Questions: Group Theory and Molecular Symmetry: Point Groups and Applications

Question 1

A vibrational mode of a centrosymmetric molecule transforms as the A₁g irreducible representation. Without calculating any integrals, a chemist concludes this mode is IR-inactive. What is the correct reasoning?

• A. A₁g representations always have zero character under all symmetry operations
• B. The transition dipole moment operator transforms as ungerade (antisymmetric under inversion), while A₁g is gerade (symmetric); their direct product cannot contain the totally symmetric representation
• C. IR activity requires at least one C₃ rotation axis, which centrosymmetric molecules lack
• D. The A₁g mode has too many nodes in its displacement pattern to couple to light

Answer: (B)

IR activity requires that the direct product of the initial state, the dipole moment operator (which transforms as a translation — x, y, or z), and the final state contains the totally symmetric representation. In a centrosymmetric point group, translations are ungerade (antisymmetric under inversion), while A₁g is gerade. The product gerade × ungerade is ungerade, which cannot contain the totally symmetric (gerade) representation. The integral vanishes by symmetry — no calculation needed.

Question 2

A student finds that water (H₂O, C₂ᵥ point group) has vibrational modes of symmetry 2A₁ + B₂. She concludes only the B₂ mode is IR-active, reasoning that A₁ is the 'totally symmetric' representation and symmetric modes cannot produce a changing dipole. Is she correct?

• A. Yes — only antisymmetric modes produce a changing dipole moment and are IR-active
• B. No — in C₂ᵥ, the z-translation transforms as A₁, so A₁ modes are also IR-active; all three of water's modes are IR-active
• C. No — none of water's modes are IR-active because the molecule has a permanent dipole
• D. No — B₂ is Raman-active only in C₂ᵥ symmetry

Answer: (B)

IR activity requires the mode to transform as the same irreducible representation as at least one translational coordinate (x, y, or z). In C₂ᵥ, the character table shows that z transforms as A₁ and (x, y) transform as B₁ and B₂. So A₁ modes ARE IR-active through the z-component of the dipole. The student's intuition that A₁ 'totally symmetric' modes can't change the dipole is wrong — stretching the O–H bonds symmetrically (A₁) changes the bond length and the dipole magnitude along z.

Question 3

For a molecule with an inversion center, no vibrational mode can be simultaneously IR-active and Raman-active.

• T. True
• F. False

Answer: True

This is the mutual exclusion rule. In centrosymmetric point groups, IR-active modes must transform as ungerade representations (like translations), while Raman-active modes must transform as gerade representations (like quadratic functions x², xy, etc.). Since gerade and ungerade are mutually exclusive — a representation cannot be both symmetric and antisymmetric under inversion — no mode can satisfy both conditions simultaneously.

Question 4

Group theory is most useful for molecules with high symmetry (Td, Oh); for low-symmetry molecules in the C₁ or Cs point group, the character table provides no useful information and direct calculation is required.

• T. True
• F. False

Answer: False

Group theory applies to all point groups, including low-symmetry ones. A molecule in C₁ has only the identity operation, so all vibrational modes transform as the totally symmetric A representation — meaning all modes are both IR- and Raman-active. This is still useful information derived from group theory, even though fewer simplifications arise. The formalism is general; the degree of simplification scales with the number of symmetry operations.

Question 5

Explain why group theory can determine that a quantum mechanical integral equals zero without evaluating it explicitly.

Model answer: If the integrand does not transform as (or contain) the totally symmetric irreducible representation when decomposed into symmetry species, the integral over all space must be zero. Any function that is antisymmetric under a symmetry operation of the molecule must integrate to zero, because the positive and negative contributions cancel exactly. Group theory identifies this by checking the direct product of the integrand's component representations: if the totally symmetric representation is absent from the reduction, the integral vanishes — and no numerical calculation is needed.

This is why group theory is so powerful: orthogonality of irreducible representations provides a vanishing theorem that eliminates most integrals before any computation. For a molecule like SF₆ with 48 symmetry operations, this eliminates the vast majority of possible interaction integrals and makes otherwise intractable problems tractable.