Problem 18 - Chemistry and Chemical Reactivity

Question

Oxygen, \(\mathrm{O}_{2},\) can acquire one or two electrons to give \(\mathrm{O}_{2}^{-}\) (superoxide ion) or \(\mathrm{O}_{2}^{2-}\) (peroxide ion). Write the electron configuration for the ions in molecular orbital terms, and then compare them with the \(\mathrm{O}_{2}\) molecule on the following bases. (a) magnetic character (b) net number of \(\sigma\) and \(\pi\) bonds (c) bond order (d) oxygen-oxygen bond length

Step-by-Step Solution

Answer

\(\mathrm{O}_{2}\) and \(\mathrm{O}_{2}^{-}\) are paramagnetic. Bond order decreases from \(\mathrm{O}_{2}\) to \(\mathrm{O}_{2}^{2-}\), so bond length increases.

Step 1: Determine the Molecular Orbital Configuration of \(\mathrm{O}_{2}\)

The molecular orbital configuration for \(\mathrm{O}_{2}\) is determined by the ordering of molecular orbitals and the number of electrons. \(\mathrm{O}_{2}\) has 16 electrons: \((\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1\). The unpaired electrons in the \(\pi^*\) orbitals indicate that \(\mathrm{O}_{2}\) is paramagnetic.

Step 2: Determine the Configuration for \(\mathrm{O}_{2}^{-}\) (Superoxide Ion)

\(\mathrm{O}_{2}^{-}\) has 17 electrons, one more than \(\mathrm{O}_{2}\). Adding one electron fills one of the \(\pi^*\) orbitals: \((\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^2\). With one unpaired electron, \(\mathrm{O}_{2}^{-}\) remains paramagnetic.

Step 3: Determine the Configuration for \(\mathrm{O}_{2}^{2-}\) (Peroxide Ion)

\(\mathrm{O}_{2}^{2-}\) has 18 electrons, filling both \(\pi^*\) orbitals with electrons: \((\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2\). All electrons are paired, making \(\mathrm{O}_{2}^{2-}\) diamagnetic.

Step 4: Compare Magnetic Character

\(\mathrm{O}_{2}\) and \(\mathrm{O}_{2}^{-}\) are paramagnetic due to unpaired electrons, while \(\mathrm{O}_{2}^{2-}\) is diamagnetic because all electrons are paired.

Step 5: Determine the Net Number of \(\sigma\) and \(\pi\) Bonds and Bond Order

The bond order is calculated by: \(\text{Bond Order} = \frac{1}{2}[(\text{electrons in bonding orbitals}) - (\text{electrons in antibonding orbitals})]\).- \(\mathrm{O}_{2}\): Bond order = \(\frac{1}{2}[10 - 6] = 2\). It has 1 \(\sigma\) and 1 \(\pi\) bond.- \(\mathrm{O}_{2}^{-}\): Bond order = \(\frac{1}{2}[10 - 7] = 1.5\).- \(\mathrm{O}_{2}^{2-}\): Bond order = \(\frac{1}{2}[10 - 8] = 1\). It has 1 \(\sigma\) bond.

Step 6: Compare Bond Lengths

Bond length is inversely proportional to bond order. - \(\mathrm{O}_{2}\) has the shortest bond length (highest bond order), followed by \(\mathrm{O}_{2}^{-}\) (intermediate bond order), and \(\mathrm{O}_{2}^{2-}\) has the longest bond length (lowest bond order).

Key Concepts

Paramagnetism

Imagine a magnet that works because of unpaired electrons. That's what paramagnetism is all about. When molecules have unpaired electrons in their outer orbitals, such as found in the molecular orbitals of \( \mathrm{O}_{2} \) and \( \mathrm{O}_{2}^{-} \), they tend to be attracted to magnetic fields.
This property arises due to these unpaired electrons creating magnetic moments. So, both \( \mathrm{O}_{2} \) and \( \mathrm{O}_{2}^{-} \) exhibit paramagnetism as they both contain unpaired electrons in their antibonding \( \pi^* \) orbitals. This is why these species behave like tiny magnets in the presence of external magnetic fields.

Diamagnetism

Diamagnetism is what we observe when all electrons in a molecule are paired. This causes the molecules to have no net magnetic moment and they are slightly repelled by magnetic fields. Even though this repulsion is usually weak, it's quite significant in terms of understanding molecular behavior.
The example of \( \mathrm{O}_{2}^{2-} \) illustrates diamagnetism very well. In this molecule, all electrons are paired in their molecular orbitals, resulting in zero unpaired electrons. Consequently, the \( \mathrm{O}_{2}^{2-} \) ion does not display magnetism in the presence of a magnetic field, showing diamagnetic properties instead.

Bond Order

Bond order tells us something essential about a bond: its strength. It's calculated using the formula: \( \text{Bond Order} = \frac{1}{2}[(\text{electrons in bonding orbitals}) - (\text{electrons in antibonding orbitals})] \). The concept helps us understand how strong or weak a bond within a molecule is.

• For \( \mathrm{O}_{2} \), the bond order comes out to be 2, indicating a double bond.
• For \( \mathrm{O}_{2}^{-} \), the bond order is 1.5, which suggests a bond strength weaker than a double bond, due to additional electrons in antibonding orbitals.
• Finally, \( \mathrm{O}_{2}^{2-} \)'s bond order is 1, indicating a single bond and showing that extra electrons in the antibonding orbitals reduce bond strength.

Understanding bond order helps predict how molecules will behave from reactions to stability.

Bond Length

The length of a bond between two atoms is inversely related to the bond order—higher bond orders generally mean shorter bonds. Why? Because a higher bond order means more shared electrons pulling the atoms closer together.
In our example compounds:

• \( \mathrm{O}_{2} \) has the shortest bond length due to its higher bond order of 2, indicating a strong and tight bond between the oxygen atoms.
• In \( \mathrm{O}_{2}^{-} \), the bond order drops to 1.5, resulting in a slightly longer bond as there are fewer electrons holding the atoms tightly together.
• Finally, \( \mathrm{O}_{2}^{2-} \) has the longest bond length with a bond order of 1. This single bond lacks the strength to keep the atoms close.

This relationship helps chemists understand molecule size and stability, as well as predict molecular geometry.