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Structure of Matter · Bonding and properties

Ionic bonding and Born–Haber cycles

In winter, grit salt is spread on Irish roads to stop ice forming. It looks like a pile of tiny cubes, yet no "molecules of NaCl" are in it. It is a giant lattice of ions held together so strongly that you cannot measure the energy needed to pull it apart directly. How do chemists find that energy anyway? By the end of this page you will see how.

An ionic crystal is not a molecule

In an ionic compound such as sodium chloride, sodium atoms lose electrons to form $\mathrm{Na^+}$ ions and chlorine atoms gain them to form $\mathrm{Cl^-}$ ions. Oppositely charged ions attract each other in every direction, so they pack into a regular three-dimensional pattern.

Two-dimensional sketch of a rock salt lattice with large anions and small cations alternating in a regular pattern Large circles: anions. Small circles: cations. Each ion is surrounded by ions of opposite charge.

Because the ions are held firmly in fixed positions, ionic solids are hard, brittle solids at room temperature with high melting points, and they do not conduct electricity when solid. Many ionic compounds, such as sodium chloride, are soluble in water, because water molecules are attracted to the ions (ion–dipole attraction) and pull them out of the lattice. When molten or dissolved, the ions are free to move, so they do conduct.

Bonding is a continuum from ionic through polar covalent to pure covalent. The larger the difference in electronegativity between the two atoms, the more ionic the bond.

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Sodium chloride, $\mathrm{NaCl}$, crystallises forming a three-dimensional rock salt structure (two interpenetrating ionic sublattices), whereas hydrogen chloride,…

Sodium chloride, $\mathrm{NaCl}$, crystallises forming a three-dimensional rock salt structure (two interpenetrating ionic sublattices), whereas hydrogen chloride, $\mathrm{HCl}$, forms discrete diatomic molecules in the gas phase at room temperature. Justify, on the basis of the type of bonding and structure, why it makes no physical sense to speak of a "$\mathrm{NaCl}$ molecule" in the solid state, stating how the empirical formula of this ionic compound is defined.

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Lattice enthalpy: charge and ionic radius

$\mathrm{NaCl}$ melts at $801\,^\circ\mathrm{C}$, $\mathrm{NaF}$ at $996\,^\circ\mathrm{C}$ and $\mathrm{MgO}$ at $2{,}852\,^\circ\mathrm{C}$. What makes one lattice so much harder to pull apart than another?

Comparison of attraction between small, close ions and large, distant ions small ions, close together strong attraction large ions, further apart weaker attraction
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Consider the lithium halides $\mathrm{LiF}$ and $\mathrm{LiI}$. Both compounds crystallise with the same ionic lattice geometry ($\mathrm{NaCl}$ type). Given that the…

Consider the lithium halides $\mathrm{LiF}$ and $\mathrm{LiI}$. Both compounds crystallise with the same ionic lattice geometry ($\mathrm{NaCl}$ type). Given that the ionic radii are $r(\mathrm{F^-}) = 133\,\mathrm{pm}$ and $r(\mathrm{I^-}) = 220\,\mathrm{pm}$, and considering the Born–Landé equation or Coulomb's law for the lattice energy (taken here as the exothermic process of formation of the lattice from gaseous ions, $\Delta H_{\mathrm{ret}} < 0$): a) Deduce, with justification, which of the two solids has the lattice enthalpy of greater modulus $(|\Delta H_{\mathrm{ret}}|)$. b) Justify which of the two compounds will have the higher melting point.

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Potassium bromide, $\mathrm{KBr(s)}$, dissociates completely into its gaseous ions according to the endothermic process:…

Potassium bromide, $\mathrm{KBr(s)}$, dissociates completely into its gaseous ions according to the endothermic process: $\mathrm{KBr(s)} \to \mathrm{K^+(g)} + \mathrm{Br^-(g)}$. For this process, the lattice dissociation enthalpy is $\Delta H_{\mathrm{ret,dis}} = +689\,\mathrm{kJ\times mol^{-1}}$. Determine the standard lattice enthalpy of formation of the lattice, $\Delta H_{\mathrm{ret}}$, corresponding to the reaction $\mathrm{K^+(g)} + \mathrm{Br^-(g)} \to \mathrm{KBr(s)}$, and briefly explain the meaning of the thermodynamic sign obtained.

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The ionic compounds magnesium fluoride, $\mathrm{MgF_2}$, and calcium fluoride, $\mathrm{CaF_2}$, are compared. Both have an identical cation/anion charge ratio (+2/−1).…

The ionic compounds magnesium fluoride, $\mathrm{MgF_2}$, and calcium fluoride, $\mathrm{CaF_2}$, are compared. Both have an identical cation/anion charge ratio (+2/−1). Ionic radii data: $r(\mathrm{Mg^{2+}}) = 72\,\mathrm{pm}$, $r(\mathrm{Ca^{2+}}) = 100\,\mathrm{pm}$, and $r(\mathrm{F^-}) = 133\,\mathrm{pm}$. Experimentally, the moduli of the lattice enthalpies of formation $(\Delta H_{\mathrm{ret}} < 0)$ for both solids are $\approx 2957\,\mathrm{kJ\times mol^{-1}}$ and $\approx 2630\,\mathrm{kJ\times mol^{-1}}$. a) Assign, with justification, each lattice enthalpy value to its corresponding compound using arguments of electrostatic attraction. b) If $\mathrm{MgF_2}$ is now compared with magnesium oxide, $\mathrm{MgO}$ (where $r(\mathrm{O^{2-}}) = 140\,\mathrm{pm}$ and its $|\Delta H_{\mathrm{ret}}|$ is approximately $3791\,\mathrm{kJ\times mol^{-1}}$), explain the role of the greater charge of the oxide anion in the electrostatic attraction. State why a rigorous quantitative comparison also requires considering the stoichiometry and the structure of the lattices.

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The Born–Haber cycle

You can measure the enthalpy of formation of sodium chloride in a calorimeter, but you cannot turn solid salt into separate gaseous ions and measure the energy. So how do we find the lattice enthalpy?

The direct route is the enthalpy of formation: $\mathrm{Na(s)} + \tfrac{1}{2}\mathrm{Cl_2(g)} \to \mathrm{NaCl(s)}$. The step-by-step route (in the order shown in the diagram) is:

  • Sublimation of sodium: $\mathrm{Na(s)} \to \mathrm{Na(g)}$.

  • First ionisation energy of sodium: $\mathrm{Na(g)} \to \mathrm{Na^+(g)} + \mathrm{e^-}$.

  • Atomisation of chlorine: $\tfrac{1}{2}\mathrm{Cl_2(g)} \to \mathrm{Cl(g)}$ (the diagram shows it after ionisation; the order of these two steps does not affect the total).

  • Electron affinity of chlorine: $\mathrm{Cl(g)} + \mathrm{e^-} \to \mathrm{Cl^-(g)}$.

  • Lattice formation: $\mathrm{Na^+(g)} + \mathrm{Cl^-(g)} \to \mathrm{NaCl(s)}$.

Born–Haber enthalpy level diagram for sodium chloride linking the elements, gaseous atoms, gaseous ions and the solid lattice H Na(s) + ½Cl₂(g) Na(g) + ½Cl₂(g) Na⁺(g) + e⁻ + ½Cl₂(g) Na⁺(g) + e⁻ + Cl(g) Na⁺(g) + Cl⁻(g) NaCl(s) sublimation ionisation atomisation electronaffinity lattice enthalpy formation
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To calculate the lattice enthalpy of formation of sodium chloride, $\Delta H_{\mathrm{ret}}$ (defined according to…

To calculate the lattice enthalpy of formation of sodium chloride, $\Delta H_{\mathrm{ret}}$ (defined according to $\mathrm{Na^+(g)} + \mathrm{Cl^-(g)} \to \mathrm{NaCl(s)}$), the Born–Haber energy diagram shown below is available.

Enthalpy, H (kJ/mol) NaCl(s) Na(s) + ½ Cl₂(g) Na(g) + ½ Cl₂(g) Na⁺(g) + e⁻ + ½ Cl₂(g) Na⁺(g) + e⁻ + Cl(g) Na⁺(g) + Cl⁻(g) ΔHf° = −411 kJ/mol +108 (ΔHsub) +496 (Ei1) +121 (ΔHat) −349 (ΔHae) ? (ΔHlatt)

The thermodynamic values at $298\,\mathrm{K}$ are: enthalpy of sublimation of sodium, $\Delta H_{\mathrm{sub}}[\mathrm{Na}] = +108\,\mathrm{kJ\times mol^{-1}}$; first ionisation energy of sodium, $E_{\mathrm{i1}}[\mathrm{Na}] = +496\,\mathrm{kJ\times mol^{-1}}$; enthalpy of atomisation of molecular chlorine (formation of 1 mol of $\mathrm{Cl(g)}$ atoms), $\Delta H_{\mathrm{at}}[\mathrm{Cl_2}] = \frac{1}{2}\Delta H_{\mathrm{enl}}(\mathrm{Cl-Cl}) = +121\,\mathrm{kJ\times mol^{-1}}$; electron affinity of chlorine (exothermic process $\mathrm{Cl(g)} + \mathrm{e^-} \to \mathrm{Cl^-(g)}$), $\Delta H_{\mathrm{ae}}[\mathrm{Cl}] = -349\,\mathrm{kJ\times mol^{-1}}$; and standard enthalpy of formation of the solid, $\Delta H_{\mathrm{f}}^\circ[\mathrm{NaCl(s)}] = -411\,\mathrm{kJ\times mol^{-1}}$. a) Set up the enthalpy balance of the Born–Haber cycle and determine the value of $\Delta H_{\mathrm{ret}}$. b) Justify the sign obtained for the lattice enthalpy. The level diagram is schematic and is not drawn to an energy scale.

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Cycles with ions of higher charge

Energy levels showing that the first electron affinity of oxygen is exothermic and the second is endothermic O(g) + 2e⁻ O⁻(g) + e⁻ O²⁻(g) first: down second: up
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From the following standard thermochemical data at $298\,\mathrm{K}$: enthalpy of sublimation of solid potassium,…

From the following standard thermochemical data at $298\,\mathrm{K}$: enthalpy of sublimation of solid potassium, $\Delta H_{\mathrm{sub}}[\mathrm{K}] = +89\,\mathrm{kJ\times mol^{-1}}$; dissociation energy of the $\mathrm{F-F}$ bond in $\mathrm{F_2(g)}$, $\Delta H_{\mathrm{enl}}[\mathrm{F_2}] = +158\,\mathrm{kJ\times mol^{-1}}$; first ionisation energy of potassium, $E_{\mathrm{i1}}[\mathrm{K}] = +419\,\mathrm{kJ\times mol^{-1}}$; lattice enthalpy of formation of potassium fluoride, $\Delta H_{\mathrm{ret}}[\mathrm{KF(s)}] = -821\,\mathrm{kJ\times mol^{-1}}$ (corresponding to $\mathrm{K^+(g)} + \mathrm{F^-(g)} \to \mathrm{KF(s)}$); and standard enthalpy of formation of $\mathrm{KF(s)}$, $\Delta H_{\mathrm{f}}^\circ[\mathrm{KF(s)}] = -567\,\mathrm{kJ\times mol^{-1}}$: a) Calculate the electron affinity of fluorine, expressed as the enthalpy of the process $\mathrm{F(g)} + \mathrm{e^-} \to \mathrm{F^-(g)}$, $\Delta H_{\mathrm{ae}}[\mathrm{F}]$. b) State whether the gain of an electron by gaseous atomic fluorine is an endothermic or exothermic process.

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The following thermochemical data are available for the formation of magnesium chloride, $\mathrm{MgCl_2(s)}$, at $298\,\mathrm{K}$: Standard enthalpy of formation,…

The following thermochemical data are available for the formation of magnesium chloride, $\mathrm{MgCl_2(s)}$, at $298\,\mathrm{K}$:

  • Standard enthalpy of formation, $\Delta H_{\mathrm{f}}^\circ[\mathrm{MgCl_2(s)}] = -641\,\mathrm{kJ\times mol^{-1}}$

  • Enthalpy of sublimation of magnesium, $\Delta H_{\mathrm{sub}}[\mathrm{Mg}] = +148\,\mathrm{kJ\times mol^{-1}}$

  • First ionisation energy of magnesium, $E_{\mathrm{i1}}[\mathrm{Mg}] = +738\,\mathrm{kJ\times mol^{-1}}$

  • Second ionisation energy of magnesium, $E_{\mathrm{i2}}[\mathrm{Mg}] = +1451\,\mathrm{kJ\times mol^{-1}}$

  • Bond dissociation energy of $\mathrm{Cl_2(g)}$, $\Delta H_{\mathrm{enl}}[\mathrm{Cl_2}] = +242\,\mathrm{kJ\times mol^{-1}}$

  • Enthalpy of the process $\mathrm{Cl(g)} + \mathrm{e^-} \to \mathrm{Cl^-(g)}$, $\Delta H_{\mathrm{ae}}[\mathrm{Cl}] = -349\,\mathrm{kJ\times mol^{-1}}$

  1. a) Write out in detail each of the elementary thermochemical equations that make up the Born–Haber cycle for $\mathrm{MgCl_2(s)}$ and determine its standard lattice enthalpy of formation, $\Delta H_{\mathrm{ret}} (\mathrm{Mg^{2+}(g)} + 2\,\mathrm{Cl^-(g)} \to \mathrm{MgCl_2(s)})$.
  2. b) Consider a hypothetical monoionic compound $\mathrm{MgCl(s)}$ whose calculated lattice enthalpy is $\Delta H_{\mathrm{ret}}[\mathrm{MgCl}] = -753\,\mathrm{kJ\times mol^{-1}}$. Determine the standard enthalpy of formation of this $\mathrm{MgCl(s)}$ from its elements, calculate the enthalpy of the reaction $\mathrm{MgCl(s)} + \frac{1}{2}\mathrm{Cl_2(g)} \to \mathrm{MgCl_2(s)}$ and explain which formation is favoured from the enthalpy point of view: $\mathrm{MgCl_2}$ or $\mathrm{MgCl}$. Explain why enthalpy alone is not enough to determine spontaneity at any temperature.
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Calcium oxide, $\mathrm{CaO(s)}$, is a solid of high thermal stability. The following data at $298\,\mathrm{K}$ are known: Enthalpy of sublimation of calcium:…

Calcium oxide, $\mathrm{CaO(s)}$, is a solid of high thermal stability. The following data at $298\,\mathrm{K}$ are known:

  • Enthalpy of sublimation of calcium: $\Delta H_{\mathrm{sub}}[\mathrm{Ca}] = +178\,\mathrm{kJ\times mol^{-1}}$

  • Sum of ionisation energies: $E_{\mathrm{i1}}[\mathrm{Ca}] + E_{\mathrm{i2}}[\mathrm{Ca}] = 590 + 1145 = +1735\,\mathrm{kJ\times mol^{-1}}$

  • Bond dissociation enthalpy of $\mathrm{O_2(g)}$: $\Delta H_{\mathrm{enl}}[\mathrm{O_2}] = +498\,\mathrm{kJ\times mol^{-1}}$

  • Formation of gaseous oxide ions in two successive steps:

$\mathrm{O(g)} + \mathrm{e^-} \to \mathrm{O^-(g)}$, $\quad \Delta H_{\mathrm{ae1}} = -141\,\mathrm{kJ\times mol^{-1}}$

$\mathrm{O^-(g)} + \mathrm{e^-} \to \mathrm{O^{2-}(g)}$, $\quad \Delta H_{\mathrm{ae2}} = +798\,\mathrm{kJ\times mol^{-1}}$

  • Lattice enthalpy of formation of $\mathrm{CaO(s)}$: $\Delta H_{\mathrm{ret}} = -3401\,\mathrm{kJ\times mol^{-1}}$
  1. a) Explain why the second electron affinity process $(\Delta H_{\mathrm{ae2}})$ has a positive (endothermic) sign.
  2. b) Calculate the standard enthalpy of formation, $\Delta H_{\mathrm{f}}^\circ$, of $\mathrm{CaO(s)}$ and justify how it is possible for a stable compound to form containing the $\mathrm{O^{2-}}$ ion even though the addition of the second electron to oxygen is thermodynamically unfavourable in the gas phase.
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Partial covalent character: Fajans' rules

The pure ionic model treats ions as hard, perfectly spherical charges. Real ions are not always like that, and the lattice enthalpy from a Born–Haber cycle (the experimental value) can be larger than the value calculated from the pure ionic model (the theoretical value, for example from the Born–Landé equation).

A small cation distorting the electron cloud of a large anion, giving partial covalent character polarising cation distorted anion cloud
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The thermodynamic data at $298\,\mathrm{K}$ for the silver halides $\mathrm{AgF}$ and $\mathrm{AgCl}$ indicate that their experimental lattice enthalpies (determined by…

The thermodynamic data at $298\,\mathrm{K}$ for the silver halides $\mathrm{AgF}$ and $\mathrm{AgCl}$ indicate that their experimental lattice enthalpies (determined by Born–Haber cycles, in modulus) are: $|\Delta H_{\mathrm{ret}}^{\mathrm{exp}}(\mathrm{AgF})| = 958\,\mathrm{kJ\times mol^{-1}}$ and $|\Delta H_{\mathrm{ret}}^{\mathrm{exp}}(\mathrm{AgCl})| = 905\,\mathrm{kJ\times mol^{-1}}$.

In addition, purely electrostatic calculations based on the theoretical Born–Landé equation (assuming a pure ionic model with spherical, non-polarisable ions) give a value of $|\Delta H_{\mathrm{ret}}^{\mathrm{teor}}(\mathrm{AgCl})| = 833\,\mathrm{kJ\times mol^{-1}}$ for $\mathrm{AgCl}$.

  1. a) Calculate the difference between the experimental and theoretical values for $\mathrm{AgCl}$ and explain the cause of this quantitative discrepancy, referring to Fajans' rules and the polarisability of the ions.
  2. b) Predict, with justification, whether the percentage discrepancy from the pure ionic model will be greater or smaller in $\mathrm{AgI}$ than in $\mathrm{AgCl}$, considering the radii of $\mathrm{Cl^-} (181\,\mathrm{pm})$ and $\mathrm{I^-} (220\,\mathrm{pm})$. As a qualitative model, Fajans' rules relate greater distortion of the anion's electron cloud to greater covalent character; larger anions tend to be more polarisable. The prediction requested is qualitative; a percentage cannot be calculated without further data.
Solve

Back to the start: the grit salt on the road is a lattice, not a pile of molecules. Nobody can measure directly the energy needed to pull one mole of $\mathrm{NaCl}$ apart into gaseous ions. The Born–Haber cycle lets us deduce it: it is $+787\ \mathrm{kJ}$, the same amount that is released when the lattice forms ($\Delta H_{\mathrm{latt}} = -787\ \mathrm{kJ\ mol^{-1}}$). This is why salt is a stable crystalline solid with a high melting point, and why a Born–Haber cycle is the way to find that energy.

Summary

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