Question

Compare the bond enthalpy of \(\mathrm{F}_{2}\) with the energy change for this process: $$ \mathrm{F}_{2}(g) \longrightarrow \mathrm{F}^{+}(g)+\mathrm{F}^{-}(g) $$ Which is the preferred dissociation for \(\mathrm{F}_{2},\) energetically speaking?

Short answer

After comparing the bond enthalpy of \(F2\) and the energy change for the production of \(\mathrm{F}^+\) and \(\mathrm{F}^-\) ions, the preferred dissociation pathway is the one that involves less energy.

Step-by-step solution

Determine the bond enthalpy of \(F2\)

To start, it is necessary to find the bond enthalpy (often given in tables in chemistry textbooks or online resources) of the molecule \(F2\). This value represents the energy required to break the bond between the two fluorine atoms.

Calculate the energy for the production of ions

Now, calculate the energy change for the dissociation. This involves considering two processes: the ionization of a neutral fluorine atom to form \(\mathrm{F}^+\), which requires energy (ionization energy), and the gaining of an electron by a neutral fluorine atom to form \(\mathrm{F}^-\), which releases energy (electron affinity energy). The net energy required for this process is the difference between the ionization energy and the electron affinity energy.

Compare the energies

With the bond enthalpy and the ionization/electron affinity energies known, compare the two values. The process that requires less energy is the preferred dissociation pathway, energetically speaking, because it is more stable and easier for the molecule to undergo.

Key concepts

F2 Molecule

The \( \mathrm{F}_{2} \) molecule, commonly known as a fluorine molecule, consists of two fluorine atoms connected by a single covalent bond. This bond forms because each fluorine atom shares one unpaired electron with the other, resulting in a stable electron arrangement.
This type of bond is characterized as a diatomic molecule, meaning it contains two atoms.

• Fluorine is a halogen with a very high electronegativity, meaning it strongly attracts electrons.
• The bond between the fluorine atoms in \( \mathrm{F}_{2} \) is relatively weak compared to other halogen bonds.
• This bond weakness is because both atoms are highly electronegative, leading to repulsion between the shared electrons.

Due to this weak bond, the bond enthalpy of \( \mathrm{F}_{2} \) is lower, making it easier to dissociate the molecule into individual fluorine atoms.

Ionization Energy

Ionization energy refers to the amount of energy needed to remove an electron from an atom or molecule to form a cation. In the context of \( \mathrm{F}_{2} \), when one fluorine atom loses an electron to form \( \mathrm{F}^+ \), energy is required to overcome the attraction between the positively charged nucleus and the negatively charged electron.

• For fluorine, the ionization energy is quite high because it requires removing one electron from its stable electronic configuration.
• A high ionization energy indicates strong binding between electrons and the nucleus, making electron removal difficult and energy-demanding.
• In the process \( \mathrm{F}_{2}(g) \rightarrow \mathrm{F}^{+}(g)+\mathrm{F}^{-}(g) \), only one of the fluorine atoms undergoes ionization.

Understanding ionization energy helps in analyzing dissociation pathways, especially when comparing energy changes during bond breaking and ion formation.

Electron Affinity

Electron affinity describes the energy change that occurs when an electron is added to a neutral atom, forming an anion. For fluorine, it signifies the energy released when a fluorine atom gains an electron to become \( \mathrm{F}^- \).

• Fluorine has one of the highest electron affinities, meaning it releases a lot of energy when gaining an electron.
• The high electron affinity of fluorine is due to its small size and high electronegativity, allowing it to pull an electron tightly to achieve a stable electron configuration.
• During \( \mathrm{F}_{2}(g) \rightarrow \mathrm{F}^{+}(g)+\mathrm{F}^{-}(g) \), the electron affinity plays a crucial role in balancing the high ionization energy required.

Therefore, electron affinity is important in determining the stability of the resulting ion and in evaluating the net energy change during molecular dissociation.

Dissociation Pathways

Dissociation pathways refer to the different methods by which a molecule can break into smaller components or ions. For \( \mathrm{F}_{2} \), two possible dissociation pathways exist:
1. **Breaking into Neutral Atoms:**

• This occurs by simply breaking the bond between the two fluorine atoms, resulting in neutral fluorine atoms.
• The energy required here is equal to the bond enthalpy of \( \mathrm{F}_{2} \).

2. **Formation of Ions (\( \mathrm{F}^+ \) and \( \mathrm{F}^- \))**

• This involves removing an electron from one fluorine atom (requiring ionization energy) and gaining an electron for the other atom (releasing energy via electron affinity).
• The net energy change for this process is the difference between the ionization energy and electron affinity.

Comparing these dissociation pathways helps in understanding the energetically preferred method. For fluorine, breaking the molecule into neutral atoms tends to require less energy than forming ions, emphasizing the importance of considering bond enthalpy and energy changes in dissociation processes.