Once the symmetries of a molecule’s vibrational normal modes are known, group theory provides a direct way to determine which vibrational transitions are spectroscopically allowed. Infrared and Raman spectroscopy probe vibrations through different physical interactions, and therefore obey different selection rules.
1. Infrared activity and the electric dipole moment
An infrared transition is observed when a vibration causes a change in the molecular dipole moment. The transition probability is proportional to the square of the transition dipole moment:
\[ \langle \psi_{v'} \,|\, \hat{\mu} \,|\, \psi_v \rangle, \]
where \(\hat{\mu}\) is the electric dipole moment operator. Using symmetry, this integral is nonzero only if
\[ \Gamma(\psi_v)\times\Gamma(\hat{\mu})\times\Gamma(\psi_{v'}) \supset A_1. \]
Because vibrational transitions typically involve adjacent vibrational levels, this condition simplifies to a requirement on the symmetry of the normal mode. A vibrational mode is IR active if its symmetry species transforms like one of the Cartesian coordinates \(x\), \(y\), or \(z\).
Example (IR activity): formaldehyde, \(\mathrm{H_2CO}\) (\(C_{2v}\))
\[ \begin{array}{c|cccc|cc|c} C_{2v} & E & C_2(z) & \sigma_v(xz) & \sigma_v'(yz) & \text{Linear} & \text{Rotations} & \text{Quadratic} \\ \hline A_1 & 1 & 1 & 1 & 1 & z & & x^2,\;y^2,\;z^2 \\ A_2 & 1 & 1 & -1 & -1 & & R_z & xy \\ B_1 & 1 & -1 & 1 & -1 & x & R_y & xz \\ B_2 & 1 & -1 & -1 & 1 & y & R_x & yz \\ \end{array} \]
Assuming that you have already determined the full \(3N\) representation for formaldehyde (4 atoms, so \(3N=12\)) and reduced it to
\[ \Gamma_{3N} = 4A_1 + A_2 + 4B_1 + 3B_2. \]
This representation includes all motions (translations, rotations, and vibrations), so we must remove translations and rotations to obtain \(\Gamma_{\text{vib}}\). :contentReference[oaicite:0]{index=0}
Step 1: Subtract translations and rotations
In \(C_{2v}\), the translational motions transform like \(x\), \(y\), and \(z\), so
\[ \Gamma_{\text{trans}} = A_1 + B_1 + B_2. \]
The rotational motions transform like \(R_x\), \(R_y\), and \(R_z\), giving
\[ \Gamma_{\text{rot}} = A_2 + B_1 + B_2. \]
Therefore,
\[ \Gamma_{\text{vib}} = \Gamma_{3N} - \Gamma_{\text{trans}} - \Gamma_{\text{rot}} = (4A_1 + A_2 + 4B_1 + 3B_2) - (A_1 + B_1 + B_2) - (A_2 + B_1 + B_2). \]
\[ \Gamma_{\text{vib}} = 3A_1 + 2B_1 + B_2. \]
Sanity check: formaldehyde is nonlinear with \(N=4\), so it should have \(3N-6=6\) vibrational modes, and \(3 + 2 + 1 = 6\). :contentReference[oaicite:1]{index=1}
Step 2: Determine which modes are IR active
A vibrational mode is IR active if it transforms like one of the dipole moment components (\(\mu_x,\mu_y,\mu_z\)), i.e., like \(x\), \(y\), or \(z\). In \(C_{2v}\):
- \(z \sim A_1\)
- \(x \sim B_1\)
- \(y \sim B_2\)
Therefore, the IR-active symmetries in \(C_{2v}\) are \(A_1\), \(B_1\), and \(B_2\).
Conclusion for formaldehyde
Since \(\Gamma_{\text{vib}} = 3A_1 + 2B_1 + B_2\), all six vibrational modes are IR active (they all transform like at least one of \(x\), \(y\), or \(z\)).
Big picture: once you have \(\Gamma_{\text{vib}}\), IR activity in \(C_{2v}\) is determined by a quick symmetry match to \(x\), \(y\), and \(z\).
Example (IR activity): carbon dioxide, \(\mathrm{CO_2}\) (\(D_{\infty h}\))
Carbon dioxide is a linear, centrosymmetric molecule with point group \(D_{\infty h}\). It is an ideal example because it has both IR-active and IR-inactive vibrational modes, and it sets up the Raman discussion nicely.
\[ \begin{array}{c|c|c|c} D_{\infty h} & \text{Typical motion} & \text{Linear} & \text{Quadratic} \\ \hline \Sigma_g^+ & \text{symmetric stretch} & & z^2,\;x^2+y^2 \\ \Sigma_u^+ & \text{antisymmetric stretch} & z & \\ \Pi_u & \text{bend (doubly degenerate)} & (x,y) & \\ \Pi_g & \text{(rare vibrational modes)} & & (xz,\;yz) \\ \Delta_g & \text{(higher degeneracy)} & & (x^2-y^2,\;xy) \\ \end{array} \]
Step 1: Count the vibrational modes
\(\mathrm{CO_2}\) has \(N=3\) atoms, so \(3N=9\) total degrees of freedom. Because it is linear, it has 3 translational and 2 rotational degrees of freedom, leaving
\[ \text{Number of vibrational modes} = 3N-5 = 9-5 = 4. \]
These 4 vibrational degrees of freedom correspond to three fundamental vibrations, because one of the modes (the bend) is doubly degenerate.
Given result (symmetry of the normal modes)
Assume the student has already determined the symmetry species of the normal modes. For \(\mathrm{CO_2}\), the fundamental vibrations are:
- Symmetric stretch: \(\Sigma_g^+\)
- Bend (doubly degenerate): \(\Pi_u\) (counts as 2 vibrational degrees of freedom)
- Antisymmetric stretch: \(\Sigma_u^+\)
Step 2: Determine which modes are IR active
A vibrational mode is IR active if it produces a changing dipole moment; equivalently, its symmetry must match one of the dipole moment components \(\mu_x,\mu_y,\mu_z\).
For linear centrosymmetric molecules:
- The component along the molecular axis (\(\mu_z\)) transforms as \(\Sigma_u^+\).
- The components perpendicular to the axis (\(\mu_x,\mu_y\)) transform as \(\Pi_u\).
Therefore, only modes with ungerade symmetry (subscript \(u\)) can be IR active in a molecule with an inversion center.
Conclusion for \(\mathrm{CO_2}\)
- \(\Sigma_u^+\) (antisymmetric stretch): IR active (changes \(\mu_z\))
- \(\Pi_u\) (bend, doubly degenerate): IR active (changes \(\mu_x,\mu_y\))
- \(\Sigma_g^+\) (symmetric stretch): IR inactive (no dipole moment change)
Big picture: for centrosymmetric molecules like \(\mathrm{CO_2}\), IR activity requires ungerade symmetry. The symmetric stretch is \(g\) and is IR inactive, while the bend (\(\Pi_u\)) and antisymmetric stretch (\(\Sigma_u^+\)) are IR active. This sets up the Raman selection rules, which favor gerade modes.
2. Raman activity and the polarizability
Raman spectroscopy probes vibrations through changes in the molecular polarizability, which describes how the electron cloud distorts in an applied electric field. The relevant quantity is the polarizability tensor \(\alpha\).
A vibrational mode is Raman active if the transition moment involving the polarizability is nonzero:
\[ \langle \psi_{v'} \,|\, \hat{\alpha} \,|\, \psi_v \rangle \neq 0. \]
Using group theory, this condition becomes
\[ \Gamma(\psi_v)\times\Gamma(\hat{\alpha})\times\Gamma(\psi_{v'}) \supset A_1. \]
The components of the polarizability tensor transform as quadratic functions of the coordinates: \(x^2\), \(y^2\), \(z^2\), \(xy\), \(xz\), and \(yz\). A vibrational mode is therefore Raman active if its symmetry species appears among these quadratic functions in the character table.
Example (Raman activity): methane, \(\mathrm{CH_4}\) (\(T_d\))
Methane is a highly symmetric, non-centrosymmetric molecule belonging to the point group \(T_d\). It is an excellent example for illustrating Raman activity, because several of its vibrational modes are Raman active, and some are both IR and Raman active.
\[ \begin{array}{c|ccccc|cc|c} T_d & E & 8C_3 & 3C_2 & 6S_4 & 6\sigma_d & \text{Linear} & \text{Rotations} & \text{Quadratic} \\ \hline A_1 & 1 & 1 & 1 & 1 & 1 & & & x^2+y^2+z^2 \\ A_2 & 1 & 1 & 1 & -1 & -1 & & R_x,R_y,R_z & \\ E & 2 & -1 & 2 & 0 & 0 & & & (2z^2-x^2-y^2,\;x^2-y^2) \\ T_1 & 3 & 0 & -1 & 1 & -1 & & (R_x,R_y,R_z) & \\ T_2 & 3 & 0 & -1 & -1 & 1 & (x,y,z) & & (xy,\;xz,\;yz) \\ \end{array} \]
Step 1: Vibrational modes of \(\mathrm{CH_4}\)
Methane has \(N=5\) atoms and is nonlinear, so it has
\[ 3N - 6 = 3(5) - 6 = 9 \]
vibrational degrees of freedom. Group-theoretical analysis (assumed complete) gives the vibrational representation:
\[ \Gamma_{\text{vib}} = A_1 + E + 2T_2. \]
This corresponds to four fundamental vibrations:
- \(A_1\) (nondegenerate)
- \(E\) (doubly degenerate)
- \(T_2\) (triply degenerate, appears twice)
Step 2: Raman selection rule
A vibrational mode is Raman active if it produces a change in the molecular polarizability. Using group theory, this means the symmetry species of the mode must match one of the quadratic functions of the Cartesian coordinates:
\[ x^2,\; y^2,\; z^2,\; xy,\; xz,\; yz. \]
From the \(T_d\) character table, these quadratic functions transform as:
- \((x^2 + y^2 + z^2) \sim A_1\)
- \((2z^2 - x^2 - y^2,\; x^2 - y^2) \sim E\)
- \((xy,\; xz,\; yz) \sim T_2\)
Step 3: Identify Raman-active modes
Comparing the vibrational symmetries with the quadratic functions:
- \(A_1\) mode: Raman active
- \(E\) mode: Raman active
- \(T_2\) modes: Raman active
Therefore, all vibrational modes of methane are Raman active.
Comparison with IR activity
In \(T_d\), the dipole moment components \((x,y,z)\) transform as \(T_2\). As a result:
- \(T_2\) modes are IR active,
- \(A_1\) and \(E\) modes are IR inactive.
This means methane has:
- modes that are Raman only (\(A_1\), \(E\))
- modes that are both IR and Raman active (\(T_2\))
Big picture: Raman spectroscopy is especially powerful for highly symmetric molecules like methane, where vibrations that do not change the dipole moment can still be observed through changes in polarizability.
3. Complementarity of IR and Raman spectroscopy
Infrared and Raman spectroscopy are often complementary techniques, providing different information about the same molecule. This complementarity is especially clear for molecules that possess a center of inversion.
For centrosymmetric molecules, the mutual exclusion principle applies:
- Vibrational modes that are IR active are Raman inactive.
- Vibrational modes that are Raman active are IR inactive.
This occurs because the dipole moment operator is antisymmetric with respect to inversion, while the polarizability tensor is symmetric. As a result, no vibrational mode in a centrosymmetric molecule can satisfy both selection rules simultaneously.
In molecules without an inversion center, a vibrational mode may be both IR active and Raman active.
Example (complementarity IR ↔ Raman): ethene, \( C_2H_4 \) (\( D_{2h} \))
Ethene is a planar, centrosymmetric molecule with point group \(D_{2h}\). It has \(N = 6\) atoms → \(3N = 18 \) degrees of freedom, and because it is nonlinear it has \( 3N-6 = 12 \) vibrational normal modes.
Complementarity in a centrosymmetric group
For molecules that have an inversion center (like ethene), the mutual-exclusion principle applies:
- Gerade (g) vibrational modes transform as even under inversion and can be Raman active (they appear among the quadratic functions).
- Ungerade (u) vibrational modes transform as odd under inversion and can be IR active (they transform like the Cartesian coordinates x,y,z).
Thus in \(D_{2h}\) we expect the 12 normal modes to split into g- and u-type species; the g-type modes will be Raman-active, the u-type modes (if they transform like x,y,z) will be IR-active, and no mode can be both.
Useful \(D_{2h}\) character table (condensed)
\[ \begin{array}{c|cccccccc|c|c} D_{2h} & E & C_{2}(z) & C_{2}(y) & C_{2}(x) & i & \sigma_{xy} & \sigma_{xz} & \sigma_{yz} & \text{Linear} & \text{Quadratic} \\ \hline A_g & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\ B_{1g}& 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & & xy \\ B_{2g}& 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & xz \\ B_{3g}& 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & & yz \\ \hline A_u & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\ B_{1u}& 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\ B_{2u}& 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\ B_{3u}& 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \end{array} \]
How to read this table: x,y,z appear in the “Linear” column for the irreps that transform like the corresponding dipole components (hence those irreps can be IR active). The quadratic functions listed in the “Quadratic” column show which irreps can be Raman active.
Applying the rules to ethene (practical notes)
- Ethene’s 12 vibrational modes will be distributed among the eight \( D_{2h} \) irreps (some possibly with multiplicity). Modes labeled with any of the \( A_g, B_{1g}, B_{2g}, B_{3g} \) species are Raman active.
- Modes labeled with \( B_{1u}, B_{2u}, B_{3u} \) are the ones that transform like \( z, y, x \) respectively and are therefore IR active. The \( A_u \) species is inactive in both IR and Raman (silent) in this point group.
- Example (typical, student-level): the symmetric C=C stretch in ethene transforms as \( A_g \) (a gerade representation) and so is strongly Raman active but IR inactive. In contrast, an antisymmetric combination that changes the dipole along the molecular axis will be an ungerade species (e.g. one of the \(B_{iu}\) and will be IR active.
Practical consequence
When you examine ethene experimentally:
- Raman spectra will show the intense \(A_g\) bands such as the symmetric C=C stretch.
- IR spectra will show bands corresponding to the \(B_{1u}, B_{2u}, B_{3u}\) modes (the antisymmetric motions).
- Together the two techniques give a complete picture of the vibrational manifold because no single mode is both strongly IR and Raman active in this centrosymmetric system.
Big picture: ethene (and other centrosymmetric molecules) nicely illustrates complementarity: gerade ↔ Raman, ungerade ↔ IR, and modes that are silent by one technique often appear in the other — so using both methods together yields a fuller vibrational assignment.
Big idea: group theory provides a unified framework for predicting IR and Raman activity by analyzing how vibrational modes transform under symmetry, revealing when transitions are allowed and how different spectroscopic methods complement one another.