# Why is the transition (0,0,0) -> (1,0,1) observed in a gas phase IR spectrum of CO2?

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

$$\Gamma_{(0,0,0)}\times\Gamma_{z}\times\Gamma_{(1,0,1)}=\Sigma_g^+\times\Sigma_u^+\times\Sigma_u^+=\Sigma_g^+\ni\Gamma_{tot.sym.}$$

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

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