Question

(a) From the following data calculate the bond enthalpy of the \(\mathrm{F}_{2}^{-}\) ion. $$ \begin{array}{ll} \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=156.9 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}^{-}(g) \longrightarrow \mathrm{F}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=333 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}_{2}^{-}(g) \longrightarrow \mathrm{F}_{2}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=290 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ (b) Explain the difference between the bond enthalpies of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{-}\).

Short answer

The bond enthalpy for \( \mathrm{F}_{2}^{-} \) is 446.9 kJ/mol and is higher than \( \mathrm{F}_2 \) due to added electron repulsion.

Step-by-step solution

Understand the Given Data

The provided data includes the reaction enthalpy for three processes: the dissociation of \( \mathrm{F}_{2}(g) \) into two \( \mathrm{F}(g) \) atoms, the ionization of \( \mathrm{F}^{-}(g) \) to \( \mathrm{F}(g) \) plus an electron, and the ionization of \( \mathrm{F}_{2}^{-}(g) \) to \( \mathrm{F}_{2}(g) \) plus an electron. These reactions have respective enthalpies of \( 156.9 \mathrm{~kJ/mol} \), \( 333 \mathrm{~kJ/mol} \), and \( 290 \mathrm{~kJ/mol} \).

Break Down the Process for Calculating \( \mathrm{F}_{2}^{-} \) Bond Enthalpy

To find the bond enthalpy of \( \mathrm{F}_{2}^{-} \), we need to determine the energy required for the dissociation reaction \( \mathrm{F}_{2}^{-}(g) \rightarrow 2 \mathrm{~F}(g) + e^{-} \). We'll do this using Hess's law by combining the given data.

Apply Hess's Law to Entropic Changes

Using Hess's Law, we can write the following equation to relate the reactions: \[\Delta H_{\mathrm{F}_{2}^{-} \rightarrow 2\mathrm{~F}(g) + e^{-}} = \Delta H_{\mathrm{F}_{2}^{-}(g) \rightarrow \mathrm{F}_{2}(g) + e^{-}} + \Delta H_{\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{~F}(g)}\]Substitute the given values:\[\Delta H_{\mathrm{F}_{2}^{-} \rightarrow 2\mathrm{~F}(g) + e^{-}} = 290 \text{ kJ/mol} + 156.9 \text{ kJ/mol} = 446.9 \text{ kJ/mol}\]

Explain the Difference in Bond Enthalpies

The bond enthalpy of \( \mathrm{F}_{2}^{-} \) is different from that of \( \mathrm{F}_{2} \) due to the interaction of the extra electron in \( \mathrm{F}_{2}^{-} \) with the \( \mathrm{F}_2 \) molecule. This added charge increases electron-electron repulsion within the molecule, causing a weaker bond compared to neutral \( \mathrm{F}_2 \).

Key concepts

Hess's Law

Hess's Law is a foundational principle in thermodynamics that helps us calculate the total enthalpy change in a chemical reaction by summing up the enthalpy changes of individual steps. Essentially, it tells us that the total energy change is the same, regardless of the path taken, as long as the initial and final states are the same. In the context of our exercise, we utilize Hess's Law to find the bond enthalpy of the \( \mathrm{F}_2^- \) ion.
By adding the enthalpy changes of the given reactions, students can calculate the desired bond enthalpy effectively. For example, we take the reaction for the ionization of \( \mathrm{F}_2^- \), which transforms it into \( \mathrm{F}_2 \) and an electron, and sum it with the dissociation of \( \mathrm{F}_2 \) into two \( \mathrm{F} \) atoms.
These steps combined provide the complete picture of the transition from \( \mathrm{F}_2^- \) to isolated \( \mathrm{F} \) atoms, and using Hess's Law allows us to sum the enthalpy changes, leading us to the final value of bond enthalpy.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom or molecule in the gas phase. It plays a crucial role in determining the stability and reactivity of ions. In our given exercise, the ionization process of \( \mathrm{F}^- \) to \( \mathrm{F} \) includes an energy change of 333 kJ/mol. This energy reflects how strongly electrons are held in the ion.
Understanding ionization energy is important because it highlights the forces holding the electrons within an ion. The greater the ionization energy, the more strongly an electron is bounded to its nucleus. This principle helps to explain why ions are stable and why certain reactions may favor the formation of ions over neutral atoms.
In the context of our example with \( \mathrm{F}_2^- \), the ionization energy required to convert it back into \( \mathrm{F}_2 \) plus an electron (290 kJ/mol) is a critical part of calculating changes in enthalpy during transformations. It provides insight into how much energy is involved in the initial steps of chemical changes.

Electron-Electron Repulsion

Electron-electron repulsion is a key concept in understanding molecular bonding and stability. It describes the repulsive interaction that occurs between electrons due to their like charges. In molecules, this repulsion plays a significant role in determining molecular geometry and bond strength.
In the context of \( \mathrm{F}_2^- \) ion, the additional electron compared to the neutral \( \mathrm{F}_2 \) molecule increases the repulsion between the electrons. This increased repulsion weakens the overall bond strength as it causes the electrons to spread out more, thus decreasing attraction between the nuclei.
Hence, the bond enthalpy of \( \mathrm{F}_2^- \) is lower compared to \( \mathrm{F}_2 \), reflecting a weaker bond due to these increased repulsive forces. Understanding this principle allows students to appreciate why changes in electronic configuration can lead to variations in molecular properties and reactivity.