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

Expert verified
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

01

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.
02

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.
03

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.

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