2.12: Absorption Spectrum of Formaldehyde
- Page ID
- 366113
We can how turn to \(\ce{H2CO}\) again and look at the measured spectrum in the laboratory.
Two UV absorption bands are observed experimentally (Figure \(\PageIndex{1}\)):
- 270 nm with \(ε=100\) and
- 185 nm with \(ε>10,000\)
As discussed earlier, the lowest energy transitions are \(n_{b} \rightarrow \pi^{*}\) and \(\pi \rightarrow \pi^{*}\). As far as the states are concerned, the transitions are
\[{ }^{1} A_{2}\left(n, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\nonumber \]
and
\[{ }^{1} A_{1}\left(\pi, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\nonumber \]
respectively.
The Cartesian translational coordinates, \(x\), \(y\), \(z\) form a bases for the \(B_1\), \(B_2\) and \(A_1\) representations in \(C_{2v}\). Or course, \(x\), \(y\), \(z\) have the same transformation properties as \(\hat{M}_{x}\), \(\hat{M}_{y}\) and \(\hat{M}_{z}\), respectively. Hence we can represent as
\[\hat{M} =\left(\begin{array}{c}
\hat{M}_{x} \\
\hat{M}_{y} \\
\hat{M}_{z}
\end{array}\right)\nonumber \]
or in the \(C_{2v}\) point group as
\[\left(\begin{array}{c}
M_{x} \\
M_{y} \\
M_{z}
\end{array}\right) = \left(\begin{array}{c}
B_{1} \\
B_{2} \\
A_{1}
\end{array}\right).\nonumber \]
Class, please confirm this with the character table!
Let’s look at \({ }^{1} A_{2}\left(n, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\)
Evaluate this integrand for all components of \(\hat{M}\):
\[A_{2} \otimes\left(\begin{array}{l}
B_{1} \\
B_{2} \\
A_{1}
\end{array}\right) \otimes A_{1}=\left(\begin{array}{l}
B_{2} \\
B_{1} \\
A_{2}
\end{array}\right)\nonumber \]
And looking at each component:
- For \(\hat{M}_{x}\): \[A_{2} \otimes B_{1} \otimes A_{1}=B_{2} \nonumber \]
- For \(\hat{M}_{y}\): \[ A_{2} \otimes B_{2} \otimes A_{1}=B_{1} \nonumber \]
- For \(\hat{M}_{z}\): \[A_{2} \otimes A_{1} \otimes A_{1}=A_{2}\nonumber \]
Therefore the \({ }^{1} A_{2} \leftarrow{ }^{1} A_{1}\) transition is forbidden by electronic symmetry since there the symmetry is ODD for all the three \(\hat{M}\) components. That is, integration over all space with an integrand of non-\(A_1\) symmetry will always results in zero!
However, in reality, other coupling elements like vibronic coupling and spin-orbit coupling will make this a weakly allow transition (Figure \(\PageIndex{1}\)) instead of explicitly forbidden.
Let’s look at \({ }^{1} A_{1}\left(\pi, \pi^{*}\right) \leftarrow{ }^{1} A_{1}\)
Evaluate this integrand for all components of \(\hat{M}\):
\[A_{1} \otimes\left(\begin{array}{l} B_{1} \\ B_{2} \\ A_{1} \end{array}\right) \otimes A_{1}=\left(\begin{array}{l} B_{1} \\ B_{2} \\ A_{1} \end{array}\right)\nonumber \]
Therefore \({ }^{1} A_{1} \leftarrow{ }^{1} A_{1}\) is allowed by since
- For \(\hat{M}_{x}\): \[A_{1} \otimes B_{1} \otimes A_{1}=B_{1} \nonumber \]
- For \(\hat{M}_{y}\): \[ A_{1} \otimes B_{2} \otimes A_{1}=B_{2} \nonumber \]
- For \(\hat{M}_{z}\): \[A_{1} \otimes A_{1} \otimes A_{1}=A_{1}\nonumber \]
when \(\hat{M}_{z}\) the dipole moment operation (\(A_1\)) is used we get an integrand that has A1 symmetry. This electronic transition is thus polarized (preferential absorption of one polarization vs. another) and is only allowed when the oscillating electric field vector, is aligned along the molecular z-axis (along the C=O bond direction).
Let's consider higher energy transitions can occur such as
\[{ }^{1} B_{1}\left(\pi, \sigma^{*}\right) \leftarrow{ }^{1} A_{1}\nonumber \]
\[B_{1} \otimes\left(\begin{array}{c}
B_{1} \\
B_{2} \\
A_{1}
\end{array}\right) \otimes A_{1}=\left(\begin{array}{c}
A_{1} \\
A_{2} \\
B_{1}
\end{array}\right)\nonumber \]
So this transition is \(\hat{M}_x\) polarization allowed and is expected to lie further in the vacuum ultraviolet.