LCAO and Molecular Orbitals
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You should be comfortable with the quantum mechanical description of atomic orbitals and the concept of wavefunction superposition. Familiarity with the previous lecture on sigma and pi bonding is essential to understand the symmetry requirements for orbital overlap.
Why this matters
LCAO provides the mathematical framework that explains why some molecules are stable while others are not.
Valence bond theory successfully describes many covalent bonds but struggles with molecules like oxygen, which is experimentally known to be paramagnetic. The Linear Combination of Atomic Orbitals (LCAO) method treats electrons as delocalised over the entire molecule rather than localised between two atoms. By combining atomic wavefunctions, we generate molecular orbitals that extend over the whole system. This approach naturally yields both bonding and antibonding states, allowing us to calculate bond orders and predict magnetic properties with high accuracy.
The Superposition Principle in Chemistry
Imagine two waves on a string. If they meet in phase, their amplitudes add up to create a larger wave. If they meet out of phase, they cancel each other out. In quantum mechanics, electrons are described by wavefunctions. When two atoms approach each other, their atomic orbitals do not simply sit side-by-side; they mix. This mixing is governed by the principle of superposition. The resulting molecular orbitals are not new, independent entities, but rather linear combinations of the original atomic orbitals. This process is analogous to how musical chords are formed by combining individual notes, creating new harmonic properties that the individual notes did not possess in isolation.
The key insight is that the "mixing" is not arbitrary. It is constrained by symmetry. Only orbitals with compatible symmetry can combine effectively. This symmetry matching ensures that the resulting molecular orbitals have the correct spatial distribution of electron density, which determines whether the interaction is stabilising (bonding) or destabilising (antibonding).
Mathematical Formulation of LCAO
For a diatomic molecule composed of atoms A and B, we approximate the molecular orbital \(\psi\) as a linear combination of the atomic orbitals \(\phi_A\) and \(\phi_B\). The simplest approximation assumes that only one orbital from each atom contributes to a given molecular orbital. The general form is:
Here, \(c_A\) and \(c_B\) are coefficients that determine the contribution of each atomic orbital. Solving the Schrödinger equation for this system yields two primary solutions. The first is the bonding molecular orbital, where the wavefunctions add constructively between the nuclei:
The second is the antibonding molecular orbital, where the wavefunctions subtract, creating a node between the nuclei:
The bonding orbital \(\psi_+\) has lower energy than the constituent atomic orbitals because the electron density is concentrated in the internuclear region, attracting both nuclei and stabilising the system. Conversely, the antibonding orbital \(\psi_-\) has higher energy due to the node, which reduces electron density between the nuclei and increases nuclear repulsion.
Bond Order and Stability
The stability of a molecule is quantified by its bond order. In the LCAO-MO framework, bond order is defined as half the difference between the number of electrons in bonding orbitals and the number of electrons in antibonding orbitals. This metric provides a direct link between electron configuration and molecular stability.
A positive bond order indicates a stable molecule. A bond order of zero implies no net bonding interaction, and the molecule will not exist as a stable entity. For example, if a hypothetical diatomic molecule had two electrons in a bonding orbital and two in an antibonding orbital, the bond order would be \((2-2)/2 = 0\). This explains why certain noble gas dimers, such as \(\mathrm{He_2}\), are not stable under standard conditions. The stabilising effect of the bonding electrons is exactly cancelled by the destabilising effect of the antibonding electrons.
Explaining Paramagnetism in Oxygen
One of the most significant successes of MO theory is its ability to explain the magnetic properties of molecules. Lewis structures and valence bond theory predict that oxygen (\(\mathrm{O_2}\)) should have all electrons paired, making it diamagnetic. However, experimental evidence shows that \(\mathrm{O_2}\) is paramagnetic, meaning it is attracted to a magnetic field. This discrepancy is resolved by MO theory.
When we construct the molecular orbital diagram for \(\mathrm{O_2}\), we fill the orbitals according to the Aufbau principle and Hund's rule. Oxygen has 12 valence electrons. The filling proceeds into the \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\sigma_{2p_z}\), \(\pi_{2p_x}\), and \(\pi_{2p_y}\) orbitals. The final two electrons occupy the degenerate \(\pi^*_{2p}\) antibonding orbitals. According to Hund's rule, these two electrons occupy separate orbitals with parallel spins rather than pairing up in the same orbital. This results in two unpaired electrons, which accounts for the observed paramagnetism. This prediction is a direct consequence of the LCAO approach and cannot be derived from simple Lewis structures.
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 antibonding orbitals matter only when they are occupied. Empty antibonding orbitals are where incoming electrons go: in a reaction a nucleophile donates into a \(\pi^*\) or \(\sigma^*\) orbital, and when light is absorbed an electron is promoted into one. Occupied antibonding orbitals cancel bonding ones, which is why \(\mathrm{He_2}\) does not form, and why adding an electron to \(\mathrm{O_2}\) to give \(\mathrm{O_2^-}\) lowers the bond order from 2 to 1.5 and lengthens the bond. C
Worked example
Problem. Consider the hypothetical diatomic molecule \(\mathrm{B_2}\). Boron has 3 valence electrons. Determine the bond order and predict whether \(\mathrm{B_2}\) is paramagnetic or diamagnetic, assuming the standard MO energy ordering for second-row diatomics where \(\pi_{2p}\) orbitals are lower in energy than \(\sigma_{2p}\).
- Calculate the total number of valence electrons: \(3 \times 2 = 6\) electrons.
- Fill the molecular orbitals in order of increasing energy: \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\pi_{2p_x}\), \(\pi_{2p_y}\). The first two electrons fill \(\sigma_{2s}\), the next two fill \(\sigma^*_{2s}\). This leaves two electrons to place in the \(\pi_{2p}\) orbitals.
- Apply Hund's rule: The two remaining electrons occupy the degenerate \(\pi_{2p_x}\) and \(\pi_{2p_y}\) orbitals singly with parallel spins.
- Count the electrons in bonding orbitals: 2 in \(\sigma_{2s}\) and 2 in \(\pi_{2p}\). Total \(n_{\text{bonding}} = 4\).
- Count the electrons in antibonding orbitals: 2 in \(\sigma^*_{2s}\). Total \(n_{\text{antibonding}} = 2\).
- Calculate the bond order: \(\text{BO} = \frac{4 - 2}{2} = 1\).
- Determine magnetic property: Since there are two unpaired electrons in the \(\pi_{2p}\) orbitals, the molecule is paramagnetic.
Units check. Bond order is a dimensionless quantity, and magnetic properties are qualitative descriptors, so no units are required for the final answer.
Three quick tests
- What is the primary difference between a bonding and an antibonding molecular orbital in terms of electron density?
Answer
Bonding orbitals have electron density concentrated between the nuclei, while antibonding orbitals have a node between the nuclei. - If a molecule has 4 electrons in bonding orbitals and 2 in antibonding orbitals, what is its bond order?
Answer
The bond order is 1. - Why does MO theory predict that \(\mathrm{O_2}\) is paramagnetic?
Answer
Because it has two unpaired electrons in degenerate \(\pi^*\) antibonding orbitals.
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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.