Two molecular orbitals are being evaluated for a potential bonding interaction in a C₂v molecule. Orbital A transforms as the A₁ irreducible representation; Orbital B transforms as B₂. Can they form a bonding interaction?
AYes — any two orbitals that overlap in space can interact regardless of symmetry
BNo — orbitals can only mix if they belong to the same irreducible representation
CYes — energy matching is the only requirement for orbital interaction
DNo — only orbitals on the same atom can interact within a given point group
Group theory dictates that two orbitals can interact only if they belong to the same irreducible representation. A₁ and B₂ are distinct symmetry species in C₂v; their overlap integral is identically zero by symmetry — not just small, but exactly zero. No amount of energy matching can overcome a symmetry-forbidden interaction. Option 0 is the common naive misconception: spatial proximity is necessary but not sufficient; symmetry compatibility is the gating criterion.
Question 2 Multiple Choice
A centrosymmetric molecule (one with an inversion center, like CO₂) has a vibrational mode that is observed to be Raman active. Can this same mode also be IR active?
AYes, if the mode has sufficient energy to absorb infrared radiation
BNo — the mutual exclusion rule forbids any mode from being both IR and Raman active in a centrosymmetric molecule
CYes, as long as the mode transforms like x, y, or z in the character table
DIt depends on whether the molecule is linear or bent
The mutual exclusion rule states that in a molecule possessing a center of inversion (i), no vibrational mode can be simultaneously IR active and Raman active. IR activity requires a mode to transform like a translational vector (x, y, or z — ungerade under inversion); Raman activity requires transformation like a quadratic function (x², xy — gerade under inversion). These two sets belong to mutually exclusive irreducible representations in any centrosymmetric point group. This is a direct consequence of character table structure, not an empirical observation.
Question 3 True / False
Water (H₂O) and sulfur dioxide (SO₂) both belong to the C₂v point group, yet they have different numbers of atoms and different bond angles. Because they share a point group, all their vibrational modes are IR active.
TTrue
FFalse
Answer: True
Both molecules have 3 atoms and 3 vibrational modes. For C₂v, the character table shows that the irreducible representations of vibrational modes for both molecules are 2A₁ + B₂ (the same symmetry species). In C₂v, A₁ transforms like z, and B₂ transforms like y — both are translational symmetry species, so all three modes are IR active. This illustrates the predictive power of point groups: knowing the point group immediately tells you the spectroscopic activity without computing any integrals.
Question 4 True / False
The point group of a molecule is determined primarily by the number and types of atoms it contains, so two molecules with the same molecular formula typically belong to the same point group.
TTrue
FFalse
Answer: False
Point group is determined by the complete set of symmetry operations the molecule possesses — its rotation axes, mirror planes, inversion center, and improper axes — which depend critically on molecular geometry, not just composition. BF₃ (trigonal planar) belongs to D₃h, while NF₃ (trigonal pyramidal) belongs to C₃v, even though both have 4 atoms. Similarly, H₂O₂ (C₂) and H₂O (C₂v) share a formula class but differ in symmetry. Molecular geometry, not atom count, determines the point group.
Question 5 Short Answer
Explain how group theory allows a chemist to determine whether a vibrational mode is IR active without computing the dipole transition integral.
Think about your answer, then reveal below.
Model answer: A vibrational mode is IR active if and only if it transforms as the same irreducible representation as one of the translational vectors (x, y, or z). The character table of the molecular point group lists which irreducible representations correspond to these vectors. By decomposing the 3N degrees of freedom into irreducible representations using the reduction formula, you identify each vibrational mode's symmetry species and check whether it appears in the x, y, or z row of the character table. If it does, the transition dipole integral is necessarily nonzero by symmetry — no numerical integration required.
This is the central payoff of group theory: symmetry arguments replace integral computation. The transition moment integral ⟨ψ_vib | μ̂ | 0⟩ is nonzero only if the integrand (the product of the mode's symmetry species and the dipole operator symmetry) contains the totally symmetric representation. The dipole operator transforms like x, y, or z. If the vibrational mode's representation matches any of these, the integral is guaranteed nonzero by symmetry. The same logic applies to Raman activity (the mode must transform like a quadratic function) and to orbital interactions (both orbitals must share the same irreducible representation).