Questions: Molecular Orbital Diagrams for Polyatomic Molecules

Question 1

In the MO diagram for water (H₂O), oxygen's 2px orbital ends up as a nonbonding orbital. Why?

• A. The 2px orbital has too high an energy to interact with hydrogen orbitals
• B. No combination of the two hydrogen 1s orbitals has the same symmetry as oxygen's 2px, so no overlap is possible
• C. The 2px orbital is fully occupied and cannot accept electrons from hydrogen
• D. Oxygen's 2px is antibonding and its interaction with hydrogen is forbidden

Answer: (B)

The two hydrogen 1s orbitals can form only two combinations: in-phase (both positive) and out-of-phase (one positive, one negative). These two combinations have specific symmetry properties that match oxygen's 2s, 2pz, and 2py — but not 2px. Because no hydrogen combination shares the symmetry of 2px, there is no overlap integral between them, and 2px cannot mix into any bonding or antibonding MO. It remains as a pure nonbonding lone pair on oxygen. This is the central insight: symmetry mismatch, not energy mismatch, is what determines whether orbitals can combine.

Question 2

For methane (CH₄) in the Td point group, the four hydrogen 1s orbitals form symmetry combinations labeled a₁ and t₂. Which carbon orbitals interact with each set?

• A. Carbon 2s interacts with a₁; carbon 2p orbitals interact with t₂
• B. Carbon 2s interacts with t₂; carbon 2p orbitals interact with a₁
• C. All four carbon valence orbitals interact with the a₁ combination only
• D. Carbon 2px, 2py, 2pz each interact with a separate hydrogen orbital individually

Answer: (A)

In the Td point group, carbon's 2s orbital has a₁ symmetry and therefore mixes only with the a₁ hydrogen combination. Carbon's three 2p orbitals (2px, 2py, 2pz together) form a triply degenerate t₂ set that mixes only with the t₂ hydrogen combinations. This produces two bonding/antibonding pairs: one a₁ pair and three degenerate t₂ pairs. Filling all four bonding MOs with 8 valence electrons accounts for methane's four equivalent C-H bonds. Option D represents a localized bonding picture (like Lewis structures) that MO theory replaces.

Question 3

In polyatomic MO theory, the rule that orbitals of different symmetry representations cannot mix is a strict selection rule, not merely a preference.

• T. True
• F. False

Answer: True

This is not a guideline — it is a rigorous mathematical result. The overlap integral between two orbitals belonging to different irreducible representations of the molecular point group is exactly zero by symmetry. This means there is no matrix element connecting them in the Hamiltonian, and they cannot mix regardless of how close in energy they are. The selection rule simplifies MO construction enormously: you only need to consider combinations within each symmetry representation separately, ignoring all cross-representation interactions.

Question 4

Most p-orbitals on different atoms in a polyatomic molecule will form bonding and antibonding MO combinations with each other.

• T. True
• F. False

Answer: False

Orbital mixing requires matching symmetry, not just matching orbital type. A p-orbital on one atom will only combine with orbitals on other atoms that belong to the same symmetry representation under the molecule's point group. If the symmetry doesn't match, the overlap integral is zero and no MO combination forms — the orbital remains nonbonding. For example, in water, oxygen's 2px remains nonbonding because no hydrogen orbital combination has the same symmetry. The common misconception that all orbitals of the same type must combine comes from the diatomic case, where symmetry automatically matches.

Question 5

Explain why MO theory for polyatomic molecules naturally handles resonance without needing multiple Lewis structures, using O₃ or benzene as an example.

Model answer: In MO theory, atomic orbitals from all atoms are combined simultaneously into molecular orbitals that are delocalized over the entire molecule. For O₃ or benzene, the π MOs extend over all atoms from the start — there is no single-bond/double-bond distinction at the MO level. The delocalization is built into the basis of the calculation, so there is no need to invoke multiple resonance contributors as a conceptual fix. The 'true structure' is just the electron density described by the filled MOs.

Lewis structures force electrons into localized bonds between pairs of atoms. When the actual electron distribution is delocalized, you need multiple resonance structures to hint at this. MO theory sidesteps this entirely by using a basis that allows electrons to occupy orbitals spanning multiple atoms. For benzene, the three π MOs (one bonding, two weakly bonding) are each spread over all six carbons — no bond is intrinsically single or double. The resonance-hybrid picture is an approximation that MO theory makes unnecessary. This is one reason MO theory is more powerful than Lewis/resonance descriptions for understanding reactivity, UV-Vis spectra, and magnetic properties.