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 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\(\rm\sigma_{v}^{\prime}\)
The problem asks us to decompose a given reducible representation, $\Gamma$, into a sum of irreducible representations for the C2v point group using the provided character table.
C2v Point Group Character Table
The character table provides the characters for the irreducible representations (A1, A2, B1, B2) under each symmetry operation (E, C2, $\sigma_v$, $\sigma_v'$) of the C2v point group. The characters for the reducible representation $\Gamma$ are also given.
| C2v | E | C2 | $\sigma_v$ | $\sigma_v^{\prime}$ |
|---|---|---|---|---|
| A1 | 1 | 1 | 1 | 1 |
| A2 | 1 | 1 | -1 | -1 |
| B1 | 1 | -1 | 1 | -1 |
| B2 | 1 | -1 | -1 | 1 |
| $\Gamma$ | 8 | -2 | -6 | 4 |
Reduction Formula for Representations
To find how many times each irreducible representation ($i$) is contained in the reducible representation ($\Gamma$), we use the reduction formula:
\(n_i = \frac{1}{h} \sum_R \chi_\Gamma(R) \chi_i(R)\)
Where:
- \(n_i\) is the number of times the irreducible representation $i$ appears in $\Gamma$.
- \(h\) is the order of the point group (total number of symmetry operations). For C2v, h = 4.
- \(R\) represents a symmetry operation of the group (E, C2, $\sigma_v$, $\sigma_v'$).
- \(\chi_\Gamma(R)\) is the character of the reducible representation $\Gamma$ for operation R.
- \(\chi_i(R)\) is the character of the irreducible representation $i$ for operation R.
- The sum is taken over all symmetry operations R.
Calculating Coefficients for Irreducible Representations
Let's calculate the coefficients \(n_i\) for each irreducible representation of the C2v point group:
For A1 (\(n_{A1}\)):
\(n_{A1} = \frac{1}{4} [ \chi_\Gamma(E)\chi_{A1}(E) + \chi_\Gamma(C2)\chi_{A1}(C2) + \chi_\Gamma(\sigma_v)\chi_{A1}(\sigma_v) + \chi_\Gamma(\sigma_v')\chi_{A1}(\sigma_v') ]\)
\(n_{A1} = \frac{1}{4} [ (8)(1) + (-2)(1) + (-6)(1) + (4)(1) ]\)
\(n_{A1} = \frac{1}{4} [ 8 - 2 - 6 + 4 ]\)
\(n_{A1} = \frac{4}{4} = 1\)
For A2 (\(n_{A2}\)):
\(n_{A2} = \frac{1}{4} [ \chi_\Gamma(E)\chi_{A2}(E) + \chi_\Gamma(C2)\chi_{A2}(C2) + \chi_\Gamma(\sigma_v)\chi_{A2}(\sigma_v) + \chi_\Gamma(\sigma_v')\chi_{A2}(\sigma_v') ]\)
\(n_{A2} = \frac{1}{4} [ (8)(1) + (-2)(1) + (-6)(-1) + (4)(-1) ]\)
\(n_{A2} = \frac{1}{4} [ 8 - 2 + 6 - 4 ]\)
\(n_{A2} = \frac{8}{4} = 2\)
For B1 (\(n_{B1}\)):
\(n_{B1} = \frac{1}{4} [ \chi_\Gamma(E)\chi_{B1}(E) + \chi_\Gamma(C2)\chi_{B1}(C2) + \chi_\Gamma(\sigma_v)\chi_{B1}(\sigma_v) + \chi_\Gamma(\sigma_v')\chi_{B1}(\sigma_v') ]\)
\(n_{B1} = \frac{1}{4} [ (8)(1) + (-2)(-1) + (-6)(1) + (4)(-1) ]\)
\(n_{B1} = \frac{1}{4} [ 8 + 2 - 6 - 4 ]\)
\(n_{B1} = \frac{0}{4} = 0\)
For B2 (\(n_{B2}\)):
\(n_{B2} = \frac{1}{4} [ \chi_\Gamma(E)\chi_{B2}(E) + \chi_\Gamma(C2)\chi_{B2}(C2) + \chi_\Gamma(\sigma_v)\chi_{B2}(\sigma_v) + \chi_\Gamma(\sigma_v')\chi_{B2}(\sigma_v') ]\)
\(n_{B2} = \frac{1}{4} [ (8)(1) + (-2)(-1) + (-6)(-1) + (4)(1) ]\)
\(n_{B2} = \frac{1}{4} [ 8 + 2 + 6 + 4 ]\)
\(n_{B2} = \frac{20}{4} = 5\)
Superposition of Irreducible Representations
The reducible representation $\Gamma$ is therefore the sum of the irreducible representations multiplied by their respective coefficients:
\(\Gamma = n_{A1}A1 + n_{A2}A2 + n_{B1}B1 + n_{B2}B2\)
\(\Gamma = 1A1 + 2A2 + 0B1 + 5B2\)
This can be written as:
\(\Gamma = A_1 \oplus 2A_2 \oplus 5B_2\)
Comparing this result with the given options, we find that it matches option 2.
Important Questions from Character Tables & Selection Rules
For the formaldehyde molecule, H2CO having C2v symmetry with the character table as given below,
C2v 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 the reducible representation Γ3N (or Γtot) is Γ3N = 4A1 + A2 + 4B1 + 3B2. The reducible representation for the vibrational modes alone, namely Γvib will be
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 :
