# 2.15: Breaking Symmetries

- Page ID
- 366323

## Local Symmetry

Small modification of the molecule that reduces symmetry will have small effects on the spectra (e.g. \(\ce{H2CO}\) to \(\ce{CH3CHO}\)).

This gives same spectra patterns even though the point group of acetaldehyde is Cs, containing only two representations \(A’\) and \(A’’\) with \(x\), \(y\) = \(A’\) and \(z = A’’\). Thus, *all transitions are symmetry allowed*. Because of this, the idea of **local symmetry** is important. Local symmetry is that associated with the **electronically active component** of the molecule only, which is the \(\ce{C=O}\) moiety.

Independent of what \(R\) and \(R'\) are, provided that they not interact strongly with the C=O bond through conjugation etc. Hence this introduces the concept of a ”correlation or coherence length” of the excitation. Hence the UV-Vis spectra of all compounds of this type are similar: acetone, acetaldehyde, cyclopentanone, etc).

Acetone:

- \(π \rightarrow π{*}\) with \(λ_{max} = 188\, nm\) and \(ε= 1860\)
- \(n \rightarrow π^{*}\) with \(λ_{max} = 279 \,nm\) and \(ε= 15\)

vs. formaldehyde

- \(π \rightarrow π{*}\) with \(λ_{max} = 185\, nm\) and \(ε>10,000\)
- \(n \rightarrow π^{*}\) with \(λ_{max} = 270 nm\) and \(ε=100\)

## Vibronic Coupling

The \({ }^{1} A_{2} \leftarrow{ }^{1} A_{1}\) transition is forbidden by symmetry selection rules for formaldehyde. Why do we see it in the \(\ce{H2CO}\) molecule at all? Answer= **Vibronic coupling. **The Born-Oppenheimer approximation is not absolute. There is always some interaction between vibrational and electronic motion in a molecule. This means that the separation

\[|q, r\rangle \approx |r; q\rangle_{e l}|q\rangle_{v i b} \nonumber \]

is not always exact. Thus we must consider the transition moment integral in its entirety.

\[\langle q, r|\hat{M}| q, r\rangle^{ex} \nonumber \]

Let’s assume \(|q, r\rangle\) is in its ground state for all vibrations (\(k_bT\) < all vibrational frequencies). The irreducible representation of all vibrations in non-excited state is \(A_1\). Thus \(|q, r\rangle\) has the symmetry of the electron wavefunction itself \(|r\rangle\). In general, the symmetry is the direct product of the symmetry representations of \(|r\rangle\) and \(|q\rangle\).

Now, for the *excited* state wavefunction (\(|q, r\rangle^{e x}\)), the symmetry is that of the electronic wavefunction (\(|r\rangle^{e x}\)) **only **if the vibrational state is totally symmetric (i.e. \(v’=0\)) for all normal modes or only \(A_1\) vibrations are excited. However, if one-quantum of a non-symmetric vibrational state is excited along with the electronic state (upon absorption of light), then the excited state representation will be:

\[\Gamma_{total} = \underbrace{\Gamma\left(|r\rangle^{r x}\right)}_{\text{vibration symmetry}} \otimes \underbrace{\Gamma\left(|q\rangle^{\jmath^{e x}}\right)}_{\text{electronic symmetry}} \nonumber \]

**If \(\Gamma_{total}\) contains the same representation as a component of \(\hat{M}\) then the transition becomes weakly allowed vibrationally. Important, the appropriate vibrational mode is exited simultaneously with electronic transition.**

\(\hat{M}\) in the \(C_{2v}\) point group has the following representations for \(x\), \(y\), and \(z\) polarizations:

\[\left(\begin{array}{c}

M_{x} \\

M_{y} \\

M_{z}

\end{array}\right) = \left(\begin{array}{c}

B_{1} \\

B_{2} \\

A_{1}

\end{array}\right).\nonumber \]

Hence, the \({ }^{1} A_{2}\left(n, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\) transition in \(\ce{H2CO}\) transition can be vibronically coupled (allowed) if the quantum of a normal mode with symmetry:

- \(a_1\) giving a z-polarization
- \(b_1\) giving a y-polarization
- \(b_2\), giving a x-polarization

is *simultaneously *excited by the absorption of the photon. We can use this to predict the symmetries of the normal modes of vibration that a molecule has. A mode active in inducing an electronic transition (vibronically) is called an **inducing mode **or **promoting mode**.

## Symmetry

Assume the electronic (lowest vibrational level) ground state is totally symmetry (e.g., closed shell) is usually the case for organic molecules. A transition may be vibronically induced by exciting one quantum of a vibrational mode of symmetry, , provided the direct produce

\[\Gamma_{q} \otimes \Gamma_{\hat{M}} \otimes \Gamma_{e x}=A_{1} \text { or } A_{1 g} \nonumber \]

where \(\Gamma_{\hat{M}}\) is the symmetry of the \(\hat{M}\) operator and \(\Gamma_{e x}\) is the symmetry of the electronic excited state. Of course, the allowed vibrations must contain one of the symmetry \(\Gamma_{v i b}\).

The \(\ce{0 -> 0}\) band does not appear in a vibrationally-induced electronic transition! This is one of its characteristics other than its reduced oscillator strength.

## Possible vibrational modes for H_{2}CO

A nonlinear molecule has \(3N-6\) internal degrees of vibrational freedom. These are these many vibrational frequencies, each of which has a symmetry of an irreducible representation for the molecular point group (https://www.chem.purdue.edu/jmol/vibs/form.html). For formaldehyde, with \(N=4\), we have \(3 \times 4-6=6\) vibrational degrees of freedom which are distributed as normal modes comprising of a superposition of local modes. These turn out to (we will discuss later) transform as: \({3a_1, b_1, 2b_2}\).

Thus only \(b_1\) and \(b_2\) unsymmetrical modes are possible? We can thus have no Z-polarized vibronic transitions (no \(a_2\) mode in H_{2}CO), but we can have x- and y-polarized vibronic transition with \(b_2\) and \(b_1\) inducing modes, respectively. Hence, we can use polarization to identify inducing modes for vibronic transitions.

The \(\ce{d -> d}\) transitions in octahedral complexes are electronically forbidden (Laporte forbidden since they are \(\ce{g <-> g}\)), but they can be weakly allowed vibrationally by vibrations that remove the inversion center. Let’s consider the \(\ce{(NH3)6Co^{III}}\) complex that has a \(d^6\) configuration (loss of two 4s electrons and one d electron to make the ion).

The ground state is \(t_{2g}\) and the excited state \(e_{g}\). Is a d-d transition allowed?

The ground state symmetry is and has a \(O_h\) symmetry of ground state configuration due to the octahedral ligand field. Let’s consider the potential electronic \(\ce{d -> d}\) transition from the \(t_{2g}\) orbitals to the \(e_{g}\) orbitals. The excited electronic state symmetries are obtained from the orbital direct product:

\[\Gamma_{t_{2 g}} \otimes \Gamma_{e_{g}}=T_{1 g}+T_{2 g} \nonumber \]

(From direct product table for O_{h})

Both \({ }^{1} T_{2 g} \leftarrow{ }^{1} A_{1 g}\) and \({ }^{1} T_{1 g} \leftarrow{ }^{1} A_{1 g}\) transitions are forbidden. Now, \(\hat{M}\) or \(\{x,\,y,\,z\}\) form the basis for a triply degenerate representation \(T_{1u}\) in \(O_h\).

O_{h} |
E | 8C_{3} |
6C_{2} |
6C_{4} |
3C_{2} =(C_{4})^{2} |
i | 6S_{4} |
8S_{6} |
3σ_{h} |
6σ_{d} |
linear functions, rotations |
quadratic functions |
cubic functions |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

A_{1g} |
+1 | +1 | +1 | +1 | +1 | +1 | +1 | +1 | +1 | +1 | - | x^{2}+y^{2}+z^{2} |
- |

A_{2g} |
+1 | +1 | -1 | -1 | +1 | +1 | -1 | +1 | +1 | -1 | - | - | - |

E_{g} |
+2 | -1 | 0 | 0 | +2 | +2 | 0 | -1 | +2 | 0 | - | (2z^{2}-x^{2}-y^{2}, x^{2}-y^{2}) |
- |

T_{1g} |
+3 | 0 | -1 | +1 | -1 | +3 | +1 | 0 | -1 | -1 | (R_{x}, R_{y}, R_{z}) |
- | - |

T_{2g} |
+3 | 0 | +1 | -1 | -1 | +3 | -1 | 0 | -1 | +1 | - | (xz, yz, xy) | - |

A_{1u} |
+1 | +1 | +1 | +1 | +1 | -1 | -1 | -1 | -1 | -1 | - | - | - |

A_{2u} |
+1 | +1 | -1 | -1 | +1 | -1 | +1 | -1 | -1 | +1 | - | - | xyz |

E_{u} |
+2 | -1 | 0 | 0 | +2 | -2 | 0 | +1 | -2 | 0 | - | - | - |

T_{1u} |
+3 | 0 | -1 | +1 | -1 | -3 | -1 | 0 | +1 | +1 | (x, y, z) | - | (x^{3}, y^{3}, z^{3}) [x(z^{2}+y^{2}), y(z^{2}+x^{2}), z(x^{2}+y^{2})] |

T_{2u} |
+3 | 0 | +1 | -1 | -1 | -3 | +1 | 0 | +1 | -1 | - | - | [x(z^{2}-y^{2}), y(z^{2}-x^{2}), z(x^{2}-y^{2})] |

For instance, for the purely electronic transition \({ }^{1}T_{1 g} \leftarrow { }^{1} A_{1 g}\) component,

\[T_{1 g} \otimes T_{1 u} \otimes A_{1 g}=A_{1 u}+E_{u}+T_{1 g}+T_{2 u} \nonumber \]

this **does not contain** the \(A_{1g}\) symmetry representation (i.e., Laporte forbidden) and is therefore, this is a forbidden electronic transition.

What vibrations will couple these two electronic states? An octahedral complex (7 atoms so there are \(3 \times 7-6=15\) normal modes) has the following vibrational symmetries (discussed later):

\[\Gamma_{vib}=a_{1 g}+e_{g}+2 t_{1 u}+t_{2 g}+t_{2 u} \nonumber \]

(15 modes here, count them up)

The potential promoting modes are the \( 2t_{1u}\) and \(t_{2u}\) normal modes. The \(t_{1u}\) mode will serve as a promoting mode provided

\[\begin{aligned}

&T_{1 u} \otimes\left\{T_{1 g} \otimes T_{1 u} \otimes A_{1 g}\right\}= \\

&T_{1 u} \otimes\left[A_{1 u}+E_{u}+T_{1 u}+T_{2 u}\right]=A_{1 g}

\end{aligned} \nonumber \]

This turns out to contain \(A_{1g}\), so the \(t_{1u}\)’s can serve as a promoting mode for a vibrationally allowed transition. So it turns out, also can the \(t_{2u}\) mode.