Question

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}{ }^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p} \mathrm{MO}\). (c) The \(\mathrm{O}_{2}{ }^{2+}\) ion has a stronger \(\mathrm{O}-\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

Short answer

(a) The peroxide ion (\(O_2^{2-}\)) has a bond order of 2 due to the completely filled \(\pi_{2p}^{*}\) orbitals, while the superoxide ion (\(O_2^{-}\)) has a bond order of 2. This results in a weaker and longer O-O bond in the peroxide ion compared to the superoxide ion. (b) The magnetic properties of \(B_2\) indicate that there are two unpaired electrons in the \(\pi_{2p}\) orbitals, making it paramagnetic and consistent with the observed MO configuration (\(\sigma_{2s}, \sigma_{2s^{*}}, \pi_{2p}, \sigma_{2p}\)). (c) The \(O_2^{2+}\) ion has a higher bond order (2) than the \(O_2\) molecule due to the absence of electrons in the \(\pi_{2p}^{*}\) orbitals, resulting in a stronger O-O bond in \(O_2^{2+}\).

Step-by-step solution

Understand bond order and strength

In MO theory, the bond order is calculated as follows: \(Bond\ Order = \frac{1}{2}(Bonding\ Electrons - Antibonding\ Electrons)\) A higher bond order indicates a stronger bond between two atoms, and a shorter bond length. Conversely, a lower bond order indicates a weaker bond and a longer bond length. Now let's analyze the given observations.

Examine O-O bond in \(O_2^{2-}\) and \(O_2^-\)

(a) Firstly, we need to determine the electron configurations of \(O_2^{2-}\) and \(O_2^{-}\) in their molecular orbitals. For peroxide ion, \(O_2^{2-}\): It has 2 additional electrons compared to an oxygen molecule (\(O_2\)). The molecular orbitals in an oxygen molecule are filled in the following order: \(\sigma_{2s}, \sigma_{2s}^{*}, \sigma_{2p}, \pi_{2p}, \pi_{2p}^{*}\) (Note: * indicates antibonding orbitals). Since the electrons in \(O_2\) occupy all orbitals up to the \(\pi_{2p}^{*}\), adding 2 electrons to form the peroxide ion will completely fill the \(\pi_{2p}^{*}\) orbitals. Hence, the bond order is: \(Bond\ Order \ \mathrm{(O_2^{2-})} = \frac{1}{2}(10 - 6) = 2\) For superoxide ion, \(O_2^{-}\): It has 1 additional electron compared to an oxygen molecule. The molecular orbitals for \(O_2^{-}\) will be filled up to the \(\pi_{2p}^{*}\) orbital, but with only one unpaired electron. The bond order is: \(Bond\ Order \ \mathrm{(O_2^{-})} = \frac{1}{2}(9 - 5) = 2\) Due to the lower bond order of the peroxide ion, its O-O bond is weaker and longer than that of the superoxide ion.

Examine magnetic properties of \(B_2\)

(b) The magnetic properties of a molecule are determined by the presence of unpaired electrons in its molecular orbitals. In the boron diatomic molecule (\(B_2\)), we note that both boron atoms have three valence electrons. Therefore, the total number of electrons in molecular orbitals of \(B_2\) is six. The molecular orbitals are filled in the following order: \(\sigma_{2s}, \sigma_{2s^{*}}, \pi_{2p}, \sigma_{2p}\) (Note: * indicates antibonding orbitals). Thus, the six electrons in \(B_2\) will fill the orbitals up to, but not including the \(\sigma_{2p}\) orbital. In this configuration, there are two unpaired electrons in the \(\pi_{2p}\) orbitals, making \(B_2\) paramagnetic. This is consistent with the observation that the magnetic properties of \(B_2\) indicate that the \(\pi_{2p}\) MOs are lower in energy than the \(\sigma_{2p}\) MO.

Examine O-O bond in \(O_2^{2+}\) and \(O_2\)

(c) As we mentioned earlier, the bond order in an oxygen molecule (\(O_2\)) is 2. Its molecular orbitals are filled as follows: \(\sigma_{2s}, \sigma_{2s}^{*}, \sigma_{2p}, \pi_{2p}, \pi_{2p}^{*}\) (Note: * indicates antibonding orbitals). For \(O_2^{2+}\), two electrons are removed from the highest occupied molecular orbital, which is the \(\pi_{2p}^{*}\) orbital. As a result, there will be no electrons in the \(\pi_{2p}^{*}\) orbitals, leading to an increased bond order: \(Bond\ Order \ \mathrm{(O_2^{2+})} = \frac{1}{2}(8 - 4) = 2\) The bond order of \(O_2^{2+}\) is higher than that of the \(O_2\) molecule, so the O-O bond is stronger in \(O_2^{2+}\) than in \(O_2\).

Key concepts

Bond Order Calculation

Understanding bond order is crucial in predicting the strength and length of bonds in molecules. Bond order is calculated using the formula:

\[Bond\ Order = \frac{1}{2}(Bonding\ Electrons - Antibonding\ Electrons)\]
In this equation, bonding electrons refer to the total number of electrons in bonding molecular orbitals (MOs), and antibonding electrons to those in antibonding MOs. A higher bond order value indicates a stronger and shorter bond, as it reflects a greater balance of electrons in bonding versus antibonding orbitals. For example, with the diatomic oxygen molecule (\(O_2\)), the bond order is calculated by taking into account both the bonding and antibonding electrons. The larger the difference between these two sets of electrons, the higher the bond order, and the more robust and compact the bond between the atoms will be.

Paramagnetism and Diamagnetism

The magnetic properties of molecules can be explained by the presence or absence of unpaired electrons in their molecular orbitals. Paramagnetic substances contain one or more unpaired electrons and are attracted to a magnetic field. Diamagnetic substances, on the other hand, have all their electrons paired up and slightly repel a magnetic field.

For instance, the molecule \(B_2\) contains unpaired electrons due to its electron configuration, which makes it paramagnetic and therefore attracted to magnetic fields. This attribute helps us understand the relative energies of molecular orbitals, as the occupancy of these unpaired electrons in specific orbitals indicates their increasing energy levels within the MO framework.

Molecular Orbital Configuration

The molecular orbital configuration provides a schematic of how electrons are distributed in the molecular orbitals of a molecule. Electrons fill the orbitals in a specific order based on their energies; the lowest energy orbitals fill up first. For example, in diatomic molecules like \(O_2\), the electron filling order typically follows: \(\sigma_{2s}, \sigma_{2s}^*, \sigma_{2p}, \pi_{2p}, \pi_{2p}^*\). The asterisk (*) indicates an antibonding molecular orbital, which has a higher energy compared to its bonding counterpart. By knowing the electron configuration, we can predict various molecular properties such as magnetism and bond length.

Electron Configurations in Ions

When atoms form ions, they lose or gain electrons, altering their molecular orbital configurations. This influences the chemical properties and reactivity of the ion compared to its neutral atom. For example, the ion \(O_2^{2+}\) is formed by removing two electrons from the neutral \(O_2\) molecule. These electrons are removed from the highest occupied antibonding molecular orbitals, which increases the bond order and strengthens the bond. Understanding these changes in electron configurations is essential for predicting the stability, reactivity, and properties of ions.

Bond Length Comparison

Bond length is directly influenced by the bond order between two atoms; a higher bond order generally correlates with a shorter bond length. This is because a greater number of bonding electrons relative to antibonding electrons pulls the nuclei closer together. For example, comparing the bond lengths of \(O_2^{2-}\) and \(O_2^-\), we see that the presence of more antibonding electrons in the peroxide ion (\(O_2^{2-}\)) leads to a lower bond order and therefore a longer bond length than in the superoxide ion (\(O_2^-\)). Thus, by calculating bond order, one can predict the relative lengths of bonds in different molecules or ions.