Question

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 \(\mathrm{O}_{2} \mathrm{~F}_{2}\) ?

Short answer

The Lewis structure for \(\mathrm{O}_{2}\mathrm{F}_{2}\) is given as: \[ \mathrm{F} - [\mathrm{O}(::)O] - \mathrm{F} \] Oxidation states in \(\mathrm{O}_{2}\mathrm{F}_{2}\): - Central oxygen atoms: +1 each - Fluorine atoms: -1 each Formal charges in \(\mathrm{O}_{2}\mathrm{F}_{2}\): - Central oxygen atoms: 0 each - Fluorine atoms: 0 each Oxidation states better account for the oxidizing and fluorinating properties of \(\mathrm{O}_{2}\mathrm{F}_{2}\) as they explain the molecule's potential to be reduced and transfer fluoride atoms during chemical reactions.

Step-by-step solution

Write the Lewis structure

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 and formal charges

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

Discuss the usefulness of oxidation states and formal charges

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 \(\mathrm{O}_{2}\mathrm{F}_{2}\).

Key concepts

oxidation states

Understanding oxidation states is crucial in grasping the concept of electron transfer in chemical reactions. These states represent the hypothetical charge an atom would have if all bonds were ionic. In the compound - Oxygen usually exhibits an oxidation state of -2. However, in for each central oxygen atom, the oxidation state is +1, as it is bonded to both oxygen and fluorine atoms. - Fluorine consistently holds an oxidation state of -1, being the most electronegative element. Determining these states helps to understand why is a strong oxidizing agent. The increase in oxidation state from the typical -2 to +1 for oxygen suggests a willingness to accept electrons. This property aligns with its behavior as a potent oxidizer, making it capable of receiving electrons from other substances, consequently oxidizing them. Thus, oxidation states are excellent tools for predicting the electron-accepting tendencies of molecules like . By understanding oxidation states, we gain insight into the reaction capabilities of a molecule, specifically in the context of oxidation and reduction processes.

formal charges

Formal charges are a way to estimate the distribution of electrical charge in a molecule, based on valence electrons. They provide a model for understanding which atoms might have slightly more positive or negative characteristics within a molecule.
- To calculate formal charges, the formula is:\[\text{Formal charge} = \text{(Valence electrons)} - \text{(Non-bonding electrons)} - \frac{1}{2}\text{(Bonding electrons)}\]For :- Oxygen: Each central oxygen atom has a formal charge of zero, calculated as \(6\) valence electrons - \(4\) non-bonding electrons - \(2\) bonding electrons, which equals zero.- Fluorine: Each fluorine atom also maintains a formal charge of zero, calculated as \(7\) valence electrons - \(6\) non-bonding electrons - \(1\) bonding electron, resulting in zero.While formal charges are helpful in predicting charge distribution for complex reactions, in the case of , they do not contribute as much to our understanding of its reactivity as compared to oxidation states. Formal charges provide a simple way of checking whether a Lewis structure is correctly drawn, ensuring the charge is evenly distributed in the molecule. Understanding formal charges helps prevent drawing errors and adds an additional layer of analysis when interpreting molecular geometry and reactivity.

valence electrons

Valence electrons are those that reside in the outermost electron shell of an atom and are the primary electrons involved in chemical bonding. Understanding the count and arrangement of valence electrons is a fundamental step in drawing Lewis structures, which visually depict molecules.For , the total number of valence electrons involved is calculated as follows:

• Oxygen: Each atom has 6 valence electrons.
• Fluorine: Each atom has 7 valence electrons.

Adding them up:\[2(6) + 2(7) = 26 \text{ valence electrons}\]These 26 electrons must be distributed around the atoms to complete the structure, maintaining adherence to the octet rule where possible. Finding the correct allocation ensures minimal formal charge and the identification of possible double or lone pairs, creating a balanced and stable molecule. By mastering the accuracy of valence electron allocation, students can draw correct Lewis structures, which are crucial for predicting molecular shapes, bond angles, and potential reactivity. Properly accounting for valence electrons ensures a solid foundation not only in drawing Lewis structures but also in understanding the behavior and properties of different molecules.