The first ionization energy of the oxygen molecule is the energy required for the following process: $$\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-}$$ The energy needed for this process is 1175 \(\mathrm{kJ} / \mathrm{mol}\) , very similar to the first ionization energy of \(\mathrm{Xe} .\) Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2} ?\) If so, suggest a product or products of this reaction.

Given that the first ionization energy for O₂ is 1175 kJ/mol, we can notice that it is very similar to the first ionization energy for Xe. This similarity suggests that O₂ and Xe might have comparable chemical properties and reactivity, which may hint at the possibility of O₂ reacting with F₂.

As O₂ has a comparable first ionization energy to Xe, and considering that Xe is an inert noble gas and generally does not react with other elements, this makes it less likely for O₂ to react with F₂. However, it is essential to note that O₂ still has a higher reactivity compared to noble gases like Xe due to its double bond character and oxygen's high electronegativity. Also, F₂ is a highly reactive element among halogens. Thus, the chances of O₂ reacting with F₂ cannot be ruled out.

Given the possibility of O₂ reacting with F₂, we can predict the products that can form during this process. Oxygen, with an electronegativity of 3.44, doesn't generally form ionic compounds with other non-metals, so we can expect the formation of a covalent compound in this case. The most likely product resulting from the reaction of O₂ and F₂ is dioxygen difluoride, which has the formula O₂F₂. This compound can be formed as follows: \( \mathrm{O}_{2}(g) + 2\,\mathrm{F}_{2}(g) \longrightarrow 2\,\mathrm{O}_{2}\mathrm{F}_{2}(s) \) In conclusion, it is possible for O₂ to react with F₂ due to the higher reactivity of both the elements, and we can expect the formation of dioxygen difluoride (O₂F₂) as the product of this reaction.