Question

Using molecular orbital theory, explain why the removal of an electron from \(\mathrm{O}_{2}\) strengthens bonding, whereas the removal of an electron from \(\mathrm{N}_{2}\) weakens bonding.

Short answer

In O\(_2\), removing an electron from the highest occupied molecular orbital (HOMO), which is \(\pi_{2p*}\) (antibonding), results in a higher bonding order (\(\frac{3}{2}\) vs 2) and thus stronger bonding. In N\(_2\), removing an electron from the HOMO, which is \(\pi_{2p}\) (bonding), leads to a lower bonding order (\(\frac{5}{2}\) vs 3) and weaker bonding. This difference occurs due to the different electron configurations in the molecular orbitals.

Step-by-step solution

Write down the molecular orbital configurations for O\(_2\) and N\(_2\).

To analyze the effect of removing an electron from O\(_2\) and N\(_2\), we first need to determine the molecular orbital configurations of both molecules. For this, we can use the following diagram:  From the diagram, we have the following molecular orbital configurations: O\(_2\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{2}\) N\(_2\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{0}\)

Calculate the bonding orders for O\(_2\) and N\(_2\).

We calculate the bonding order using the following formula: Bonding Order = \(\frac{1}{2}(n_{b} - n_{a})\) where \(n_{b}\) is the number of bonding electrons, and \(n_{a}\) is the number of anti-bonding electrons. For O\(_2\), Bonding Order = \(\frac{1}{2}(10 - 6) = 2\) For N\(_2\), Bonding Order = \(\frac{1}{2}(10 - 4) = 3\)

Remove one electron from O\(_2\) and N\(_2\) and write down their new molecular orbital configurations.

Removing one electron from O\(_2\), we get O\(_2^+\) with the molecular orbital configuration: O\(_2^+\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{1}\) Removing one electron from N\(_2\), we get N\(_2^+\) with the molecular orbital configuration: N\(_2^+\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{3}\pi_{2p*}^{0}\)

Calculate the bonding orders for O\(_2^+\) and N\(_2^+\)

For O\(_2^+\), Bonding Order = \(\frac{1}{2}(9 - 6) = \frac{3}{2}\) For N\(_2^+\), Bonding Order = \(\frac{1}{2}(9 - 4) = \frac{5}{2}\)

Compare the bonding orders to determine whether the bonding is strengthened or weakened.

Comparing the bonding orders, we can conclude: For O\(_2\) bonding is strengthened upon removal of an electron: O\(_2\): Bonding Order = 2 O\(_2^+\): Bonding Order = \(\frac{3}{2}\) For N\(_2\) bonding is weakened upon removal of an electron: N\(_2\): Bonding Order = 3 N\(_2^+\): Bonding Order = \(\frac{5}{2}\) Therefore, using molecular orbital theory, we have established that the removal of an electron from O\(_2\) strengthens bonding, whereas the removal of an electron from N\(_2\) weakens bonding.

Key concepts

Bonding Order

The concept of bonding order is central to understanding how molecules behave when electrons are added or removed. It provides a measure of the stability of a molecule. The bonding order is calculated using the formula:

• Bonding Order = \(\frac{1}{2}(n_{b} - n_{a})\)

Where \(n_{b}\) represents the number of bonding electrons and \(n_{a}\) denotes the number of anti-bonding electrons. A higher bonding order indicates a stronger, more stable bond. For instance, a bonding order of 3 suggests a triple bond, indicating a strong connection between atoms, while a bonding order of 1 implies a weaker, single bond.
When electrons are removed or added, it affects the number of bonding and anti-bonding electrons, hence altering the bonding order and the molecule's stability. Understanding this concept is key when analyzing molecules like \(O_2\) and \(N_2\) and predicting how they will respond to changes in their electron compositions.

Electron Removal

Removing an electron from a molecule can result in a change in its bonding and stability. When an electron is taken from a molecule, it affects the bonding and anti-bonding orbitals differently.
For example, when an electron is removed from an anti-bonding orbital, the bonding order increases, strengthening the bond. Conversely, removing an electron from a bonding orbital decreases the bonding order, weakening the bond.
In the case of \(O_2\), removing an electron strengthens the bond. This is because the electron removed comes from an anti-bonding orbital, thereby increasing the bonding order from 2 to 1.5. This indicates that the molecule becomes more stable. On the other hand, for \(N_2\), the electron removal from a bonding orbital weakens the bond, decreasing the bonding order from 3 to 2.5. This results in a less stable molecule.

O2 Molecule

The \(O_2\) molecule is a classic example used to demonstrate molecular orbital theory. Oxygen molecules are characterized by a unique configuration of bonded electrons which contribute to its stability.
For \(O_2\), the molecular orbital configuration is \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{2}\). It has a total of 10 bonding electrons and 6 anti-bonding electrons, resulting in a bonding order of 2, corresponding to a double bond.
When an electron is removed from \(O_2\) to form \(O_2^+\), the configuration changes, decreasing the number of anti-bonding electrons. This change results in a new bonding order of 1.5, which suggests a strengthened molecular bond. Hence, the overall increase in bond stability when an electron is removed is explained by a reduction in repulsive anti-bonding interactions.

N2 Molecule

The \(N_2\) molecule, on the other hand, behaves differently when subjected to electron removal. Known for its triple bond, \(N_2\) is bonded strongly and is very stable, a reflection of its high bonding order.The molecular orbital configuration of \(N_2\) is \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{0}\). This results in 10 bonding electrons and 4 anti-bonding electrons, leading to a bonding order of 3.
Upon removal of one electron, forming \(N_2^+\), the lost electron affects the bonding orbitals directly. The bonding order decreases to 2.5, indicating a weaker bond and thus a less stable molecule. This weakening occurs because the electron removed reduces electron pair interactions that are crucial to the triple bond's strength, demonstrating how electron removal can reduce molecular stability in certain cases.