Let ($v_1$,$v_2$,$v_3$) denote the vibrational state of $CO_2$.

Why is the transition $(0,0,0)\rightarrow (1,0,1)$ observed when the trasition $(0,0,0)\rightarrow(1,0,0)$ (asymmetric stretch) is not observed because it is IR inactive (not allowed)?

Is the following procedure correct for showing that the $(0,0,0)\rightarrow(1,0,1)$ transition is allowed?

The criteria for the transition being allowed is

$$\Gamma_{tot. sym.}\in\Gamma_{initial} \times \Gamma_{x,y,z}\times\Gamma_{final}$$

Where $\Gamma$ is the "irrep" of the group $D_{\infty h}$ and $\Gamma_{tot. sym.}=\Sigma_g^+$. The initial state is $(0,0,0)$ and the irrep is $\Gamma_{initial}=\Gamma_{(0,0,0)}=\Sigma_g^+$

The irreps of the three normal modes $v_1$, $v_2$ and $v_3$ are

$$\Gamma_1=\Sigma_g^+\text{, }\Gamma_2=\Pi_u\text{, }\Gamma_3=\Sigma_u^+$$

From a character table we get that

$$\Gamma_{x,y}=\Pi_u\text{ and }\Gamma_z=\Sigma_u^+$$

Now, this is the step I'm not sure about; we say that the irrep of the state $(1,0,1)$ is $$\Gamma_{(1,0,1)}=\Gamma_1\times\Gamma_3=\Sigma_g^+\times\Sigma_u^+=\Sigma_u^+$$

And finally, using a product table, we get


And thus the transition is allowed and we observe it as a band in an IR spectrum.

enter image description here

  • $\begingroup$ Everything is allowed in real life, which is what you're seeing. Selection/transition rules are based on approximations. $\endgroup$ – Todd Minehardt Apr 18 '20 at 17:53
  • $\begingroup$ Your approach seems reasonable $\endgroup$ – Tyberius Apr 18 '20 at 17:53
  • 1
    $\begingroup$ You seem to have answered your own question, (1 0 1) is allowed and observed, (1 0 0) is not and is not. $\endgroup$ – porphyrin Apr 18 '20 at 19:30

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