Bond Order, Magnetism, and the O₂ Problem
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You should be comfortable drawing Lewis structures for diatomic molecules and recalling the basic principles of atomic orbital hybridization and electron configuration.
Why this matters
Classical bonding theories fail to explain why oxygen gas is paramagnetic, a discrepancy resolved only by molecular orbital theory.
For decades, chemists relied on Lewis structures and valence bond theory to describe chemical bonds. These models successfully predicted the geometry and stability of many molecules, but they encountered a significant hurdle with the oxygen molecule, \(\mathrm{O_2}\). Experimental evidence clearly showed that \(\mathrm{O_2}\) is attracted to magnetic fields, indicating the presence of unpaired electrons. However, the standard Lewis structure for oxygen, featuring a double bond with all electrons paired, predicted that the molecule should be diamagnetic. This contradiction highlighted the limitations of localized electron models and necessitated a more sophisticated approach that considers the delocalization of electrons across the entire molecule.
The Concept of Bond Order
In molecular orbital theory, we do not simply add electrons to bonds between two specific atoms. Instead, we fill molecular orbitals that extend over the entire molecule. The stability of a molecule is quantified by a value called bond order. Intuitively, bond order represents the net number of chemical bonds between two atoms. If you have more electrons in orbitals that hold the nuclei together (bonding orbitals) than in orbitals that push them apart (antibonding orbitals), the molecule is stable. The bond order is calculated by taking the difference between the number of electrons in bonding and antibonding orbitals and dividing by two. A bond order of zero implies no net bond and the molecule will not exist, while a bond order of one, two, or three corresponds to single, double, or triple bonds, respectively. This metric provides a direct link between the electronic configuration and the physical strength and length of the bond.
Calculating Bond Order
The formal definition of bond order is derived from the electron population of the molecular orbitals. As established in the molecular orbital theory framework, the bond order is given by the expression:
Here, \(n_{\text{bonding}}\) is the total number of electrons occupying bonding molecular orbitals, and \(n_{\text{antibonding}}\) is the total number of electrons occupying antibonding molecular orbitals. It is crucial to note that only electrons in these specific types of orbitals contribute to the bond order. Electrons in non-bonding orbitals, which are typically associated with lone pairs or orbitals that do not significantly overlap between the nuclei, do not affect the bond order calculation. This formula allows us to predict relative bond strengths; for instance, a molecule with a bond order of 2.5 will generally have a shorter and stronger bond than one with a bond order of 2.0.
Magnetism and Electron Pairing
The magnetic properties of a molecule are determined by the pairing of its electrons. According to Hund's rule, electrons occupy degenerate orbitals singly with parallel spins before pairing up. If all electrons in a molecule are paired, the molecule is diamagnetic and is weakly repelled by a magnetic field. If a molecule contains one or more unpaired electrons, it is paramagnetic and is attracted to a magnetic field. This distinction provides a powerful experimental test for molecular orbital configurations. For example, the nitrogen molecule, \(\mathrm{N_2}\), has all electrons paired in its molecular orbitals and is diamagnetic. In contrast, the oxygen molecule, \(\mathrm{O_2}\), possesses two unpaired electrons in its highest occupied molecular orbitals, making it paramagnetic. This observation is a direct consequence of the energy ordering of the \(\pi\) and \(\sigma\) orbitals in second-period diatomics.
The Oxygen Anomaly
The case of \(\mathrm{O_2}\) is the classic example demonstrating the superiority of molecular orbital theory over valence bond theory. In the Lewis structure, oxygen is depicted with a double bond and two lone pairs on each atom, resulting in a closed-shell system with no unpaired electrons. This model predicts diamagnetism, which contradicts experimental data. Molecular orbital theory resolves this by considering the filling of the \(\pi_{2p}\) and \(\pi^*_{2p}\) orbitals. Oxygen has 12 valence electrons. After filling the lower energy bonding orbitals, the remaining two electrons must occupy the degenerate antibonding \(\pi^*_{2p}\) orbitals. Following Hund's rule, these two electrons occupy separate orbitals with parallel spins. This results in two unpaired electrons, correctly predicting the paramagnetic behaviour of oxygen. The bond order for \(\mathrm{O_2}\) is calculated as \((8 - 4) / 2 = 2\), consistent with the double bond character observed in its bond length and energy.
The result
Molecular orbital theory
Two atomic orbitals \(\phi_A,\phi_B\) combine linearly into a bonding molecular orbital \(\psi_+\propto\phi_A+\phi_B\) (in-phase, lower energy, electron density concentrated between the nuclei) and an antibonding orbital \(\psi_-\propto\phi_A-\phi_B\) (out-of-phase, higher energy, with a node between the nuclei). Filling these molecular orbitals by the same aufbau and Hund's-rule procedure used for atoms gives a bond order \(\text{BO}=\tfrac{n_{\text{bonding}}-n_{\text{antibonding}}}{2}\), and correctly predicts molecular magnetism — including the paramagnetism of \(\text{O}_2\), which the Lewis/valence-bond picture (this unit's earlier results) cannot explain.
A common misconception is that a bonding electron and an antibonding electron cancel exactly. The antibonding orbital is raised more than the bonding orbital is lowered, so filling both a bonding orbital and its antibonding partner is slightly destabilising overall. That is why \(\mathrm{He_2}\) and \(\mathrm{Ne_2}\) do not form chemical bonds, even though the bond-order formula gives exactly zero. Bond order is a counting index that tracks bond strength and length well; it is not an exact energy. C
Worked example
Problem. Determine the bond order and magnetic character of the dioxygen cation, \(\mathrm{O_2^+}\), given that it has 11 valence electrons.
- Identify the electron configuration. \(\mathrm{O_2}\) has 12 valence electrons. Removing one electron to form \(\mathrm{O_2^+}\) leaves 11 electrons. The configuration is \((\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^1\).
- Count the bonding electrons. The bonding orbitals are \(\sigma_{2s}\), \(\sigma_{2p}\), and \(\pi_{2p}\). Total bonding electrons \(n_{\text{bonding}} = 2 + 2 + 4 = 8\).
- Count the antibonding electrons. The antibonding orbitals are \(\sigma^*_{2s}\) and \(\pi^*_{2p}\). Total antibonding electrons \(n_{\text{antibonding}} = 2 + 1 = 3\).
- Calculate the bond order using the formula \(\text{BO} = \frac{n_{\text{bonding}}-n_{\text{antibonding}}}{2}\).
- Determine magnetism. Since there is one unpaired electron in the \(\pi^*_{2p}\) orbital, the species is paramagnetic.
Units check. Bond order is a dimensionless quantity representing the net number of bonds.
Three quick tests
- What is the bond order of the nitrogen molecule, \(\mathrm{N_2}\), and is it paramagnetic or diamagnetic?
Answer
Bond order is 3. It is diamagnetic because all electrons are paired. - If a diatomic molecule has a bond order of 1.5, what can you infer about its stability compared to a molecule with a bond order of 1.0?
Answer
The molecule with a bond order of 1.5 is more stable, has a shorter bond length, and a higher bond dissociation energy. - Why does the Lewis structure fail to predict the magnetic properties of \(\mathrm{O_2}\)?
Answer
The Lewis structure depicts all electrons as paired in a double bond, predicting diamagnetism, whereas the actual molecule has two unpaired electrons in degenerate antibonding orbitals, making it paramagnetic.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.