Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{~F}_{2}\left(\mathrm{O}_{2} \mathrm{~F}_{2}\right.\) exists as \(\left.\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F}\right)\). Assign oxidation states and formal charges to the atoms in \(\mathrm{O}_{2} \mathrm{~F}_{2}\). This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{~F}_{2}\) ?

To write the Lewis structure for \(\mathrm{O}_{2}\mathrm{F}_{2}\), we follow these steps: 1. Calculate the total number of valence electrons available: For oxygen, O, there are 6 valence electrons. For fluorine, F, there are 7 valence electrons. Since there are 2 oxygen and 2 fluorine atoms: Total valence electrons = 2(6)+2(7) = 12+14 = 26 valence electrons. 2. Choose the central atom(s): In \(\mathrm{O}_{2}\mathrm{F}_{2}\), we are given the atoms arrangement as \(\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F}\), so the oxygen atoms will be the central atoms. 3. Place the atoms and distribute the electrons around them: \[ \underbrace{\cdot \mathrm{F} \cdot}_{2 \ \mathrm{electrons}} - (\cdot \mathrm{O} \cdot)\overset{8 \ \mathrm{electrons}}{-} (\cdot \mathrm{O} \cdot) - \underbrace{\cdot \mathrm{F} \cdot}_{2 \ \mathrm{electrons}} \] 4. Place the remaining electrons on the central atoms: We still have (26 - 12 = 14) electrons to distribute on the oxygen atoms. We can place 7 on each oxygen atom. \[ \mathrm{F} - \underbrace{[\cdot \mathrm{O} \cdot]\cdots [\cdot \mathrm{O} \cdot]}_{14 \ \mathrm{electrons}} - \mathrm{F} \] So, the final Lewis structure for \(\mathrm{O}_{2}\mathrm{F}_{2}\) is: \[ \mathrm{F} - [\mathrm{O}(::)O] - \mathrm{F} \]

Assign oxidation states: Use the oxidation state rules to determine the oxidation states of oxygen and fluorine in \(\mathrm{O}_{2}\mathrm{F}_{2}\) Oxygen: Usually has an oxidation state of -2. Fluorine: Always has an oxidation state of -1 since it's the most electronegative element. For \(\mathrm{O}_{2}\mathrm{F}_{2}\): - The oxidation state of central oxygen atoms is +1 each (because they are bonded with oxygen and fluorine atoms). - The oxidation state of fluorine atoms is -1 each. Assign formal charges: - Oxygen: The formal charge for each central oxygen atom = (6 - 4 - 2) = 0 - Fluorine: The formal charge for each fluorine atom = (7 - 1 - 6) = 0

Here, it's important to consider the distinction between oxidation state and formal charge. - Oxidation states provide an insight into the loss or gain of electrons due to a change in the element's valence state during a chemical reaction. - Formal charges, on the other hand, provide information about the distribution of electrons among the atoms in a molecule. In the case of \(\mathrm{O}_{2}\mathrm{F}_{2}\), the oxidation states (+1 for central oxygen atoms and -1 for fluorine atoms) give a better indication of the compound's oxidizing and fluorinating properties. This is because the oxidation states explain the potential of the molecule to get reduced (accept electrons) by other elements, making it a vigorous oxidizing agent, and aid in the transfer of fluoride atoms to other substrates, bestowing it with fluorinating properties. In summary, oxidation states are more useful than formal charges in accounting for the oxidizing and fluorinating properties of \(\mathrm{O}_{2}\mathrm{F}_{2}\).