Heteronuclear Diatomics
~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the basic construction of molecular orbitals from atomic orbitals and the calculation of bond order for homonuclear diatomics. Familiarity with the concept of electronegativity and its qualitative effect on electron density is also assumed.
Why this matters
Most real-world chemical bonds are between different atoms, making heteronuclear diatomics the standard case rather than the exception.
While homonuclear diatomics like \(\mathrm{N_2}\) or \(\mathrm{O_2}\) are useful for establishing the fundamental rules of Molecular Orbital (MO) theory, they represent a special case where the atomic orbitals (AOs) have identical energies. In heteronuclear diatomics, such as \(\mathrm{CO}\), \(\mathrm{NO}\), or \(\mathrm{HF}\), the atoms have different nuclear charges and different atomic radii. This leads to a mismatch in AO energies, which fundamentally alters the shape, energy, and electron distribution of the resulting molecular orbitals. Understanding this asymmetry is crucial for predicting bond polarity, dipole moments, and the reactivity of molecules that constitute the bulk of organic and biological chemistry.
The Asymmetry of Mixing
In a homonuclear diatomic molecule, the two atomic orbitals contributing to a bond have exactly the same energy. When they combine, they mix equally. The resulting bonding molecular orbital contains 50% of the character from atom A and 50% from atom B. This is a symmetric situation.
In a heteronuclear diatomic molecule, the atomic orbitals have different energies. The orbital on the more electronegative atom is lower in energy (more stable) because the nucleus holds the electron more tightly. The orbital on the less electronegative atom is higher in energy. When these two orbitals combine, they do not mix equally. The lower-energy atomic orbital contributes more to the bonding molecular orbital, while the higher-energy atomic orbital contributes more to the antibonding molecular orbital. This is often described as "unequal mixing." The result is that the bonding electron density is skewed towards the more electronegative atom, creating a permanent dipole moment. The molecular orbitals are no longer symmetric with respect to the midpoint of the bond.
Energy Level Diagrams and Unequal Mixing
The interaction between two atomic orbitals of different energies can be visualized using an energy level diagram. Let \(\phi_A\) be the AO on the more electronegative atom (lower energy, \(E_A\)) and \(\phi_B\) be the AO on the less electronegative atom (higher energy, \(E_B\)), such that \(E_A < E_B\).
The resulting molecular orbitals are linear combinations: \(\psi_+ = c_A \phi_A + c_B \phi_B\) (bonding) and \(\psi_- = c_A \phi_A - c_B \phi_B\) (antibonding). Because \(E_A\) is closer to the energy of the bonding MO, the coefficient \(c_A\) will be larger than \(c_B\) in the bonding orbital. Conversely, in the antibonding orbital, the coefficient of the higher-energy AO (\(\phi_B\)) will be larger. This means the bonding MO is "localized" more on atom A, and the antibonding MO is "localized" more on atom B.
The energy gap between the bonding and antibonding orbitals is determined by the resonance integral \(\beta\). However, the position of the bonding MO relative to the atomic orbitals is shifted. The bonding MO is closer in energy to the lower-energy AO (\(E_A\)) than to the higher-energy AO (\(E_B\)). This is a direct consequence of the variational principle and the fact that the lower-energy component dominates the wavefunction of the lower-energy state.
In this simple approximation, the bonding orbital is lowered and the antibonding orbital raised by the same small amount, \(\beta^2/(E_B - E_A)\). The larger the energy mismatch, the weaker the covalent mixing, and the more the bond behaves as ionic. This has significant implications for bond strength and length in highly polar bonds.
Bond Polarity and Dipole Moments
The unequal mixing of atomic orbitals leads directly to bond polarity. Since the bonding molecular orbital has greater electron density on the more electronegative atom, that atom carries a partial negative charge (\(\delta^-\)), and the other atom carries a partial positive charge (\(\delta^+\)).
The magnitude of this charge separation is quantified by the bond dipole moment \(\mu\). For a diatomic molecule, the dipole moment is simply the product of the magnitude of the partial charge \(Q\) and the bond length \(d\):
The direction of the dipole moment vector points from the positive to the negative charge. In MO theory, we can estimate the degree of ionic character by comparing the observed dipole moment to the theoretical dipole moment if the bond were 100% ionic (i.e., a full electron transfer). The percentage ionic character is given by:
where \(e\) is the elementary charge. This metric allows us to bridge the gap between the purely covalent MO picture and the purely ionic picture, showing that most heteronuclear bonds exist on a continuum between these two extremes.
Application to Carbon Monoxide (CO)
Carbon monoxide is a classic example of a heteronuclear diatomic with a triple bond. It is isoelectronic with \(\mathrm{N_2}\), having 10 valence electrons. However, the MO diagram for CO is more complex due to the energy mismatch between C and O orbitals.
Oxygen is more electronegative than carbon (\(\chi_{\mathrm{O}} \approx 3.44\), \(\chi_{\mathrm{C}} \approx 2.55\)). Therefore, the O 2s and 2p orbitals are lower in energy than the C 2s and 2p orbitals. The bonding \(\sigma\) and \(\pi\) orbitals will have greater contribution from the O atomic orbitals. The antibonding orbitals will have greater contribution from the C atomic orbitals.
Despite the polarity, the dipole moment of CO is surprisingly small (\(\approx 0.11\ \mathrm{D}\)) and points in the "unexpected" direction: the carbon end is slightly negative. This is because the lone pair on the carbon atom (which resides in an orbital with significant C character) contributes significantly to the dipole, opposing the dipole created by the bonding electrons. This subtle interplay between bonding electron density and lone pair density is a hallmark of heteronuclear diatomics and cannot be predicted by simple electronegativity arguments alone.
The results
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.
Electronegativity and bond polarity
When two bonded atoms differ in electronegativity \(\chi\) (their intrinsic pull on shared bonding electrons), the bonding electron pair is shared unequally, producing partial charges \(\pm\delta\) and a bond dipole moment \(\mu=Qd\) (charge separation \(Q\) times bond length \(d\)). The molecule's overall dipole moment is the vector sum of every bond dipole, weighted by the molecular geometry (VSEPR) — so a molecule can have strongly polar bonds yet be nonpolar overall if those bond dipoles cancel by symmetry.
A common misconception is that the more electronegative atom always contributes more to the *antibonding* orbital. While it is true that the higher-energy AO contributes more to the antibonding MO, the *lower*-energy AO (more electronegative) contributes more to the *bonding* MO. Confusing which orbital is stabilised and which is destabilized relative to the atomic levels leads to incorrect predictions about electron density distribution and reactivity. The key is to remember that the bonding MO is always closer in energy to the lower-energy AO, and thus has greater character from that AO. C
Worked example
Problem. Consider a hypothetical heteronuclear diatomic molecule AB. The atomic orbital energy of A is \(-10.0\ \mathrm{eV}\) and of B is \(-6.0\ \mathrm{eV}\). The resonance integral \(\beta\) is \(2.0\ \mathrm{eV}\). Calculate the approximate energy of the bonding molecular orbital using the perturbation formula. If the bond length is \(1.0\ \text{\AA}\) and the observed dipole moment is \(1.0\ \mathrm{D}\), calculate the percentage ionic character. (Use \(1\ \mathrm{D} = 3.336 \times 10^{-30}\ \mathrm{C\,m}\) and \(e = 1.602 \times 10^{-19}\ \mathrm{C}\).)
- Identify the lower energy AO: \(E_A = -10.0\ \mathrm{eV}\) (more electronegative). The higher energy AO is \(E_B = -6.0\ \mathrm{eV}\). The energy difference is \(\Delta E = E_B - E_A = 4.0\ \mathrm{eV}\).
- Calculate the bonding MO energy using \(E_{\text{bonding}} \approx E_A - \frac{\beta^2}{\Delta E}\). Substituting values: \(E_{\text{bonding}} \approx -10.0 - \frac{(2.0)^2}{4.0} = -10.0 - 1.0 = -11.0\ \mathrm{eV}\).
- Convert the bond length to meters: \(d = 1.0\ \text{\AA} = 1.0 \times 10^{-10}\ \mathrm{m}\).
- Calculate the theoretical ionic dipole moment: \(\mu_{\text{ionic}} = ed = (1.602 \times 10^{-19}\ \mathrm{C})(1.0 \times 10^{-10}\ \mathrm{m}) = 1.602 \times 10^{-29}\ \mathrm{C\,m}\).
- Convert the observed dipole moment to \(\mathrm{C\,m}\): \(\mu_{\text{obs}} = 1.0\ \mathrm{D} \times 3.336 \times 10^{-30}\ \mathrm{C\,m/D} = 3.336 \times 10^{-30}\ \mathrm{C\,m}\).
- Calculate the percentage ionic character: \(\%\,\text{ionic} = \frac{3.336 \times 10^{-30}}{1.602 \times 10^{-29}} \times 100 \approx 20.8\%.\)
Units check. The energy is in electron-volts, consistent with the input data. The percentage is a dimensionless ratio, as expected for ionic character.
Three quick tests
- Why is the bonding molecular orbital in a heteronuclear diatomic closer in energy to the lower-energy atomic orbital?
Answer
Because the lower-energy AO contributes more to the bonding MO due to unequal mixing, stabilizing the MO closer to its energy level. - If atom A is more electronegative than atom B, which MO (bonding or antibonding) has greater character from atom A?
Answer
The bonding MO has greater character from atom A. - A molecule has a bond length of \(2.0\ \text{\AA}\) and an observed dipole moment of \(2.0\ \mathrm{D}\). What is the approximate percentage ionic character?
Answer
Approximately 21% (using similar conversion factors as the worked example).
Back to CU-102 →
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.