Question

Use the following to calculate \(\Delta H_{\text {latice }}^{\circ}\) of \(\mathrm{MgF}_{2}\) : \(\begin{array}{llr}\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}(g) & \Delta H^{\circ}= & 148 \mathrm{~kJ} \\ \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g) & \Delta H^{\circ}= & 159 \mathrm{~kJ} \\ \mathrm{Mg}(g) \longrightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-} & \Delta H^{\circ}= & 738 \mathrm{~kJ} \\ \mathrm{Mg}^{+}(g) \longrightarrow \mathrm{Mg}^{2+}(g)+\mathrm{e}^{-} & \Delta H^{\circ}= & 1450 \mathrm{~kJ} \\\ \mathrm{~F}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{F}^{-}(g) & \Delta H^{\circ}= & -328 \mathrm{~kJ} \\ \mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s) & \Delta H^{\circ}=-1123 \mathrm{~kJ}\end{array}\) Compared with the lattice energy of LiF ( \(1050 \mathrm{~kJ} / \mathrm{mol}\) ) or the lattice energy you calculated for \(\mathrm{NaCl}\) in Problem \(9.30,\) does the relative magnitude of the value for \(\mathrm{MgF}_{2}\) surprise you? Explain.

Short answer

The calculated lattice enthalpy of \text{MgF}_{2} is 716 kJ/mol. This is lower than that of LiF (1050 kJ/mol) due to different ion charges and ionic radii.

Step-by-step solution

- Sublimation of Magnesium

Convert solid magnesium (\text{Mg}(s)) to gas (\text{Mg}(g)). This step requires 148 kJ/mol: \[\text{Mg}(s) \rightarrow \text{Mg}(g) \quad \Delta H^{\circ} = 148 \text{ kJ/mol} \]

- Bond Dissociation of Fluorine

Convert \text{F}_{2}(g) to 2 \text{F}(g). This step requires 159 kJ/mol: \[\text{F}_{2}(g) \rightarrow 2\text{F}(g) \quad \Delta H^{\circ} = 159 \text{ kJ/mol} \]

- First Ionization Energy of Magnesium

Remove the first electron from \text{Mg}(g) to form \text{Mg}^{+}(g). This step requires 738 kJ/mol: \[\text{Mg}(g) \rightarrow \text{Mg}^{+}(g) + e^{-} \quad \Delta H^{\circ} = 738 \text{ kJ/mol} \]

- Second Ionization Energy of Magnesium

Remove the second electron from \text{Mg}^{+}(g) to form \text{Mg}^{2+}(g). This step requires 1450 kJ/mol: \[\text{Mg}^{+}(g) \rightarrow \text{Mg}^{2+}(g) + e^{-} \quad \Delta H^{\circ} = 1450 \text{ kJ/mol} \]

- Electron Affinity of Fluorine

Add one electron each to two \text{F}(g) atoms to form two \text{F}^{-}(g) ions. This releases 328 kJ/mol for each fluorine atom: \[\text{2F}(g) + 2e^{-} \rightarrow 2\text{F}^{-}(g) \quad \Delta H^{\circ} = 2 \times (-328) = -656 \text{ kJ/mol} \]

- Formation Energy of Magnesium Fluoride

Convert \text{Mg}(s) and \text{F}_{2}(g) to solid \text{MgF}_{2}(s). This releases 1123 kJ/mol: \[\text{Mg}(s) + \text{F}_{2}(g) \rightarrow \text{MgF}_{2}(s) \quad \Delta H^{\circ} = -1123 \text{ kJ/mol} \]

- Calculation of the Lattice Enthalpy

Apply Hess's Law and sum the enthalpies of all the steps. The lattice enthalpy is the energy required minus the energy released: \[\Delta H_\text{lattice}^{\circ} = 148 + 159 + 738 + 1450 - 656 - 1123 \quad \Delta H_\text{lattice}^{\circ} = 716 \text{ kJ/mol} \]

- Comparison with LiF and NaCl

Compare the calculated lattice enthalpy of MgF_{2} (716 kJ/mol) with known values. The lattice enthalpy for LiF is 1050 kJ/mol, significantly higher than MgF_{2}. The calculated lattice enthalpy for NaCl should be compared similarly. This difference can be explained by considering the different ionic radii and charges of the ions involved. Mg^{2+} has a higher charge than Li^{+} and Na^{+}, affecting the lattice energy.

Key concepts

Hess's Law

Hess's Law is a fundamental principle used to calculate the total enthalpy change for a reaction. The law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction path. This means you can add up the enthalpy changes of individual steps to find the overall enthalpy change of the reaction.
The significance of Hess's Law in this context is to simplify the calculation of the lattice energy of MgF2. By breaking down the reaction into smaller steps for which the enthalpy changes are known, you can apply Hess's Law to find the total lattice energy.
This method is particularly useful for complex reactions where direct measurement of enthalpy change is challenging.

Sublimation Energy

Sublimation energy is the energy required to convert a substance from a solid to a gas phase without passing through the liquid phase. For magnesium (\text{Mg}(s)), the sublimation energy is the amount of energy needed to convert solid magnesium into gaseous magnesium.
In our calculation, the sublimation energy for Mg is 148 kJ/mol:
\text{Mg}(s) \rightarrow \text{Mg}(g) \tag{\(\text{ΔH}^{\text{°}} = 148 \text{kJ/mol}\)}
This step is essential because you need Mg in the gaseous state before you can ionize it to form \text{Mg}^{2+} ions, which are necessary for the formation of MgF2.

Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion in the gas phase. For magnesium, you need two ionization steps to form a \text{Mg}^{2+} ion.

• The first ionization energy removes the first electron:
  \text{Mg}(g) \rightarrow \text{Mg}^{+}(g) + \text{e}^{−} \tag{\(ΔH^{\text{°}} = 738 \text{kJ/mol}\)}


• The second ionization energy removes the second electron:
  \text{Mg}^{+}(g) \rightarrow \text{Mg}^{2+}(g) + \text{e}^{−} \tag{\(ΔH^{\text{°}} = 1450 \text{kJ/mol}\)}

Combining these two steps gives you the total energy needed to convert \text{Mg}(g) to \text{Mg}^{2+}(g). This energy is critical for calculating the lattice energy, as \text{Mg}^{2+} ions are key components of MgF2.

Bond Dissociation Energy

Bond dissociation energy is the energy required to break a bond between two atoms in a molecule. In this case, you need to dissociate a molecule of fluorine gas (F2) into two fluorine atoms.
This step requires 159 kJ/mol:
\text{F}_{2}(g) \rightarrow 2\text{F}(g) \tag{\(ΔH^{\text{°}} = 159 \text{kJ/mol}\)}
Breaking the molecule into individual atoms is necessary because, in the formation of MgF2, you need fluorine atoms to pair with the magnesium ions.

Electron Affinity

Electron affinity is the energy change that occurs when an electron is added to a neutral atom in the gas phase to form a negative ion. For fluorine, adding an electron to form \text{F}^{−} releases energy.

• The electron affinity for one fluorine atom is −328 kJ/mol:
  \text{F}(g) + \text{e}^{-} \rightarrow \text{F}^{-}(g) \tag{\(ΔH^{\text{°}} = -328 \text{kJ/mol}\)}


• Since you need two fluoride ions (\text{F}^{−}) to pair with one \text{Mg}^{2+}, the total energy release is:
  \text{2F}(g) + 2\text{e}^{-} \rightarrow 2\text{F}^{-}(g) \tag{\(ΔH^{\text{°}} = 2 \times (-328) = -656 \text{kJ/mol}\)}

Electron affinity helps explain why energy is released when fluorine atoms gain electrons to become fluoride ions, contributing to the total lattice energy calculation for MgF2.