Practice problems drawn from Housecroft, Atkins, JD Lee, and others. Select an answer or click "Show Answer" to reveal the explanation.
When two atomic orbitals on adjacent atoms combine according to the LCAO method, two molecular orbitals result. Referring to the figure above, what happens when the orbitals combine in-phase versus out-of-phase?
When two atomic orbitals combine in-phase (same sign of the wavefunction), constructive interference occurs and electron density builds up between the two nuclei, producing a bonding MO that is lower in energy than the original atomic orbitals. Out-of-phase combination (opposite signs) leads to destructive interference, creating a node between the nuclei — this is an antibonding MO, higher in energy than the original AOs. The number of molecular orbitals always equals the number of atomic orbitals combined.
The bond order of a diatomic molecule is calculated using the formula BO = (number of bonding electrons − number of antibonding electrons) / 2. A molecule with a bond order of zero is predicted to:
Bond order = (bonding e− − antibonding e−) / 2. When BO = 0, the stabilizing effect of bonding electrons is exactly cancelled by the destabilizing effect of antibonding electrons (in fact, antibonding MOs are slightly more destabilizing than bonding MOs are stabilizing). The molecule has no net bond and does not exist as a stable species. A classic example is He2: (σ1s)2(σ*1s)2, giving BO = (2 − 2)/2 = 0.
Referring to the figure, a σ bond formed from head-on overlap of two p orbitals is generally stronger than a π bond formed from lateral (side-on) overlap. Why?
In head-on (σ) overlap, the lobes of two p orbitals point directly at each other along the internuclear axis. This concentrates electron density between the nuclei and maximizes the overlap integral S, producing a large stabilization energy. In lateral (π) overlap, the p lobes are parallel and overlap above and below the axis — the overlap region is more diffuse and less effective. As a result, π bonds are individually weaker than σ bonds, though a triple bond (one σ + two π) is collectively very strong.
Molecular oxygen (O2) is experimentally observed to be paramagnetic. Looking at the MO diagram above, which feature of the electronic configuration explains this paramagnetism?
O2 has 16 electrons. After filling the bonding MOs, two electrons remain to occupy the two degenerate π*2p antibonding orbitals. By Hund's rule, these two electrons enter the two π* orbitals with parallel spins rather than pairing in one orbital. This gives O2 two unpaired electrons, making it paramagnetic — a fact that Lewis structures (which predict all electrons paired) fail to explain. The bond order is (8 − 4)/2 = 2, consistent with a double bond.
The photoelectron spectrum of N2 shown above reveals peaks at different ionization energies corresponding to electrons in different MOs. How does PES provide experimental support for MO theory?
Photoelectron spectroscopy ionizes electrons from occupied MOs. Each peak in the spectrum corresponds to a distinct MO energy level — the ionization energy equals the negative of the orbital energy (Koopmans' theorem). For N2, the PES shows that the σ2p MO is higher in energy than the π2p MOs, confirming the effect of s-p mixing. The vibrational fine structure of each peak also reveals whether the electron was in a bonding (many vibrational states, changed bond length upon ionization) or weakly bonding/non-bonding orbital. This direct experimental evidence validates the MO energy ordering predicted by theory.
The phenomenon of s-p mixing (also called s-p hybridization in MO context) alters the expected energy ordering of molecular orbitals in homonuclear diatomics. Which of the following correctly describes when s-p mixing occurs and its effect?
s-p mixing arises because the σ2s and σ2p MOs have the same symmetry (σg). When the 2s and 2p atomic orbital energies are close together (as in Li through N), these σ MOs interact and repel each other in energy: the σ2s is pushed down and the σ2p is pushed up. This reverses the expected ordering so that σ2p lies above π2p. In O2 and F2, the 2s–2p gap is large enough that s-p mixing is negligible, and the "normal" ordering (σ2p below π2p) is restored.
Calculate the bond orders of N2 (10 valence electrons) and N2+ (9 valence electrons). Which species has the shorter bond?
For N2 (10 valence electrons): (σ2s)2(σ*2s)2(π2p)4(σ2p)2. Bonding = 8, antibonding = 2, so BO = (8 − 2)/2 = 3. For N2+ (9 valence electrons), one electron is removed from the highest occupied MO (σ2p): bonding = 7, antibonding = 2, so BO = (7 − 2)/2 = 2.5. Higher bond order correlates with shorter, stronger bonds. N2 bond length is 109.8 pm; N2+ is 111.6 pm — confirming that the triple bond in N2 is shorter.
He2 does not exist as a stable molecule, yet He2+ has been detected experimentally. Using bond order arguments and the MO diagram above, explain why.
He2 has 4 electrons: (σ1s)2(σ*1s)2. BO = (2 − 2)/2 = 0 — the antibonding electrons completely cancel the bonding, and in fact antibonding MOs are slightly more destabilizing than bonding MOs are stabilizing, so the molecule is energetically unfavorable. He2+ has 3 electrons: (σ1s)2(σ*1s)1. BO = (2 − 1)/2 = 0.5 — there is a net bonding interaction, producing a weak but detectable bond. He2+ has been observed with a bond dissociation energy of about 230 kJ/mol.
In the MO diagram for HF shown above, the bonding MO has a greater contribution from the F 2p orbital than from the H 1s orbital. Why?
In heteronuclear diatomics, the bonding MO is weighted toward the more electronegative atom. Fluorine (χ = 4.0) is far more electronegative than hydrogen (χ = 2.1), so the F 2p orbitals lie at lower energy. The bonding MO, being lower in energy, has a larger coefficient on F — meaning the bonding electron density is polarized toward fluorine. Conversely, the antibonding MO is weighted toward H. This asymmetry is the MO explanation for the polar nature of the H–F bond (partial negative charge on F).
CO is isoelectronic with N2 (both have 14 electrons). However, the MO diagram for CO differs from that of N2. What is the key difference and why?
Although CO and N2 are isoelectronic, CO is heteronuclear: the oxygen AOs are lower in energy than the carbon AOs because oxygen is more electronegative. This makes the MO diagram asymmetric — unlike the symmetric N2 diagram. The bonding MOs have larger coefficients on O (polarized toward the more electronegative atom), while the antibonding MOs and the HOMO lone pair have larger coefficients on C. Despite this asymmetry, CO retains a bond order of 3 and is in fact the strongest diatomic bond known (1076 kJ/mol), even stronger than N2 (945 kJ/mol).
In carbon monoxide (CO), the highest occupied molecular orbital (HOMO) is a lone pair. On which atom is this HOMO lone pair predominantly localized, and what is the chemical significance of this?
The HOMO of CO (sometimes labeled 3σ or 5σ) is a lone pair that is largely localized on carbon. This may seem counterintuitive since oxygen is more electronegative, but the HOMO has antibonding character with respect to the 2s interaction, and antibonding MOs are weighted toward the less electronegative atom. This carbon-centered lone pair is the reason CO acts as a strong σ-donor ligand in coordination chemistry: it donates the C lone pair into empty metal d orbitals. CO always binds to metals M–C≡O, never M–O≡C (except in rare isocarbonyl cases).
Nitric oxide (NO) has 15 electrons. Why is NO described as a radical, and what is its bond order?
NO has 15 electrons — an odd number, guaranteeing at least one unpaired electron. The electronic configuration places 8 electrons in bonding MOs and 3 in antibonding MOs (the extra electron relative to N2 enters a π* orbital). Bond order = (8 − 3)/2 = 2.5. The single unpaired electron in π* makes NO a radical (a species with an unpaired electron). This radical character underlies NO's biological importance as a signaling molecule. Removing the antibonding electron to form NO+ increases the bond order to 3 and makes the ion isoelectronic with N2 and CO.
The concept of "frontier orbitals" is central to understanding chemical reactivity. What are frontier orbitals, and why are they important?
Frontier molecular orbital (FMO) theory, developed by Kenichi Fukui (Nobel Prize 1981), identifies the HOMO and LUMO as the key orbitals controlling reactivity. A nucleophile donates electrons from its HOMO into the LUMO of the electrophile. The energy gap between HOMO and LUMO determines the kinetic accessibility of a reaction — a smaller gap generally means easier reaction. In coordination chemistry, frontier orbitals explain ligand binding: a ligand's HOMO donates into a metal's LUMO (σ donation), and the metal's filled d orbitals (acting as HOMO) can donate into the ligand's LUMO (π back-bonding).
CO is one of the strongest-field ligands in coordination chemistry. Referring to the figure above, which combination of orbital interactions explains why CO binds so strongly to transition metals?
CO binds to metals through a synergistic two-way interaction. First, the carbon lone pair (the HOMO of CO, localized on C) donates into an empty metal d orbital — this is σ donation, which increases electron density on the metal. Second, filled metal d orbitals with appropriate symmetry donate electron density back into the empty π* antibonding orbitals of CO — this is π back-bonding (or back-donation). The two processes reinforce each other: σ donation increases electron density on the metal, which enhances π back-bonding, which in turn makes CO a better σ donor. A consequence of π back-bonding is that electron density enters CO's antibonding orbitals, weakening and lengthening the C≡O bond (the CO stretching frequency in IR decreases from 2143 cm−1 in free CO).
Using an MO perspective, predict the bond character of ICl. Consider the relative electronegativities of iodine (χ = 2.7) and chlorine (χ = 3.2).
ICl is a heteronuclear diatomic between two halogens with different electronegativities. Chlorine (χ = 3.2) is more electronegative than iodine (χ = 2.7), so the Cl atomic orbitals lie at lower energy. In the MO diagram, the bonding MOs have larger coefficients on Cl, meaning bonding electron density is polarized toward chlorine. This gives Cl a partial negative charge (δ−) and I a partial positive charge (δ+). The bond is polar covalent — not purely ionic (the electronegativity difference of 0.5 is moderate) and not purely covalent (the atoms are not identical). The MO approach naturally explains the polarity: no additional assumptions about "percent ionic character" are needed, as the unequal orbital coefficients directly produce the charge asymmetry.
Molecular Orbital Theory