CHEM 361 / Test Bank / MO Theory
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Test Bank — Unit 1

Molecular Orbital Theory

Practice problems drawn from Housecroft, Atkins, JD Lee, and others. Select an answer or click "Show Answer" to reveal the explanation.

LCAO & MO Fundamentals

Questions 1–3
Housecroft & Sharpe, Fig. 2.10 Housecroft
In-phase and out-of-phase combination of atomic orbitals
Fig. 2.10 — In-phase and out-of-phase combination of atomic orbitals
Q1

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?

  • a In-phase gives an antibonding MO; out-of-phase gives a bonding MO
  • b Both combinations produce bonding MOs of equal energy
  • c In-phase gives a bonding MO (constructive interference, increased electron density between nuclei); out-of-phase gives an antibonding MO (destructive interference, a node between nuclei)
  • d In-phase gives a non-bonding MO; out-of-phase gives an antibonding MO
Answer: (c)

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.

Atkins & Shriver, Chapter 2 Atkins
Q2

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:

  • a Have a single bond and be stable
  • b Be unstable and not exist as a bound species under normal conditions
  • c Have a very strong triple bond
  • d Be paramagnetic with one unpaired electron
Answer: (b)

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.

Housecroft & Sharpe, Fig. 2.12 Housecroft
Sigma and pi overlap
Fig. 2.12 — Head-on (sigma) vs. lateral (pi) overlap of p orbitals
Q3

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?

  • a Head-on overlap concentrates electron density directly along the internuclear axis, giving greater orbital overlap and a larger stabilization energy
  • b Lateral overlap places more electron density along the internuclear axis than head-on overlap
  • c Pi bonds involve s orbitals, which are smaller and overlap less effectively
  • d Sigma bonds always involve d orbitals and therefore have higher overlap integrals
Answer: (a)

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.

Homonuclear Diatomics

Questions 4–8
Housecroft & Sharpe, Fig. 2.17 Housecroft
MO diagram of O2
Fig. 2.17 — MO diagram for O2
Q4

Molecular oxygen (O2) is experimentally observed to be paramagnetic. Looking at the MO diagram above, which feature of the electronic configuration explains this paramagnetism?

  • a All electrons are paired in bonding MOs, so O2 is actually diamagnetic
  • b There is one unpaired electron in the σ*2p orbital
  • c The bond order is zero, and all electrons are in antibonding orbitals
  • d There are two unpaired electrons, one in each of the degenerate π*2p orbitals (Hund's rule)
Answer: (d)

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.

Housecroft & Sharpe, Fig. 2.11 Housecroft
Photoelectron spectrum of N2
Fig. 2.11 — Photoelectron spectrum (PES) of N2
Q5

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?

  • a PES shows that all electrons in N2 have identical energies, confirming a single bonding MO
  • b Each peak corresponds to ionization from a specific MO; the number and spacing of peaks match the predicted MO energy level diagram, including the effect of s-p mixing on the σ2p level
  • c PES only measures core electrons and tells nothing about valence MOs
  • d PES disproves MO theory by showing more peaks than predicted
Answer: (b)

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.

JD Lee, Chapter 4 JD Lee
Q6

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?

  • a It occurs in O2 and F2, raising the π2p above the σ2p
  • b It occurs in all diatomics equally and has no effect on energy ordering
  • c It is significant in Li2 through N2 (where the 2s–2p energy gap is small), and it pushes the σ2p above the π2p in energy
  • d It only occurs in heteronuclear diatomics like CO and HF
Answer: (c)

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.

Housecroft & Sharpe, Fig. 2.18 Housecroft
MO diagrams comparing N2 species
Fig. 2.18 — MO energy levels for N2 and related species
Q7

Calculate the bond orders of N2 (10 valence electrons) and N2+ (9 valence electrons). Which species has the shorter bond?

  • a N2 has BO = 3 and N2+ has BO = 2.5; N2 has the shorter bond
  • b Both have BO = 3 and identical bond lengths
  • c N2+ has BO = 3.5 and the shorter bond
  • d N2 has BO = 2 and N2+ has BO = 2.5; N2+ has the shorter bond
Answer: (a)

For N2 (10 valence electrons): (σ2s)2(σ*2s)22p)42p)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.

Housecroft & Sharpe, Fig. 2.19 Housecroft
MO diagram for He2 and He2+
Fig. 2.19 — MO diagrams for He2 and He2+
Q8

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.

  • a He2 has BO = 1 but the bond is too weak; He2+ has BO = 0
  • b He2 has BO = 0 (2 bonding, 2 antibonding electrons — no net bond); He2+ has BO = 0.5 (2 bonding, 1 antibonding electron), giving a weak but real bond
  • c Both have BO = 0, but He2+ is stabilized by its positive charge alone
  • d He2 has BO = 0.5 and He2+ has BO = 1
Answer: (b)

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.

Heteronuclear Diatomics

Questions 9–12
Housecroft & Sharpe, Fig. 2.20 Housecroft
MO diagram for HF
Fig. 2.20 — MO diagram for HF
Q9

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?

  • a Hydrogen is more electronegative than fluorine, so its orbital dominates
  • b The F 2p orbital is higher in energy and therefore contributes more to the bonding MO
  • c Fluorine is much more electronegative; its 2p orbitals are lower in energy and closer in energy to the bonding MO, so they contribute more to it
  • d Both atoms contribute equally to all MOs in heteronuclear diatomics
Answer: (c)

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).

Housecroft & Sharpe, Fig. 2.22 Housecroft
MO diagram for CO
Fig. 2.22 — MO diagram for CO
Q10

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?

  • a CO has fewer MOs because carbon has fewer orbitals than nitrogen
  • b The MO diagram is asymmetric: oxygen's AOs lie at lower energy than carbon's due to its greater electronegativity, so the bonding MOs are polarized toward O and antibonding MOs toward C
  • c CO has a bond order of 2 instead of 3 because heteronuclear bonding is always weaker
  • d There is no difference; isoelectronic species always have identical MO diagrams
Answer: (b)

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).

Atkins & Shriver, Chapter 2 Atkins
Q11

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?

  • a On carbon — despite C being less electronegative, the antibonding character of the HOMO weights it toward C; this is why CO binds to transition metals through the carbon end
  • b On oxygen — the more electronegative atom always holds the HOMO
  • c Equally shared between C and O because the bond order is 3
  • d On carbon, because carbon is more electronegative than oxygen
Answer: (a)

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).

Housecroft & Sharpe, Fig. 2.24 Housecroft
MO diagram for NO
Fig. 2.24 — MO diagram for NO
Q12

Nitric oxide (NO) has 15 electrons. Why is NO described as a radical, and what is its bond order?

  • a NO has zero unpaired electrons and a bond order of 3
  • b NO has two unpaired electrons in the π* orbitals and a bond order of 2
  • c NO has one unpaired electron in a bonding π orbital and a bond order of 3
  • d NO has one unpaired electron in a π* orbital and a bond order of 2.5; the odd electron count makes it a radical
Answer: (d)

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.

Applications & Frontier Orbitals

Questions 13–15
Atkins & Shriver, Chapter 2 Atkins
Q13

The concept of "frontier orbitals" is central to understanding chemical reactivity. What are frontier orbitals, and why are they important?

  • a They are the core 1s orbitals, which determine the atom's identity
  • b They are the HOMO (highest occupied molecular orbital) and LUMO (lowest unoccupied molecular orbital); chemical reactions involve electron transfer or sharing between the HOMO of one species and the LUMO of another
  • c They are the orbitals at the boundary between s and p subshells
  • d They are the non-bonding orbitals that do not participate in any reactions
Answer: (b)

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).

Housecroft & Sharpe, Fig. 2.23 Housecroft
CO as a ligand: sigma donation and pi back-bonding
Fig. 2.23 — CO as a ligand: σ donation and π back-bonding
Q14

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?

  • a CO donates from its π bonding orbitals into the metal's s orbital only
  • b The metal donates electrons to CO's σ bonding orbital, weakening the C–O bond
  • c CO σ-donates the carbon lone pair (HOMO) into an empty metal d orbital, and the metal π-back-donates from its filled d orbitals into the CO π* (LUMO); these two interactions are synergistic
  • d CO binds only through electrostatic attraction between the C–O dipole and the metal cation
Answer: (c)

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).

JD Lee, Chapter 4; Housecroft Fig. 2.28 JD Lee
MO diagram for ICl
Fig. 2.28 — MO approach to interhalogen bonding (ICl)
Q15

Using an MO perspective, predict the bond character of ICl. Consider the relative electronegativities of iodine (χ = 2.7) and chlorine (χ = 3.2).

  • a Polar covalent: the bonding MOs are polarized toward Cl (more electronegative), giving Cl a partial negative charge; the bond is covalent but with significant ionic character
  • b Purely ionic: I donates an electron completely to Cl, forming I+Cl
  • c Purely covalent: both atoms contribute equally to all MOs because they are both halogens
  • d Non-bonding: the electronegativity difference is too small to form a bond
Answer: (a)

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.

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Molecular Orbital Theory