Question

Use a Born-Haber cycle ( Sec. \(5-13\) ) to calculate the lattice energy of \(\mathrm{MgF}_{2}\) using these thermodynamic data. $$\begin{aligned} \Delta_{\text {sub }} H \mathrm{Mg}(\mathrm{s}) &=+146 \mathrm{~kJ} / \mathrm{mol} \\ \text { B.E. } \mathrm{F}_{2}(\mathrm{~g}) &=+158 \mathrm{~kJ} / \mathrm{mol} \\\ \text { I.E. } \mathrm{Mg}(\mathrm{g}) &=+738 \mathrm{~kJ} / \mathrm{mol} \\ \text { I.E. }_{2} \mathrm{Mg}^{+}(\mathrm{g}) &=+1450 \mathrm{~kJ} / \mathrm{mol} ; \\ \text { E.A. } \mathrm{F}(\mathrm{g}) &=-328 \mathrm{~kJ} / \mathrm{mol} \\ \Delta_{\mathrm{f}} H^{\circ} \mathrm{MgF}_{2}(\mathrm{~s}) &=-1124 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}$$ Compare this lattice energy with that of \(\mathrm{SrF}_{2}\), \(-2496 \mathrm{~kJ} / \mathrm{mol}\). Explain the difference in the values in structural terms.

Short answer

The lattice energy of \(\mathrm{MgF}_2\) is \(-2960\,\mathrm{kJ/mol}\), higher than \(\mathrm{SrF}_2\) due to Mg's smaller ion size leading to stronger attraction.

Step-by-step solution

Understanding the Born-Haber Cycle Components

To calculate the lattice energy of \(\mathrm{MgF}_2\) using a Born-Haber cycle, we'll combine various thermochemical processes that compose the formation of an ionic compound. The cycle will use the following thermodynamic data: sublimation energy of magnesium, bond dissociation energy of \(\mathrm{F}_2\), ionization energies of magnesium, electron affinity of fluorine, and the formation enthalpy of \(\mathrm{MgF}_2\).

Write the Process Equations

Start by converting the solid \(\mathrm{Mg}\) to gas: \( \text{Mg(s)} \rightarrow \text{Mg(g)} \) with \(\Delta_{\text{sub}}H = +146\, \text{kJ/mol}\). Next, dissociate the \(\text{F}_2\) molecule: \( \frac{1}{2} \text{F}_2(g) \rightarrow \text{F}(g) \) with \(\text{B.E.} = +158 \text{kJ/mol}, \text{per 2 moles, it's} +79 \text{kJ/mol per F}.\) For ionization, convert \(\text{Mg(g)} \rightarrow \text{Mg}^+(g) + e^-\) (\(\text{I.E._1 = +738\,\text{kJ/mol}}\)) and then \(\text{Mg}^+(g) + e^- \rightarrow \text{Mg}^{2+}(g) + 2e^- \) (\(\text{I.E._2 = +1450\,\text{kJ/mol}}\)). Finally, for electron affinity, convert \(\text{F}(g) + e^- \rightarrow \text{F}^- (g)\), \(\text{E.A. = -328\,\text{kJ/mol}}\) applied to two fluorines.

Calculating Enthalpy Changes

Calculate each process, sum up for magnesium: \(+146 + 738 + 1450 = +2334\,\text{kJ/mol}\). For a single \(\text{F}_2\) to two \(\text{F}^-\), calculate: \(2 \times (79 - 328) = -498\,\text{kJ/mol}\). Total as \(+2334 + (-498) = +1836\, \text{kJ/mol}.\)

Determine Lattice Energy Using Enthalpy of Formation

Using the enthalpy of formation of \(\mathrm{MgF}_2\), \( \Delta_f H^{\circ} = -1124 \text{kJ/mol},\) the cycle equation gives: \( \Delta_{\text{lat}}H = -1124 - (+1836) = -2960\, \text{kJ/mol}. \)

Comparing Lattice Energies and Structural Explanation

The calculated lattice energy of \(\mathrm{MgF}_2\) is \(-2960\,\text{kJ/mol}\), compared to \(\mathrm{SrF}_2\), which is \(-2496\,\text{kJ/mol}\). The difference is due to the smaller ionic radius of Mg compared to Sr, leading to stronger electrostatic attraction (higher lattice energy) when ions are closely packed in \(\mathrm{MgF}_2\).

Key concepts

Understanding Lattice Energy

Lattice energy is a crucial concept when studying ionic compounds. It refers to the energy required to separate one mole of a solid ionic compound into gaseous ions. This energy is significant because it indicates the strength of the bonds holding the compound together.

For example, in the case of \( \text{MgF}_2 \), the calculated lattice energy is \(-2960 \, \text{kJ/mol}\). Such a high value signifies a strong attraction between ions in the solid state.

Lattice energy can be affected by the size of the ions and their charges. Smaller ions or ions with higher charges will typically have higher lattice energies. This is due to the increased electrostatic forces as ions are closer together or have greater charge magnitudes.

Thermodynamic Data in the Born-Haber Cycle

Thermodynamic data provides essential information for calculating lattice energies through the Born-Haber cycle. Several components contribute to this process:

• Sublimation energy: The energy required to turn a solid atom into a gas, such as magnesium sublimating with \(+146\, \text{kJ/mol}\).
• Bond dissociation energy (B.E.): The energy needed to break bonds in molecules; for \(\text{F}_2\), this is \(+158\, \text{kJ/mol}\) but halved for one fluorine.
• Ionization energy (I.E.): The energy to remove electrons from gaseous atoms or ions; magnesium requires \(+738\, \text{kJ/mol}\) and further \(+1450\, \text{kJ/mol}\) to reach \(\text{Mg}^{2+}\).
• Electron affinity (E.A.): The energy released when an electron is added to a gaseous atom; fluorine gains electrons with \(-328\, \text{kJ/mol}\).
• Enthalpy of formation \((\Delta_f H^{\circ})\): The energy change when forming a compound from its elements; for \(\text{MgF}_2\), this is \(-1124\, \text{kJ/mol}\).

Using these values, the Born-Haber cycle allows us to derive the lattice energy by balancing these energy changes.

Ionic Compound Formation

The formation of ionic compounds involves transferring electrons between atoms, resulting in ions. These oppositely charged ions attract each other, forming strong ionic bonds.

In \(\text{MgF}_2\), magnesium loses two electrons to form \(\text{Mg}^{2+}\), while each fluorine gains one electron to form \(\text{F}^-\). These ions then organize into a crystal lattice, maximizing attraction and minimizing repulsion.

The structure and composition of \(\text{MgF}_2\) versus \(\text{SrF}_2\) illustrate how ionic sizes influence lattice energy. Magnesium's smaller ionic radius results in closer packing than strontium, leading to a higher lattice energy for \(\text{MgF}_2\). This highlights the relationship between ionic structure and properties, crucial for understanding material characteristics.