Problem 22 - Chemistry and Chemical Reactivity

Question

Platinum hexafluoride is an extremely strong oxidizing agent. It can even oxidize oxygen, its reaction with \(\mathrm{O}_{2}\) giving \(\mathrm{O}_{2}^{+} \mathrm{PtF}_{6}^{-} .\) Sketch the molecular orbital energy level diagram for the \(\mathrm{O}_{2}^{+}\) ion. How many net \(\sigma\) and \(\pi\) bonds does the ion have? What is the oxygen-oxygen bond order? How has the bond order changed on taking away electrons from \(\mathrm{O}_{2}\) to obtain \(\mathrm{O}_{2}^{+} ?\) Is the \(\mathrm{O}_{2}^{+}\) ion paramagnetic?

Step-by-Step Solution

Answer

The \( \mathrm{O}_{2}^{+} \) ion has a bond order of 2.5 and is paramagnetic with one unpaired electron.

Step 1: Understanding the Problem

To determine the molecular orbital diagram of the \( \mathrm{O}_{2}^{+} \) ion, we need to know that \( \mathrm{O}_{2} \) has 16 electrons. Removing one electron to form \( \mathrm{O}_{2}^{+} \) leaves us with 15 electrons. We will calculate the bond order and assess if the ion is paramagnetic.

Step 2: Constructing the Molecular Orbital Diagram

Oxygen molecule \( \mathrm{O}_{2} \) has molecular orbitals filled in the following order: \( \sigma_{2s}, \sigma_{2s}^*, \sigma_{2p_z}, \pi_{2p_x}=\pi_{2p_y}, \pi_{2p_x}^*=\pi_{2p_y}^*, \sigma_{2p_z}^* \). Place the 15 electrons of \( \mathrm{O}_{2}^{+} \) into these orbitals, filling from lowest to highest energy. \( \mathrm{O}_{2}^{+} \) removes one electron from one of the \( \pi^* \) orbitals.

Step 3: Calculating the Bond Order

The bond order can be calculated using the formula: Bond Order = \( (\text{Number of bonding electrons} - \text{Number of antibonding electrons})/2 \). For \( \mathrm{O}_{2}^{+} \), there are 9 bonding electrons and 4 antibonding electrons. Therefore, the bond order is \((9 - 4)/2 = 2.5\).

Step 4: Assessing the Change in Bond Order

The bond order of \( \mathrm{O}_{2} \) is 2, while that of \( \mathrm{O}_{2}^{+} \) is 2.5. Therefore, the bond order increases by 0.5 upon removing an electron.

Step 5: Determining Paramagnetism

Since \( \mathrm{O}_{2}^{+} \) has an unpaired electron in one of the \( \pi^* \) orbitals, it is paramagnetic due to the presence of this unpaired electron.

Key Concepts

Oxidation State

The concept of oxidation state helps us understand how many electrons an atom gains, loses, or shares when forming compounds. In the case of the reaction involving platinum hexafluoride and oxygen, the strong oxidizing agent helps oxygen undergo a change in oxidation state by removing an electron. This process results in the formation of the ion \( \mathrm{O}_{2}^{+} \), where each oxygen's oxidation state essentially increases due to the loss of an electron.

Understanding oxidation states is crucial to predicting the reactivity and properties of elements in a chemical reaction.

• Oxidation involves the loss of electrons, increasing the oxidation state.
• Reduction involves the gain of electrons, decreasing the oxidation state.

The initial oxidation state of oxygen in \( \mathrm{O}_{2} \) is 0. When it forms \( \mathrm{O}_{2}^{+} \), it indicates that an electron is removed, creating a positively charged ion. This shift not only alters its oxidation state but also affects the molecular properties of oxygen, influencing its bond order and magnetic properties.

Bond Order

Bond order is a fundamental concept in molecular orbital theory that informs us about the stability and strength of a bond. It is calculated by taking the difference between the number of bonding and antibonding electrons, divided by two. Specifically for \( \mathrm{O}_{2}^{+} \), the bond order can be calculated as: \[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]

In this system, \( \mathrm{O}_{2}^{+} \) has 9 bonding electrons and 4 antibonding electrons, resulting in a bond order of \( \frac{9 - 4}{2} = 2.5 \).

This is particularly interesting because:

• The bond order of \( \mathrm{O}_{2} \) is 2, but removing an electron to form \( \mathrm{O}_{2}^{+} \) increases it to 2.5.
• This increase suggests a stronger bond due to the reduction of antibonding electrons.

Understanding bond order is important because it gives insights into molecule stability and how reactions might proceed. A higher bond order generally means a more stable molecule with stronger bonds.

Paramagnetism

Paramagnetism in molecules arises due to the presence of unpaired electrons. When examining the \( \mathrm{O}_{2}^{+} \) ion, we find that it is paramagnetic. This is because the molecular orbital diagram for \( \mathrm{O}_{2}^{+} \) shows an unpaired electron in one of its antibonding \( \pi^* \) orbitals.

Key ideas related to paramagnetism:

• Molecules with unpaired electrons exhibit paramagnetism and are attracted to magnetic fields.
• Diamagnetic molecules, on the other hand, have all paired electrons and are not attracted to magnetic fields.

For \( \mathrm{O}_{2}^{+} \), the presence of an unpaired electron means it retains magnetic properties, similar to the neutral \( \mathrm{O}_{2} \) molecule, which also has unpaired electrons. This behavior is critical when considering the physical properties and reactivity of substances in magnetic fields.