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Concept 03 / 07Chemical Bonding and Molecular Structure

Ionic Bonding: Lattice Energy, the Born-Haber Cycle and Fajans' Rules

Conditions favouring ionic bond formation, lattice enthalpy and the factors controlling it, the Born-Haber cycle as an application of Hess's law, general properties of ionic compounds, polarisation and Fajans' rules for covalent character, and consequences such as melting points and solubility trends.

Figure 1
An ionic (electrovalent) bond forms by transfer of electrons from an atom of low ionisation enthalpy (a metal) to one of highly negative electron gain enthalpy (a non-metal), followed by electrostatic attraction between the resulting ions. Formation is favoured by low ionisation enthalpy of the metal, highly negative electron gain enthalpy of the non-metal, and, above all, high lattice enthalpy. Energetics. Forming isolated gaseous ions is actually endothermic: for NaCl, ionisation of Na costs +496+496 kJ/mol while electron gain by Cl releases only −349-349. What makes ionic compounds stable is the lattice enthalpy, the energy released when gaseous ions come together to form one mole of the crystal (about −787-787 kJ/mol for NaCl). The Born-Haber cycle (figure) applies Hess's law: the enthalpy of formation equals the sum of sublimation, dissociation, ionisation, electron gain and lattice enthalpies: ΔfH=ΔsubH+12ΔdissH+IE+ΔegH+U\Delta_{f}H = \Delta_{sub}H + \frac{1}{2}\Delta_{diss}H + \text{IE} + \Delta_{eg}H + U. Lattice enthalpy cannot be measured directly, so it is calculated from this cycle. Lattice enthalpy magnitude increases with ionic charges and decreases with ionic size: ∣U∣∝z+z−r++r−|U| \propto \frac{z_{+}z_{-}}{r_{+} + r_{-}}. So MgO (2+2+, 2−2-) has about four times the lattice energy of NaCl, and LiF more than CsI. High lattice energy explains high melting points (MgO melts near 2850∘2850^{\circ}C) and low solubility. Properties of ionic compounds: crystalline solids of high melting and boiling points; brittle; non-conducting as solids but conducting when molten or dissolved (mobile ions); usually soluble in polar solvents like water (when hydration enthalpy compensates lattice enthalpy); reactions are fast because they involve free ions in solution. Polarisation and Fajans' rules. No bond is purely ionic. A cation distorts (polarises) the electron cloud of an adjacent anion, drawing electron density into the region between them and introducing covalent character. Fajans' rules state that covalent character increases when: the cation is small (high charge density): LiCl\mathrm{LiCl} is more covalent than NaCl\mathrm{NaCl}; the cation charge is high: AlCl3\mathrm{AlCl_{3}} > MgCl2\mathrm{MgCl_{2}} > NaCl\mathrm{NaCl}; the anion is large (easily polarised): LiI\mathrm{LiI} > LiBr\mathrm{LiBr} > LiCl\mathrm{LiCl} > LiF\mathrm{LiF}; the cation has a pseudo-noble-gas configuration (ns2np6nd10ns^{2}np^{6}nd^{10}), since d electrons shield poorly: CuCl\mathrm{CuCl} is more covalent than NaCl\mathrm{NaCl} although Cu+\mathrm{Cu^{+}} and Na+\mathrm{Na^{+}} are similar in size. Consequences of covalent character: lower melting points (AlCl3\mathrm{AlCl_{3}} sublimes near 180∘180^{\circ}C while NaCl\mathrm{NaCl} melts at 801∘801^{\circ}C), solubility in organic solvents (LiCl dissolves in ethanol), and lower thermal stability of salts with large polarisable anions (carbonates of small cations decompose more easily: BeCO3\mathrm{BeCO_{3}} is unstable, BaCO3\mathrm{BaCO_{3}} very stable). Key Takeaways: - Ionic bond: low IE metal + high electron affinity non-metal + high lattice energy. - Born-Haber: ΔfH=ΔsubH+12D+IE+ΔegH+U\Delta_{f}H = \Delta_{sub}H + \frac{1}{2}D + \text{IE} + \Delta_{eg}H + U. - ∣U∣∝z+z−r++r−|U| \propto \frac{z_{+}z_{-}}{r_{+} + r_{-}}. - Fajans: small or highly charged cation, large anion, pseudo-noble-gas cation all increase covalent character. - Covalent character lowers melting point and thermal stability. - Pitfall: thinking ion formation itself is exothermic; the lattice energy is what makes it favourable. Brain-Teaser 1: Using the cycle, find the lattice enthalpy of NaCl from: ΔfH=−411\Delta_{f}H = -411, sublimation of Na +108+108, bond dissociation of Cl2\mathrm{Cl_{2}} +242+242, IE of Na +496+496, ΔegH\Delta_{eg}H of Cl −349-349 (kJ/mol). Solution: −411=108+2422+496+(−349)+U=376+U-411 = 108 + \frac{242}{2} + 496 + (-349) + U = 376 + U, so U=−787U = -787 kJ/mol. Every step except the lattice term is measurable; the cycle closes the loop. Brain-Teaser 2: Arrange in increasing covalent character: NaCl\mathrm{NaCl}, MgCl2\mathrm{MgCl_{2}}, AlCl3\mathrm{AlCl_{3}}, SiCl4\mathrm{SiCl_{4}}. Arrange LiF\mathrm{LiF}, LiCl\mathrm{LiCl}, LiBr\mathrm{LiBr}, LiI\mathrm{LiI} likewise. Solution: Cation charge rises and size falls from Na+\mathrm{Na^{+}} to Si(IV), so NaCl < MgCl2\mathrm{MgCl_{2}} < AlCl3\mathrm{AlCl_{3}} < SiCl4\mathrm{SiCl_{4}} (the last is a covalent liquid). For the lithium halides, anion size increases from F to I, so LiF < LiCl < LiBr < LiI. The two series apply two different Fajans rules. Brain-Teaser 3: Why is AgCl much less soluble in water than NaCl, although Ag+\mathrm{Ag^{+}} (115115 pm) and Na+\mathrm{Na^{+}} (102102 pm) are similar in size? Solution: Ag+\mathrm{Ag^{+}} (4d104d^{10}) has a pseudo-noble-gas configuration; its d electrons shield the nucleus poorly, so it polarises Cl−\mathrm{Cl^{-}} strongly, giving AgCl significant covalent character and a lattice stabilised by that covalency. Water, which solvates ions well, cannot break this lattice easily. The fourth Fajans rule, not size, explains the difference.

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Ionic Bonding: Lattice Energy, the Born-Haber Cycle and Fajans' Rules — DPP

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