Question

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{F}_{2}\) . Predict the bond angles and hybridization of the two central oxygen atoms. Assign oxidation states and formal charges to the atoms in $\mathrm{O}_{2} \mathrm{F}_{2} .\( The compound \)\mathrm{O}_{2} \mathrm{F}_{2}$ 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 of \(\mathrm{O}_{2} \mathrm{F}_{2}\) consists of two central oxygen atoms connected by a single bond, with each oxygen atom singly bonded to two peripheral fluorine atoms. The central oxygen atoms have \(sp^3\) hybridization and the bond angles are approximately 109°. The oxidation states for oxygen and fluorine in \(\mathrm{O}_{2} \mathrm{F}_{2}\) are -1 and +2 for the oxygen atoms, and -1 for the fluorine atoms, respectively. Formal charges are 0 for the oxygen with a double bond, -1 for the oxygen with a single bond to fluorine, and 0 for each fluorine atom. Oxidation states are more useful in accounting for the oxidizing and fluorinating properties of \(\mathrm{O}_{2} \mathrm{F}_{2}\), as they indicate the tendency of the atoms to gain or lose electrons during a reaction.

Step-by-step solution

Write the Lewis Structure for \(\mathrm{O}_{2} \mathrm{F}_{2}\)

First, calculate the total number of valence electrons in the compound: - Oxygen has 6 valence electrons, and there are 2 oxygen atoms, for a total of 12 electrons. - Fluorine has 7 valence electrons, and there are 2 fluorine atoms, for a total of 14 electrons. So the total number of valence electrons is 12 + 14 = 26. Now, follow these steps to construct the Lewis structure: 1. Place the central atoms, in this case, the two oxygen atoms, and connect them with a single bond. 2. Connect the two peripheral fluorine atoms with single bonds to each oxygen atom. 3. Distribute the remaining valence electrons as lone pairs: - Oxygen needs two more electrons to complete its octet. - Fluorine needs six more electrons to complete its octet. 4. Check if the octet rule is satisfied for each atom. The Lewis structure for \(\mathrm{O}_{2} \mathrm{F}_{2}\) can be represented as: F F | | O - O | | F F

Predict the bond angles and hybridization of the two central oxygen atoms

To predict the bond angles and hybridization, apply the VSEPR theory (Valence Shell Electron Pair Repulsion theory): 1. Count the electron groups around the oxygen atoms (a single bond, a double bond, a triple bond, or a lone pair). In this case, each oxygen has 4 electron groups (2 single bonds and 2 lone pairs). 2. Determine the hybridization: Since there are 4 electron groups, the hybridization will be \(sp^3\). 3. Determine the bond angles: The ideal bond angle for an \(sp^3\) hybridized atom is 109.5°. However, since lone pairs are more repulsive, the bond angles between the single bonds will be slightly smaller, closer to 109°. Therefore, the hybridization of the two central oxygen atoms is \(sp^3\), and the bond angles are approximately 109°.

Assign oxidation states and formal charges to the atoms in \(\mathrm{O}_{2} \mathrm{F}_{2}\)

1. Assign oxidation states: - Oxygen atom with a double bond: -1 (since it's more electronegative than the other oxygen) - Oxygen atom with a single bond to Fluorine: +2 - Fluorine atoms: -1 (most electronegative element) 2. Assign formal charges for each atom: - Formal charge on the oxygen atom with a double bond: 6 (valence electrons for O) - 6 (assigned electrons: 4 from lone pairs + 2 from the bond) = 0 - Formal charge on the oxygen atom with a single bond to Fluorine: 6 (valence electrons for O) - 7 (assigned electrons: 4 from lone pairs + 1 from the O-O bond + 2 from the O-F bond) = -1 - Formal charge on each fluorine atom: 7 (valence electrons for F) - 7 (assigned electrons: 6 from lone pairs + 1 from the O-F bond) = 0

Discuss whether oxidation states or formal charges are more useful in accounting for the properties of \(\mathrm{O}_{2} \mathrm{F}_{2}\)

Oxidation states are more useful in accounting for the chemical properties of \(\mathrm{O}_{2} \mathrm{F}_{2}\), particularly its oxidizing and fluorinating properties. This is because oxidation states provide insight into the tendency of an atom to gain or lose electrons during the reaction. The oxygen atom in \(\mathrm{O}_{2} \mathrm{F}_{2}\) with an oxidation state of +2 has a high tendency to accept electrons. This makes \(\mathrm{O}_{2} \mathrm{F}_{2}\) a strong oxidizing agent. The fluorine atoms with an oxidation state of -1 have a higher tendency to accept electrons and form a more stable F- ion, making \(\mathrm{O}_{2} \mathrm{F}_{2}\) a strong fluorinating agent as well. On the other hand, formal charges, while being useful for predicting molecular geometry, don't specifically indicate the atoms' reactivity or the chemical behavior of the molecule.