Exercises: Ionic bonding and Born–Haber cycles

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.