MO Diagrams for Homonuclear Diatomics
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You should be comfortable with the Linear Combination of Atomic Orbitals (LCAO) approximation and understand how atomic orbitals combine to form bonding and antibonding molecular orbitals.
Why this matters
Molecular orbital diagrams provide a unified framework for predicting the stability and magnetic properties of diatomic molecules.
While Lewis structures and valence bond theory are intuitive for simple molecules, they fail to explain phenomena such as the paramagnetism of oxygen. Molecular orbital (MO) theory overcomes these limitations by treating electrons as delocalised over the entire molecule. By constructing energy level diagrams for homonuclear diatomics, we can systematically determine electron configurations, calculate bond orders, and predict whether a molecule will be diamagnetic or paramagnetic. This lecture focuses on the specific ordering of orbitals for molecules like B2, C2, N2, O2, and F2.
The Energy Level Framework
In a homonuclear diatomic molecule, the two nuclei are identical, meaning the atomic orbitals on each atom have the same energy. When these orbitals interact, they form molecular orbitals that are spread over both nuclei. The key to drawing the correct MO diagram is determining the relative energy spacing between the sigma (\(\sigma\)) and pi (\(\pi\)) orbitals. For lighter diatomics (up to N2), the \(2p\) orbitals interact strongly with the \(2s\) orbitals due to small energy differences. This interaction, known as s-p mixing, raises the energy of the \(\sigma_{2p}\) orbital relative to the \(\pi_{2p}\) orbitals. Consequently, the \(\pi_{2p}\) orbitals are lower in energy than the \(\sigma_{2p}\) orbital. For heavier diatomics (O2 and F2), the energy gap between \(2s\) and \(2p\) is larger, reducing s-p mixing. In these cases, the \(\sigma_{2p}\) orbital is lower in energy than the \(\pi_{2p}\) orbitals. This distinction is crucial for correctly filling the orbitals with electrons.
Orbital Ordering and Filling
Once the energy levels are established, we fill the molecular orbitals with electrons following the Aufbau principle and Hund's rule. We start with the lowest energy orbitals and fill them in pairs before moving to higher energy levels. For B2, C2 and N2 (up to 10 valence electrons), the ordering is: \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\pi_{2p}\), \(\sigma_{2p}\). For O2 and F2 (12 and 14 valence electrons), the ordering is: \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\sigma_{2p}\), \(\pi_{2p}\). Note that the \(\sigma_{2s}\) and \(\sigma^*_{2s}\) orbitals are always filled first and, once both are full, their bonding and antibonding contributions cancel in the bond order.
Calculating Bond Order
The bond order is a quantitative measure of the stability of a molecule. It is calculated by taking the difference between the number of electrons in bonding orbitals and the number of electrons in antibonding orbitals, divided by two. A positive bond order indicates a stable molecule, while a bond order of zero suggests the molecule will not exist. The formula for bond order is consistent across all homonuclear diatomics, regardless of the specific orbital ordering. Electrons in \(\sigma\) and \(\pi\) bonding orbitals stabilise the molecule, while electrons in \(\sigma^*\) and \(\pi^*\) antibonding orbitals destabilise it. By summing these contributions, we obtain a net measure of the chemical bond strength.
Predicting Magnetic Properties
One of the most powerful predictions of MO theory is the magnetic behaviour of molecules. If all electrons in the molecular orbitals are paired, the molecule is diamagnetic and is weakly repelled by a magnetic field. If there are unpaired electrons, the molecule is paramagnetic and is attracted to a magnetic field. This is particularly evident in the case of oxygen (\(\text{O}_2\)). Using the heavy atom orbital ordering, the last two electrons occupy the degenerate \(\pi^*_{2p}\) orbitals. According to Hund's rule, these electrons occupy separate orbitals with parallel spins, resulting in two unpaired electrons. This correctly predicts the paramagnetism of \(\text{O}_2\), a fact that cannot be explained by the Lewis structure, which shows all electrons as paired.
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.
Bond order predicts bond length well within one family: \(\mathrm{O_2^+}\) (2.5) 112 pm, \(\mathrm{O_2}\) (2) 121 pm, \(\mathrm{O_2^-}\) (1.5) about 133 pm, \(\mathrm{O_2^{2-}}\) (1) about 149 pm. A common misconception is that it can be compared across different elements without allowing for atomic size. \(\mathrm{Cl_2}\) and \(\mathrm{F_2}\) both have bond order 1, yet the Cl–Cl bond (199 pm) is far longer than F–F (142 pm), because the bonding uses larger 3p orbitals. C
Worked example
Problem. Determine the bond order and magnetic character of the carbon dimer, C2. Assume the light atom orbital ordering applies. The total number of valence electrons in C2 is 8 (4 from each carbon atom).
- Identify the orbital filling order for C2: \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\pi_{2p}\), \(\sigma_{2p}\).
- Fill the orbitals with 8 valence electrons: - \(\sigma_{2s}\): 2 electrons - \(\sigma^*_{2s}\): 2 electrons - \(\pi_{2p}\): 4 electrons (2 in \(\pi_{2p,x}\), 2 in \(\pi_{2p,y}\)) - \(\sigma_{2p}\): 0 electrons
- Count the bonding and antibonding electrons: - Bonding electrons (\(n_{\text{bonding}}\)): 2 (in \(\sigma_{2s}\)) + 4 (in \(\pi_{2p}\)) = 6 - Antibonding electrons (\(n_{\text{antibonding}}\)): 2 (in \(\sigma^*_{2s}\))
- Calculate the bond order: \[ \text{BO} = \frac{6 - 2}{2} = 2 \]
- Determine magnetic character: All electrons are paired in the filled orbitals (\(\sigma_{2s}\), \(\sigma^*_{2s}\), and the two \(\pi_{2p}\) orbitals). Therefore, C2 is diamagnetic.
Units check. Bond order is a dimensionless quantity, representing the number of covalent bonds.
Three quick tests
- What is the bond order of the nitrogen dimer, N2?
Answer
3. N2 has 10 valence electrons. Filling: \(\sigma_{2s}^2\), \(\sigma^*_{2s}{}^2\), \(\pi_{2p}^4\), \(\sigma_{2p}^2\). Bonding: 2+4+2=8. Antibonding: 2. BO = (8-2)/2 = 3. - Why is O2 paramagnetic?
Answer
Because it has two unpaired electrons in the degenerate \(\pi^*_{2p}\) antibonding orbitals, following Hund's rule. - Which orbital ordering applies to F2?
Answer
The heavy atom ordering: \(\sigma_{2s} < \sigma^*_{2s} < \sigma_{2p} < \pi_{2p} < \pi^*_{2p} < \sigma^*_{2p}\). This is because s-p mixing is minimal for heavier atoms.
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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.