Question

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\(?\)

Short answer

The Lewis structure of OF2 is: ``` F \ O / F ``` The molecular geometry of OF2 is bent, and the hybridization of the oxygen atom in OF2 is sp3. While OF2 is a good oxidizing agent due to the -2 oxidation state of the oxygen atom, it is not as strong an oxidizing agent as O2F2, where oxygen has a +1 oxidation state.

Step-by-step solution

Determine the Lewis Structure of OF2

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 ```

Determine the Molecular Structure of OF2

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.

Determine the Hybridization of the Oxygen Atom in OF2

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.

Determine if OF2 is a strong oxidizing agent like O2F2

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.

Key concepts

Molecular Geometry

Molecular geometry is the 3D arrangement of atoms within a molecule. It determines many physical properties like boiling point, polarity, and reactivity. For oxygen difluoride (\(\mathrm{OF}_2\)), the molecular geometry can be deduced from the distribution of electrons around the central oxygen atom. In OF2, oxygen acts as the central atom surrounded by two fluorine atoms and has two lone pairs of electrons. This setup results in a "bent" molecular geometry. This is similar to the geometry of water (H2O), where the presence of lone electron pairs causes repulsion, thereby creating an angle between the bonded pairs.

Oxidizing Agent

An oxidizing agent is a substance that gains electrons in a chemical reaction, causing another substance to be oxidized. It's an electron acceptor. In the case of OF2, although it acts as an oxidizing agent, it's not as potent as \(\mathrm{O}_{2}\mathrm{F}_2\). This is due to the difference in oxidation states of the oxygen atom in the compounds. In OF2, oxygen has a stable oxidation state of -2, whereas in \(\mathrm{O}_{2}\mathrm{F}_{2}\), it has a +1 state making it strongly seek electrons to attain stability.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory is used to predict the shape of individual molecules. It posits that electron pairs around a central atom in a molecule will arrange themselves as far apart as possible to minimize repulsion. In \(\mathrm{OF}_2\), VSEPR theory identifies 4 electron groups (two bonding pairs and two lone pairs) around the oxygen. The repellant effect of lone pairs makes the bonding pairs closer together, resulting in a bent molecular geometry.

Hybridization

Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals suitable for pairing electrons in chemical bonds. For \(\mathrm{OF}_2\), the oxygen atom exhibits \(sp^3\) hybridization. This occurs because the central oxygen has 4 electron groups—2 from bonding with fluorines and 2 as lone pairs. The \(sp^3\) hybrid orbitals result from mixing one s orbital and three p orbitals, which gives OF2 its bent shape.

Valence Electrons

Valence electrons are the electrons present in the outermost shell of an atom and are crucial in determining how an atom bonds with others. For \(\mathrm{OF}_2\), understanding valence electrons aids in drawing its Lewis structure. Oxygen, with 6 valence electrons, bonds with two fluorine atoms, each having 7 valence electrons. This accounts for a total of 20 valence electrons, distributed in a way to satisfy the octet rule for each atom, allowing us to predict the molecule's behavior and reactivity.