Unlike diatomic molecules, polyatomic molecules possess multiple independent vibrational motions. These vibrations involve coordinated movements of atoms and are best described using the concepts of normal modes and molecular symmetry.
Number of vibrational degrees of freedom
A molecule with \(N\) atoms has a total of \(3N\) degrees of freedom, corresponding to motion in the \(x\), \(y\), and \(z\) directions for each atom.
These degrees of freedom are partitioned into:
- Translations of the entire molecule (3 degrees of freedom),
- Rotations of the entire molecule,
- Vibrations (internal motions of the atoms).
For a nonlinear molecule, there are 3 rotational degrees of freedom, leaving
\[ \text{Number of vibrational modes} = 3N - 6. \]
For a linear molecule, rotation about the molecular axis does not change the orientation of the molecule, so there are only 2 rotational degrees of freedom. In this case,
\[ \text{Number of vibrational modes} = 3N - 5. \]
Each vibrational mode corresponds to a normal mode, in which all atoms oscillate with the same frequency and a fixed phase relationship.
Why symmetry matters for vibrational modes
Not all vibrational modes behave the same way under the symmetry operations of a molecule. Some modes may stretch bonds symmetrically, others may bend angles, and still others may involve antisymmetric motion.
Group theory provides a systematic way to:
- classify vibrational modes by their symmetry,
- determine which modes are IR active or Raman active,
- understand degeneracies and mode types.
Determining vibrational symmetries using group theory
The basic procedure for analyzing vibrational modes is:
- Determine the point group of the molecule.
- Construct a representation describing the 3N Cartesian displacements of all atoms.
- Subtract the representations corresponding to overall translations and rotations.
- Decompose the remaining representation into irreducible representations, which correspond to the vibrational modes.
Each irreducible representation obtained in this way labels a vibrational mode (or set of degenerate modes) and describes how that mode transforms under the symmetry operations of the molecule.
Example: H2O
Let’s work through a complete vibrational mode analysis for water, \( \mathrm{H_2O} \), using group theory. Water is a classic example because it is small, nonlinear, and has a point group with simple symmetry elements.
Step 1: Count the number of vibrational modes
Water has \(N=3\) atoms, so it has \(3N=9\) total degrees of freedom. Because water is nonlinear, it has 3 translational and 3 rotational degrees of freedom. Therefore,
\[ \text{Number of vibrational modes} = 3N-6 = 9-6 = 3. \]
So \(\mathrm{H_2O}\) has exactly three normal modes.
Step 2: Determine the point group
Water is a bent molecule with a \(C_2\) axis through oxygen (bisecting the H–O–H angle) and two vertical mirror planes. Its point group is \(C_{2v}\).
Step 3: Build the \(3N\) representation, \(\Gamma_{3N}\)
We form a reducible representation by tracking how the Cartesian displacement vectors of each atom transform under each symmetry operation. For \(C_{2v}\), the operations are \(E,\, C_2,\, \sigma_v(xz),\, \sigma_v'(yz)\).
Use the following rules:
- If an atom is moved to a different position by an operation, it contributes 0 to the character (because its displacement vectors do not map onto themselves).
- If an atom remains in place, its three displacement directions \((x,y,z)\) may change sign; the character contribution is the trace of the transformation matrix for those three coordinates.
Choose axes so that the molecule lies in the \(yz\)-plane, with \(z\) along the \(C_2\) axis. Oxygen lies on the symmetry axis; the hydrogens are exchanged by the \(C_2\) operation and by one of the mirror planes.
Now compute characters:
- \(E\): all 3 atoms remain in place, each contributes 3 (\(x,y,z\) unchanged) → character = \(9\).
- \(C_2\): O stays; H atoms are exchanged (contribute 0). For O, rotation about \(z\): \(x\to -x\), \(y\to -y\), \(z\to z\), so trace = \((-1)+(-1)+1=-1\). → character = \(-1\).
- \(\sigma_v(xz)\): reflection in the plane perpendicular to the molecular plane exchanges the H atoms (contribute 0). O stays fixed. Reflection: \(x\to x\), \(y\to -y\), \(z\to z\), trace = \(1-1+1=1\). → character = \(1\).
- \(\sigma_v'(yz)\): reflection in the molecular plane leaves all atoms in place. For each atom: \(x\to -x\), \(y\to y\), \(z\to z\), trace = \(-1+1+1=1\). Three atoms → character = \(3\).
Therefore,
\[ \Gamma_{3N} = (9,\,-1,\,1,\,3). \]
Step 4: Subtract translations and rotations
In \(C_{2v}\), the Cartesian coordinates transform as: \(x \sim B_1\), \(y \sim B_2\), \(z \sim A_1\). Therefore,
\[ \Gamma_{\text{trans}} = A_1 + B_1 + B_2. \]
Rotations transform as: \(R_x \sim B_2\), \(R_y \sim B_1\), \(R_z \sim A_2\), so
\[ \Gamma_{\text{rot}} = A_2 + B_1 + B_2. \]
The vibrational representation is
\[ \Gamma_{\text{vib}} = \Gamma_{3N} - \Gamma_{\text{trans}} - \Gamma_{\text{rot}}. \]
To carry this out, we now decompose \(\Gamma_{3N}\) into irreducible representations.
Step 5: Reduce \(\Gamma_{3N}\) to irreducible representations
Use the \(C_{2v}\) character table:
\[ \begin{array}{c|cccc} C_{2v} & E & C_2 & \sigma_v(xz) & \sigma_v'(yz)\\ \hline A_1 & 1 & 1 & 1 & 1\\ A_2 & 1 & 1 & -1 & -1\\ B_1 & 1 & -1 & 1 & -1\\ B_2 & 1 & -1 & -1 & 1\\ \end{array} \]
Apply the reduction formula \(a_i=\frac{1}{h}\sum \chi^{(\Gamma)}(R)\chi^{(i)}(R)\) with group order \(h=4\).
For \(\Gamma_{3N}=(9,-1,1,3)\):
\[ a_{A_1}=\frac{1}{4}(9\cdot1 + (-1)\cdot1 + 1\cdot1 + 3\cdot1)=3 \]
\[ a_{A_2}=\frac{1}{4}(9\cdot1 + (-1)\cdot1 + 1\cdot(-1) + 3\cdot(-1)=1) \]
\[ a_{B_1}=\frac{1}{4}(9\cdot1 + (-1)\cdot(-1) + 1\cdot1 + 3\cdot(-1)=2) \]
\[ a_{B_2}=\frac{1}{4}(9\cdot1 + (-1)\cdot(-1) + 1\cdot(-1) + 3\cdot1=3) \]
So,
\[ \Gamma_{3N} = 3A_1 + A_2 + 2B_1 + 3B_2. \]
Step 6: Obtain \(\Gamma_{\text{vib}}\)
\[ \Gamma_{\text{vib}} = (3A_1 + A_2 + 2B_1 + 3B_2) - (A_1 + B_1 + B_2) - (A_2 + B_1 + B_2) = 2A_1 + B_2. \]
Final result: water has three vibrational normal modes with symmetries \(2A_1 + B_2\).
- \(A_1\): symmetric stretch (IR active because \(z\sim A_1\))
- \(A_1\): bend (also IR active)
- \(B_2\): antisymmetric stretch (IR active because \(y\sim B_2\))
Big idea: group theory not only confirms that \(\mathrm{H_2O}\) has three normal modes, but also labels their symmetries, which immediately predicts their IR activity.
Example: CHCl3
Here is a worked vibrational-mode symmetry analysis for chloroform, \(\mathrm{CHCl_3}\). Chloroform is a nice example because it has a \(C_3\) axis and doubly-degenerate vibrational modes.
Step 1: Count the number of vibrational modes
Chloroform has \(N=5\) atoms, so \(3N=15\) total degrees of freedom. It is nonlinear, so
\[ \text{Number of vibrational modes} = 3N-6 = 15-6 = 9. \]
Step 2: Determine the point group
\(\mathrm{CHCl_3}\) is approximately tetrahedral, but because three ligands are identical (Cl) and one is different (H), it has a \(C_3\) axis (along the C–H bond) and three vertical mirror planes. Therefore its point group is \(C_{3v}\).
Step 3: Build the \(3N\) representation, \(\Gamma_{3N}\)
We determine the characters for the reducible representation describing all Cartesian displacements (\(x,y,z\) for each atom). For \(C_{3v}\), the classes are \(E\), \(2C_3\), and \(3\sigma_v\).
- \(E\): all 5 atoms remain in place, each contributes 3 \(\Rightarrow \chi(E)=15\).
- \(C_3\): C and H lie on the rotation axis; the three Cl atoms are permuted. For an atom on the axis, the 3D trace is \(1+2\cos120^\circ = 0\), so C contributes 0 and H contributes 0; permuted atoms contribute 0. \(\Rightarrow \chi(2C_3)=0\).
- \(\sigma_v\): each mirror plane contains the C–H axis and one Cl atom. C, H, and that Cl remain in place. For an atom in the mirror plane, the trace is \(1\). Thus \(\chi(\sigma_v)=3\) for each plane. \(\Rightarrow \chi(3\sigma_v)=3\) (character per operation).
So the reducible representation is
\[ \Gamma_{3N} = (15,\;0,\;3). \]
Step 4: Reduce \(\Gamma_{3N}\) in \(C_{3v}\)
Use the \(C_{3v}\) character table:
\[ \begin{array}{c|ccc} C_{3v} & E & 2C_3 & 3\sigma_v \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & 1 & -1 \\ E & 2 & -1 & 0 \\ \end{array} \]
With group order \(h=6\), the reduction gives
\[ \Gamma_{3N} = 4A_1 + A_2 + 5E. \]
Step 5: Subtract translations and rotations
In \(C_{3v}\), translations transform as \(z\sim A_1\) and \((x,y)\sim E\), so
\[ \Gamma_{\text{trans}} = A_1 + E. \]
Rotations transform as \(R_z\sim A_2\) and \((R_x,R_y)\sim E\), so
\[ \Gamma_{\text{rot}} = A_2 + E. \]
Therefore,
\[ \Gamma_{\text{vib}} = \Gamma_{3N} - \Gamma_{\text{trans}} - \Gamma_{\text{rot}} = (4A_1 + A_2 + 5E) - (A_1+E) - (A_2+E) = 3A_1 + 3E. \]
Interpretation
Final result: chloroform has 9 vibrational degrees of freedom with symmetries \(3A_1 + 3E\). Because \(E\) is doubly degenerate, the three \(E\) species account for 6 of the 9 vibrational degrees of freedom.
IR activity: in \(C_{3v}\), modes transforming like \(z\) (\(A_1\)) or \(x,y\) (\(E\)) are IR active. Therefore, all vibrational modes of chloroform are IR active: \(3A_1 + 3E\).
Big idea: for polyatomic molecules, group theory tells you not only how many vibrational modes exist (\(3N-6\)), but also the symmetry labels of those modes, which immediately predicts IR activity and degeneracy.
Big idea: polyatomic vibrational spectra arise from a finite number of normal modes, and group theory provides the language needed to count, classify, and predict the spectroscopic behavior of those modes.