | Tetrahedral | \(\ce{CH4},\ \ce{NH3},\ \ce{H2O}\) |
| \(sp^3d\) | 5 | Trigonal bipyramidal | \(\ce{PCl5}\) |
| \(sp^3d^2\) | 6 | Octahedral | \(\ce{SF6}\) |
Steric number 2, 3, 4, 5, 6 maps directly to \(sp,\ sp^2,\ sp^3,\ sp^3d,\ sp^3d^2\). Only σ bonds and lone pairs count — π bonds do not change the hybridization.
Molecular Orbital Theory
Where VBT keeps electrons on individual atoms, molecular orbital theory lets atomic orbitals combine into molecular orbitals spread over the whole molecule. Two atomic orbitals make two MOs: a lower-energy bonding orbital and a higher-energy antibonding orbital \((\ast)\).
where \(N_b\) and \(N_a\) are electrons in bonding and antibonding MOs. A positive bond order means a stable molecule; unpaired electrons make it paramagnetic. This is MOT's triumph — it correctly predicts that \(\ce{O2}\) is paramagnetic, which Lewis structures cannot.
Polarity & Dipole Moment
When bonded atoms differ in electronegativity, the bond is polar — one end slightly negative, the other slightly positive. The dipole moment measures this separation, and being a vector, it depends on shape as much as on the bonds.
\(\ce{CO2}\) has two polar \(\ce{C=O}\) bonds yet \(\mu=0\) — being linear, the dipoles oppose and cancel. Bent \(\ce{H2O}\), with the same idea of polar bonds, does not cancel and is strongly polar. Geometry decides.
The Hydrogen Bond
When hydrogen is bonded to a small, highly electronegative atom \((\ce{N},\ \ce{O},\ \ce{F})\), the bare proton attracts a lone pair on a neighbouring electronegative atom. This hydrogen bond is far weaker than a covalent bond but strong among intermolecular forces.
Putting It to Work
Problem. In one resonance form of ozone \(\ce{O3}\), the central oxygen has one lone pair, one single bond and one double bond. Find its formal charge.
Solution. Valence \(=6\), lone-pair electrons \(=2\), bonding electrons \(=6\):
Problem. Predict the shape and bond angle of \(\ce{NH3}\).
Solution. Nitrogen has 3 bonding pairs and 1 lone pair — steric number 4, so the electron geometry is tetrahedral. The lone pair pushes the bonds together:
Problem. Find the hybridization of carbon in ethyne \(\ce{C2H2}\).
Solution. Each carbon forms 2 σ bonds (one to \(\ce{H}\), one to \(\ce{C}\)) and has no lone pair — steric number 2:
The remaining two p orbitals on each carbon form the two π bonds of the triple bond.
Problem. Using MOT, find the bond order of \(\ce{O2}\) (16 electrons) and state whether it is paramagnetic.
Solution. Filling the MOs gives \(N_b=10,\ N_a=6\):
The two \(\pi^\ast\) orbitals each hold one electron, so \(\ce{O2}\) has two unpaired electrons — it is paramagnetic.
Problem. Both \(\ce{CO2}\) and \(\ce{H2O}\) contain polar bonds. Why is one non-polar and the other polar?
Solution. \(\ce{CO2}\) is linear, so its two equal bond dipoles point in opposite directions and sum to zero — \(\mu=0\). \(\ce{H2O}\) is bent \((104.5^{\circ})\), so its bond dipoles add to a net moment \((\mu=1.85\ \text{D})\). Geometry, not the bonds alone, decides molecular polarity.
Problem. Arrange \(\ce{O2},\ \ce{O2+},\ \ce{O2-}\) in order of increasing bond length.
Solution. Removing an antibonding electron raises bond order; adding one lowers it:
Higher bond order means shorter bond, so increasing length is \(\ce{O2+}<\ce{O2}<\ce{O2-}\).
Chapter Summary
To lower energy and reach an octet; formal charge picks the best Lewis structure.
Ionic = transfer (lattice energy); covalent = sharing; Fajans' rules blend the two.
Higher bond order → shorter, stronger bond; resonance averages them.
VSEPR from electron-pair repulsion; hybridization from steric number.
\(\text{B.O.}=\tfrac12(N_b-N_a)\); unpaired electrons → paramagnetic.
\(\mu=q\times d\); symmetric shapes cancel dipoles; H-bonding shifts properties.
Problems
Draw the Lewis structure first, count σ bonds and lone pairs, then read off shape and hybridization. Difficulty rises down the list.
- Draw the Lewis structure of \(\ce{CO2}\) and assign formal charges.
- Predict the shapes and bond angles of \(\ce{BF3},\ \ce{H2O}\) and \(\ce{SF6}\).
- State the hybridization of the central atom in \(\ce{PCl5},\ \ce{SO2}\) and \(\ce{XeF4}\).
- Explain why the bond angle in \(\ce{H2O}\,(104.5^{\circ})\) is smaller than in \(\ce{NH3}\,(107^{\circ})\).
- How many σ and π bonds are present in \(\ce{C2H4}\) and in \(\ce{HCN}\)?
- Using Fajans' rules, arrange \(\ce{NaCl},\ \ce{MgCl2},\ \ce{AlCl3}\) in order of increasing covalent character.
- Calculate the bond order of \(\ce{N2}\) and \(\ce{N2+}\) and predict which has the stronger bond.
- Why is \(\ce{O2}\) paramagnetic while \(\ce{N2}\) is diamagnetic? Use MOT.
- Account for the order of boiling points \(\ce{HF}>\ce{HCl}\) despite \(\ce{HCl}\) having the larger molar mass.
- Predict whether \(\ce{NH3}\) or \(\ce{NF3}\) has the larger dipole moment, and justify it.
- Compare the bond orders of \(\ce{O2},\ \ce{O2^2-}\) and \(\ce{O2+}\), and order their bond strengths.
- The dipole moment of \(\ce{HCl}\) is \(1.03\ \text{D}\) and the bond length is \(127\ \text{pm}\). Estimate its percent ionic character.