For the formaldehyde molecule, H2CO having C2v symmetry with the character table as given below, the reducible representation Γ3N (or Γtot) is Γ3N = 4A1 + A2 + 4B1 + 3B2. The reducible representation for the vibrational modes alone, namely Γvib will beC2v E C2 σv (xz) σv (yz) A1 1 1 1 1 z A2 1 1 -1 -1 Rz B1 1 -1 1 -1 x, Ry B2 1 -1 -1 1 y, Rx
Formaldehyde Vibrational Modes
For a molecule, the total number of degrees of freedom is $3N$, where $N$ is the number of atoms. These degrees of freedom can be represented by a reducible representation, often denoted as $\Gamma_{3N}$ or $\Gamma_{tot}$. These degrees of freedom are partitioned into three types of molecular motion:
- Translational motion (movement of the entire molecule in space).
- Rotational motion (rotation of the entire molecule about its center of mass).
- Vibrational motion (relative movement of atoms within the molecule).
The relationship between these representations is given by:
\begin{equation*} \Gamma_{3N} = \Gamma_{trans} + \Gamma_{rot} + \Gamma_{vib} \end{equation*}
To find the reducible representation for vibrational modes ($\Gamma_{vib}$), we can rearrange this equation:
\begin{equation*} \Gamma_{vib} = \Gamma_{3N} - \Gamma_{trans} - \Gamma_{rot} \end{equation*}
The question provides the total reducible representation for the formaldehyde molecule ($\text{H}_2\text{CO}$) with $\text{C}_{2v}$ symmetry as $\Gamma_{3N} = 4A_1 + A_2 + 4B_1 + 3B_2$. We need to determine $\Gamma_{trans}$ and $\Gamma_{rot}$ from the provided character table.
Translational and Rotational Representation
The character table provides the irreducible representations that correspond to translational motion along the x, y, and z axes ($T_x, T_y, T_z$) and rotational motion about the x, y, and z axes ($R_x, R_y, R_z$). These are usually indicated in the rightmost columns of the character table.
| $\text{C}_{2v}$ | E | $\text{C}_2$ | $\sigma_v (\text{xz})$ | $\sigma_v (\text{yz})$ | ||
|---|---|---|---|---|---|---|
| $\text{A}_1$ | 1 | 1 | 1 | 1 | z | |
| $\text{A}_2$ | 1 | 1 | -1 | -1 | $\text{R}_z$ | |
| $\text{B}_1$ | 1 | -1 | 1 | -1 | x | $\text{R}_y$ |
| $\text{B}_2$ | 1 | -1 | -1 | 1 | y | $\text{R}_x$ |
From the character table:
- Translational modes correspond to the irreducible representations listed with x, y, z.
- $T_z$ corresponds to $\text{A}_1$.
- $T_x$ corresponds to $\text{B}_1$.
- $T_y$ corresponds to $\text{B}_2$.
Therefore, the reducible representation for translational modes is $\Gamma_{trans} = \text{A}_1 + \text{B}_1 + \text{B}_2$.
- Rotational modes correspond to the irreducible representations listed with $\text{R}_x, \text{R}_y, \text{R}_z$.
- $\text{R}_z$ corresponds to $\text{A}_2$.
- $\text{R}_y$ corresponds to $\text{B}_1$.
- $\text{R}_x$ corresponds to $\text{B}_2$.
Therefore, the reducible representation for rotational modes is $\Gamma_{rot} = \text{A}_2 + \text{B}_1 + \text{B}_2$.
Calculating Vibrational Representation
Now we can calculate $\Gamma_{vib}$ by subtracting $\Gamma_{trans}$ and $\Gamma_{rot}$ from $\Gamma_{3N}$:
\begin{align*} \Gamma_{vib} &= \Gamma_{3N} - \Gamma_{trans} - \Gamma_{rot} \\ &= (4A_1 + A_2 + 4B_1 + 3B_2) - (A_1 + B_1 + B_2) - (A_2 + B_1 + B_2) \end{align*}
Let's subtract the coefficients for each irreducible representation:
- For $\text{A}_1$: $4 - 1 - 0 = 3$
- For $\text{A}_2$: $1 - 0 - 1 = 0$
- For $\text{B}_1$: $4 - 1 - 1 = 2$
- For $\text{B}_2$: $3 - 1 - 1 = 1$
So, the reducible representation for vibrational modes is $\Gamma_{vib} = 3\text{A}_1 + 0\text{A}_2 + 2\text{B}_1 + 1\text{B}_2$.
This simplifies to $\Gamma_{vib} = 3\text{A}_1 + 2\text{B}_1 + \text{B}_2$.
Formaldehyde ($\text{H}_2\text{CO}$) is a non-linear molecule with $N=4$ atoms. The number of vibrational modes for a non-linear molecule is $3N - 6 = 3(4) - 6 = 12 - 6 = 6$. The sum of the coefficients in our calculated $\Gamma_{vib}$ is $3 + 2 + 1 = 6$, which matches the expected number of vibrational modes.
Important Questions from Character Tables & Selection Rules
The reducible representation, Γ, in the table is equal to the following superposition of the irreducible representations of C2v point group.
C2v
E
C2
σv
\(\rm\sigma_{v}^{\prime}\) A1
1
1
1
1
A2
1
1
−1
−1
B1
1
−1
1
−1
B2
1
−1
−1
1
Γ
8
−2
−6
4
The observed IR spectrum for BCl3 exhibits three bands at 995, 480, and 244 cm-1, while the Raman bands are observed at 995, 471, and 244 cm-1. Given that for BCl3, Γvib = A'1 + 2E' + A2", the frequency of A1 mode in cm-1 is
D3h E 2C3 3C2 σh 2S3 3σv A'1 1 1 1 1 1 1 x2 + y2, z2 A' 2 1 1 -1 1 1 -1 Rz E' 2 -1 0 2 -1 0 (x, y) x2 - y2 , yz A" 1 1 1 1 -1 -1 -1 A" 2 1 1 -1 -1 -1 -1 z E" 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz) The character table for the point group D3h is given below.
D3h E 2C3 (z) \(\rm 3C_{2}^{'}\) σh(xy) 2S3 3σv \(\rm A_{1}^{'}\) +1 +1 +1 +1 +1 +1 - x2 + y2, z2 \(\rm A_{2}^{'}\) +1 +1 −1 +1 +1 −1 Rz - E' +2 −1 0 +2 −1 0 (x, y) (x2 − y2, xy) \(\rm A_{1}^{''}\) +1 +1 +1 −1 −1 −1 - - \(\rm A_{2}^{''}\) +1 +1 −1 −1 −1 +1 z - E'' +2 −1 0 −2 +1 0 (Rx, Ry) (xz, yz) In the electronic ground state, BF3 has D3h symmetry. Therefore,
In the character table given below:
Td E $8C_3$ $3C_2$ $6S_4$ $6\sigma_d$ $A_1$ 1 1 1 1 1 $A_2$ 1 1 1 -1 -1 E 2 -1 2 0 0 $T_1$ 3 0 -1 1 -1 $T_2$ 3 0 -1 -1 1
The order of the point group is :
