Conditions for Formation

Ionisation energy is the minimum energy required to remove one electron from a gaseous atom: M(g) → M⁺(g) + e⁻. Metals in Groups 1 and 2 have low first ionisation energies (e.g., Na: 496 kJ mol⁻¹, Mg: 738 kJ mol⁻¹) because their outermost electrons are well-shielded from the nucleus and are far from it. A low IE means relatively little energy input is needed to create a cation, making the first step of electron transfer feasible.

Electron affinity is the energy released when a gaseous atom accepts one electron: X(g) + e⁻ → X⁻(g). Electronegative non-metals such as the halogens and oxygen have large negative electron affinities (Cl: −349 kJ mol⁻¹, F: −328 kJ mol⁻¹), meaning substantial energy is released on anion formation. A high (exothermic) EA offsets the energy cost of ionisation and makes electron transfer energetically worthwhile. Note: the second electron affinity (e.g., O²⁻) is endothermic, so the lattice energy must compensate.

Lattice enthalpy (ΔH_latt) is the enthalpy change when one mole of an ionic lattice is formed from its gaseous ions: M⁺(g) + X⁻(g) → MX(s). It is always exothermic (e.g., NaCl: −787 kJ mol⁻¹, MgO: −3791 kJ mol⁻¹). This large energy release is the principal driving force for ionic bond formation. Lattice enthalpy increases with higher ionic charges and smaller ionic radii (per Coulomb's law: U ∝ Q₊Q₋/r). The Born–Haber cycle shows that the overall standard enthalpy of formation equals the algebraic sum of all these steps; a high lattice enthalpy ensures ΔH_f° is sufficiently negative for a stable compound.

A practical shorthand for the three conditions above is the Pauling electronegativity difference: Δχ = χ(non-metal) − χ(metal). When Δχ > 1.7, the bond is considered predominantly ionic (> ~50% ionic character). For NaCl, Δχ = 3.16 − 0.93 = 2.23; for MgO, Δχ = 3.44 − 1.31 = 2.13. The large difference reflects the simultaneous presence of a low-IE metal and a high-EA non-metal, and predicts that complete electron transfer is more stable than electron sharing.

The Born–Haber cycle applies Hess's law to relate the enthalpy of formation of an ionic compound to measurable thermodynamic quantities. For sodium chloride: ΔH_f°(NaCl) = ΔH_sub(Na) + IE₁(Na) + ½D(Cl₂) + EA(Cl) + ΔH_latt(NaCl). Numerically: ΔH_f° = +108 + 496 + 122 − 349 − 787 = −411 kJ mol⁻¹. The large negative lattice enthalpy (−787 kJ mol⁻¹) is responsible for making the overall process exothermic despite the energy costs of sublimation and ionisation.

• \Delta H_f^\circ — Standard enthalpy of formation (kJ mol⁻¹): Enthalpy change when one mole of ionic compound forms from its elements in standard states.
• \Delta H_{\text{sub}} — Sublimation enthalpy of metal (kJ mol⁻¹): Energy required to convert one mole of solid metal to gaseous atoms; always endothermic.
• \text{IE}_1 — First ionisation energy (kJ mol⁻¹): Energy to remove the outermost electron from a gaseous metal atom; always endothermic.
• \frac{1}{2}D — Half bond dissociation enthalpy (kJ mol⁻¹): Energy to produce one mole of gaseous non-metal atoms from the diatomic molecule; endothermic.
• \text{EA} — Electron affinity (kJ mol⁻¹): Energy released when a gaseous non-metal atom accepts one electron; exothermic (negative value).
• \Delta H_{\text{latt}} — Lattice enthalpy (kJ mol⁻¹): Energy released when gaseous ions combine to form one mole of crystalline ionic solid; large and exothermic.

Problem. Using the data below, calculate the standard enthalpy of formation of potassium fluoride (KF) and determine whether it meets the conditions for ionic bond formation. ΔH_sub(K) = +89 kJ mol⁻¹ IE₁(K) = +419 kJ mol⁻¹ ½D(F₂) = +79 kJ mol⁻¹ EA(F) = −328 kJ mol⁻¹ ΔH_latt(KF) = −817 kJ mol⁻¹ Also calculate the electronegativity difference given χ(F) = 3.98 and χ(K) = 0.82.

Answer. ΔH_f°(KF) = −558 kJ mol⁻¹; Δχ = 3.16. KF strongly satisfies all three conditions for ionic bond formation.

• Lattice Enthalpy — Lattice enthalpy is the third and often largest condition for ionic bond formation; understanding its dependence on ionic charge and radius explains why some ionic compounds (e.g., MgO) are far more stable than others (e.g., NaCl).
• Properties of Ionic Compounds — The conditions that create ionic bonds directly determine the physical properties of ionic compounds—high melting points, brittleness, and electrical conductivity when molten—all of which arise from the strong, non-directional electrostatic forces in the lattice.
• Ionic Nomenclature — Correctly naming ionic compounds requires knowing which elements form cations and which form anions—knowledge rooted in the IE and EA conditions for ionic bond formation.
• Lewis–Langmuir Concept — The Lewis model describes ionic bonding as complete electron transfer to achieve noble-gas electron configurations (octets), directly reflecting the low IE of metals and the high EA of non-metals.
• Exceptions to the Octet Rule — Some ions (e.g., Fe²⁺/Fe³⁺, Cu⁺/Cu²⁺) involve incomplete d-subshells and do not achieve a simple octet, showing that the octet-based conditions for ionic bonding have important exceptions related to transition-metal ionisation energies.
• Dipole Moment — Electronegativity difference—the summary criterion for ionic bond formation (Δχ > 1.7)—is also the factor that determines bond polarity and dipole moment, linking ionic and polar-covalent bonding on a continuous spectrum.
• Postulates (VSEPR / Valence Bond) — Contrasting the conditions for ionic bond formation with those for covalent bond formation (similar electronegativities, shared electrons) reinforces why element combination and electronegativity difference determine bond type.