Give the Lewis structure, molecular structure, and hybridization of the oxygen atom for \(\mathrm{OF}_{2} .\) Would you expect \(\mathrm{OF}_{2}\) to be a strong oxidizing agent like \(\mathrm{O}_{2} \mathrm{F}_{2}\) discussed in Exercise 67\(?\)

To determine the Lewis structure of OF2, follow these steps: 1. Find the total number of valence electrons. Oxygen has 6 valence electrons and each fluorine atom has 7 valence. Therefore, the total number of valence electrons present in OF2 is 6 + 2(7) = 20 electrons. 2. Place oxygen in the center and fluorine atoms on either side. 3. Distribute the valence electrons: Start by creating single bonds between the central oxygen atom and both fluorine atoms by sharing 2 electrons. There are now 16 electrons left for distribution. 4. Complete the octet for the surrounding atoms (fluorine): Add 3 lone pairs (6 electrons) to each fluorine atom. The fluorine atoms now have an octet. 5. Remaining electrons, if any, will be placed on the central atom (oxygen). After completing the octet for both fluorine atoms, no electrons are remaining. The Lewis structure of OF2 is ``` F \ O / F ```

To find the molecular structure of OF2, we must understand VSEPR (Valence Shell Electron Pair Repulsion) theory, which helps us predict the geometry of molecules. 1. Count the number of electron groups surrounding the central oxygen atom: There are 2 bonded atoms (fluorine) and 2 lone pairs. 2. Determine the molecular geometry based on the VSEPR table. In this case, OF2 has 4 electron groups (2 atoms and 2 lone pairs), making it a tetrahedral electron geometry. However, only 2 atoms are bonded, making the molecular geometry of OF2 bent.

To determine the hybridization of the oxygen atom in OF2, follow these steps: 1. Count the total number of electron groups surrounding the central oxygen atom. There are 2 bonding groups (from two Fluorine atoms) and 2 lone pairs on the oxygen atom, for a total of 4 electron groups. 2. Based on the number of electron groups, assign the hybridization state: 4 electron groups correspond to sp3 hybridization. The oxygen atom in OF2 is sp3 hybridized.

To determine if OF2 is a strong oxidizing agent, we examine its oxidation number and compare it to O2F2. 1. Assign oxidation numbers to the atoms in OF2. Oxygen (-2) and Fluorine (+1). 2. Compare the oxidation numbers to O2F2: - In OF2, the oxygen atom has an oxidation state of -2, and wants to gain electrons to become more stable, making it a good oxidizing agent. - In O2F2, the oxygen atom has an oxidation state of +1, and wants to lose electrons to become more stable, making it a strong oxidizing agent. Comparing their oxidation states, OF2 is a good oxidizing agent, but it is not as strong as O2F2.