Ionic Bonding
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You should be comfortable with the concepts of ionisation energy, electron affinity, and the fact that enthalpy is a state function, as introduced in the previous lecture on the energetics of bonding.
Why this matters
Ionic bonding explains the stability and high melting points of salts, which are fundamental to biological and industrial chemistry.
While covalent bonding involves the sharing of electrons, ionic bonding arises from the electrostatic attraction between oppositely charged ions. This interaction is not a "force" in the quantum mechanical sense of a shared electron pair, but rather a macroscopic manifestation of Coulomb's law acting on point charges. Understanding ionic bonding requires us to move beyond the isolated atom and consider the collective stability of the crystal lattice. The key challenge in quantifying this stability is that we cannot isolate a single ion pair in the gas phase to measure its bond energy directly; instead, we must rely on thermodynamic cycles to deduce the energy released when the solid forms.
The Electrostatic Origin of Ionic Stability
Imagine two spheres, one positively charged and one negatively charged, held apart by a distance. If you release them, they will accelerate toward each other due to the electrostatic force. In a solid crystal, this attraction is not limited to just one pair of ions; every cation is surrounded by anions, and every anion is surrounded by cations. This creates a three-dimensional network of attractions whose total, summed over the whole crystal, is considerably larger than the attraction within a single isolated ion pair. The stability of the ionic solid is therefore a collective property. The "bond" is not a localized link between two atoms, but the entire lattice structure. This is why ionic compounds typically have high melting points and are brittle; breaking the solid requires overcoming the cumulative electrostatic attraction of the entire lattice, not just a few individual bonds.
Coulomb's Law and Lattice Energy
The magnitude of the electrostatic potential energy between two point charges is given by Coulomb's law. For a cation with charge \(+q_1\) and an anion with charge \(-q_2\) separated by a distance \(r\), the potential energy \(E\) is proportional to \(-\frac{q_1 q_2}{r}\). The negative sign indicates that the energy decreases (becomes more stable) as the ions approach each other. However, in a real crystal, the interaction is not simply between one pair of ions. The total lattice energy is the sum of all attractive and repulsive interactions in the lattice. While the repulsive terms (due to electron cloud overlap) are important at very short distances, the dominant term is the attractive Coulombic term. Consequently, the lattice energy \(U\) is strongly dependent on the charges of the ions and the distance between them. Doubling the charge of one ion doubles the lattice energy and doubling both quadruples it, while doubling the distance halves it. This explains why salts with highly charged ions (such as \(\mathrm{MgO}\)) have much higher lattice energies and melting points than those with singly charged ions (such as \(\mathrm{NaCl}\)).
The Indirect Measurement Problem
Because the lattice energy is defined as the energy change for the process \(\mathrm{M^{n+}(g) + X^{n-}(g) \to MX(s)}\), it cannot be measured directly. We cannot create a gas of pure \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) ions without them immediately recombining or reacting with the container. To determine this value, we must use a thermodynamic cycle that connects the elements in their standard states to the ionic solid via measurable steps. This approach relies on Hess's Law, which states that the total enthalpy change for a reaction is independent of the pathway taken, provided the initial and final states are the same. By constructing a cycle that goes from elements to gaseous atoms, then to gaseous ions, and finally to the solid, we can solve for the unknown lattice energy using only experimentally accessible quantities such as sublimation enthalpies, bond dissociation energies, ionisation energies, and electron affinities.
Applying Hess's Law
Consider the formation of a generic ionic compound \(\mathrm{MX}\) from its elements. The standard enthalpy of formation, \(\Delta H_f\), is the energy change for \(\mathrm{M(s) + \frac{1}{2}X_2(g) \to MX(s)}\). We can construct an alternative path: first, sublime the metal \(\mathrm{M(s) \to M(g)}\) with enthalpy \(\Delta H_{\text{sub}}\); second, break the diatomic molecule \(\frac{1}{2}\mathrm{X_2(g) \to X(g)}\) with enthalpy \(\frac{1}{2}D_{X_2}\); third, ionize the metal atom \(\mathrm{M(g) \to M^+(g) + e^-}\) with ionisation energy \(\mathrm{IE}\); fourth, add an electron to the non-metal atom \(\mathrm{X(g) + e^- \to X^-(g)}\) with electron affinity \(\mathrm{EA}\); and finally, form the solid from the gaseous ions \(\mathrm{M^+(g) + X^-(g) \to MX(s)}\) with lattice energy \(U\). Summing these steps must equal the direct formation enthalpy. Note that electron affinity is often defined as the energy released, so it enters the equation with a negative sign if we are summing enthalpy changes, or we must be careful with the sign convention in the final algebraic expression.
The result
Ionic bonding and the Born-Haber cycle
The lattice energy of an ionic solid — the energy released when gaseous ions come together to form the crystal, \(M^{n+}(g)+X^{n-}(g)\to MX(s)\) — cannot be measured directly, but it can be extracted exactly from measurable quantities by writing an indirect thermodynamic path from elements to the solid and applying Hess's law: the enthalpy change for a reaction is the same regardless of the path taken, because enthalpy is a state function.
A common misconception is that the lattice energy is always positive. In thermodynamics, the lattice energy is often defined as the energy *released* when the solid forms from gaseous ions, making it a negative value (exothermic). However, some textbooks define it as the energy *required* to separate the solid into gaseous ions, making it positive. You must always check the sign convention used in your specific course materials. The Born–Haber equation used in this unit assumes \(U\) is the energy released (negative contribution to the enthalpy sum), or equivalently, that the term is subtracted. If your definition is the reverse, the sign in the final equation will flip. C
Worked example
Problem. Use the Coulomb scaling \(U \propto |z_+ z_-|/r_0\) to estimate how much larger the lattice energy of MgO is than that of NaCl. Both have the rock-salt structure, so the geometric (Madelung) factor is the same. Nearest-neighbour distances: NaCl 282 pm, MgO 210 pm. The lattice enthalpy of NaCl is 787 kJ mol-1.
- Charge factor: \(|z_+z_-|\) is \(1 \times 1 = 1\) for NaCl and \(2 \times 2 = 4\) for MgO.
- Distance factor: \(282/210 = 1.343\).
- Predicted ratio: \(4 \times 1.343 = 5.37\).
- Estimate for MgO: \(5.37 \times 787 \approx 4.2 \times 10^3\ \mathrm{kJ\,mol^{-1}}\). The Born–Haber value is about \(3.8 \times 10^3\ \mathrm{kJ\,mol^{-1}}\) (a ratio of about 4.8); the gap comes from repulsion and from the cost of forming O2-, which the simple scaling ignores.
Units check. Charges are pure numbers and the distances cancel, so the ratio is dimensionless. Most of the factor comes from the charges (×4), only a third extra from the smaller ions.
Three quick tests
- Why does \(\mathrm{MgO}\) have a higher lattice energy than \(\mathrm{NaCl}\)?
Answer
\(\mathrm{Mg^{2+}}\) and \(\mathrm{O^{2-}}\) have higher charges than \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\), leading to stronger electrostatic attraction. - What is the sign of the lattice energy if defined as the energy released during solid formation?
Answer
Negative, because the process is exothermic. - Which thermodynamic law allows us to calculate lattice energy indirectly?
Answer
Hess's Law, which relies on enthalpy being a state function.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.