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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 \(\mathbf{O}_{2}\) to obtain \(\mathbf{O}_{2}^{+}\) ? Is the \(\mathbf{O}_{2}^{+}\) ion paramagnetic?

Short Answer

Expert verified
\( \mathrm{O_2^+} \) has a bond order of 2.5, is paramagnetic, and has one \( \sigma \) and one \( \pi \) bond.

Step by step solution

01

Draw the Molecular Orbital Diagram for \( \mathrm{O_2} \)

The molecular orbital (MO) diagram for \( \mathrm{O_2} \) involves electrons in both \( \sigma \) and \( \pi \) bonding and antibonding orbitals. In \( \mathrm{O_2} \), there are 12 valence electrons that fill the molecular orbitals as follows: \( \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi_{2p_x}^{*1}, \pi_{2p_y}^{*1} \).
02

Remove an Electron to Form \( \mathrm{O_2^+} \)

To convert \( \mathrm{O_2} \) into \( \mathrm{O_2^+} \), remove one electron from the \( \pi^* \) antibonding orbitals. Therefore, \( \mathrm{O_2^+} \) will have the configuration: \( \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi_{2p_x}^{*1} \).
03

Calculate the Bond Order for \( \mathrm{O_2^+} \)

The bond order is calculated using the formula: \( BO = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons}) \). For \( \mathrm{O_2^+} \), there are 9 bonding electrons and 4 antibonding electrons: \( BO = \frac{1}{2}(9 - 4) = 2.5 \).
04

Determine Changes in Bond Order From \( \mathrm{O_2} \) to \( \mathrm{O_2^+} \)

The bond order for \( \mathrm{O_2} \) was 2, and for \( \mathrm{O_2^+} \) it becomes 2.5. The bond order increases by 0.5 when \( \mathrm{O_2} \) loses an electron to form \( \mathrm{O_2^+} \).
05

Analyze the Paramagnetism of \( \mathrm{O_2^+} \)

\( \mathrm{O_2^+} \) has one unpaired electron in the \( \pi_{2p_x}^* \) orbital, so it is paramagnetic.
06

Count the Net \( \sigma \) and \( \pi \) Bonds

\( \mathrm{O_2^+} \) has one \( \sigma \) bond (from \( \sigma_{2p_z} \)) and one net \( \pi \) bond (considering the electrons in \( \pi_{2p_x} \) and \( \pi_{2p_y} \) orbitals).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Platinum hexafluoride
Platinum hexafluoride (\( \text{PtF}_6 \) is known to be one of the most potent oxidizing agents. An oxidizing agent is a substance that has the ability to accept electrons from another species, effectively undergoing reduction itself while the other substance is oxidized.

This property of \( \text{PtF}_6 \) is incredibly strong, to the extent that it can oxidize oxygen itself. In general, most substances cannot oxidize oxygen, which shows the remarkable power of \( \text{PtF}_6 \).

During the chemical reaction with oxygen, the compound \( \text{PtF}_6 \) transforms elemental oxygen \( \text{O}_2 \) into \( \text{O}_2^+ \) and \( \text{PtF}_6^- \). This specific reaction points to its capabilities as it influences the electronic configuration of oxygen, creating a positively charged molecule (\( \text{O}_2^+ \)). Such a level of reactivity makes \( \text{PtF}_6 \) a subject of great interest in both theoretical and practical chemistry.
Oxidizing agents
Oxidizing agents play a pivotal role in chemical reactions. They are responsible for causing the oxidation of other substances. To put it simply, an oxidizing agent accepts electrons and is reduced during the process.

This principle is not only a cornerstone of organic and inorganic chemistry but also serves key processes in energy transfer both in biological and industrial contexts.
  • Characteristics of Oxidizing Agents: They have high electronegativity and the ability to gain electrons easily.
  • Strong Oxidizers: Compounds like fluorine and \( \text{PtF}_6 \) are excellent examples due to their exceptional ability to accept electrons.

Oxidation reactions are essential in processes like combustion and metabolism. The reaction of \( \text{PtF}_6 \) with oxygen highlights how potent oxidizers can influence otherwise stable molecules, adding an electron to even robust atoms such as oxygen.
Oxygen bond order
The concept of bond order is a useful theoretical tool that helps us understand the stability of a molecule. It is calculated based on the formula: onumber \[ BO = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons}) \] For the molecule \( \text{O}_2 \), the initial bond order is 2. This reflects a stable double bond, common and naturally occurring in atmospheric oxygen.

However, when this molecule is oxidized by \( \text{PtF}_6 \) to form \( \text{O}_2^+ \), the bond order increases to 2.5. This increase signifies a strengthening of the bond between the oxygen atoms with the loss of an electron, creating a half additional bond according to molecular orbital theory.

Understanding oxygen bond order is crucial because it can affect the chemical's reactivity, stability, and how it interacts with other substances.
Paramagnetism
Paramagnetism is an interesting phenomenon occurring in materials with unpaired electrons. The presence of unpaired electrons allows molecules to be attracted to magnetic fields.

The classic example of a paramagnetic molecule is oxygen (\( \text{O}_2 \)) because it has two unpaired electrons. When oxygen is further oxidized to form \( \text{O}_2^+ \), it retains one unpaired electron in the \( \pi_{2px}^* \) orbital.
  • Behavior in Magnetic Fields: Such substances will align or be drawn towards a magnetic field, distinct from diamagnetic substances that repel magnetic fields due to all electrons being paired.
  • Practical Implication: Paramagnetism is vital in various fields, including medical imaging techniques such as MRI, which use this property for detailed body scanning.

The presence of unpaired electrons in \( \text{O}_2^+ \) and its paramagnetic nature serves as a reminder of the complex interactions between molecular electrons and external fields, crucial for advanced scientific and technological applications.

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Most popular questions from this chapter

Draw the Lewis structures of the acid \(\mathrm{HPO}_{2} \mathrm{F}_{2}\) and its anion \(\mathrm{PO}_{2} \mathrm{F}_{2}^{-}\). What is the molecular geometry and hybridization for the phosphorus atom in each species? (H is bonded to an O atom in the acid.)

Draw the Lewis structure, and then specify the electron-pair and molecular geometries for each of the following molecules or ions. Identify the hybridization of the central atom. (a) \(\mathrm{XeOF}_{4}\) (c) central \(\mathrm{S}\) in \(\mathrm{SOF}_{4}\) (b) \(\mathrm{BrF}_{5}\) (d) central Br in \(\mathrm{Br}_{3}^{-}\)

Draw the Lewis structure for carbonyl fluoride, \(\mathrm{COF}_{2}\) What are its electron-pair geometry and molecular geometry? What is the hybridization of the carbon atom? What orbitals overlap to form the \(\sigma\) and \(\pi\) bonds between carbon and oxygen?

Bromine forms a number of oxides of varying stability. (a) One oxide has \(90.90 \%\) Br and \(9.10 \%\) O. Assuming its empirical and molecular formulas are the same, draw a Lewis structure of the molecule and specify the hybridization of the central atom (O). (b) Another oxide is unstable BrO. Assuming the molecular orbital diagram in Figure 9.18 applies to BrO, write its electron configuration (where Br uses \(4 s\) and \(4 p\) orbitals). What is the HOMO for the molecule?

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