# What does the peak separation of carbon dioxide IR rotational-vibrational spectrum corresponds to and why?

Spectrum of $\ce{CO2}$ at $2349\space \mathrm{cm^{-1}}$ region

For some other molecules (for example $\ce{CO}$) the peak separation corresponds to $2B$ (where $B$ is the rotational constant wavenumber) However for $\ce{CO2}$ when I calculated the peak separation and compared it to literature value, it seems approximately to equal to $4B$? Any idea why is this?

Also, what is the explanation for $\ce{CO2}$ spectrum being asymmetric? and why does the $P$ branch have larger peak separations than the $R$ branch?

## 1 Answer

(In this answer I am using the symbol $\require{begingroup} \begingroup \newcommand{\tb}[0]{\tilde{B}} \tb$ to represent a wavenumber instead of $B$, which I choose to define as an energy – not that we will be using it.)

Why is the peak separation $4\tb$ instead of $2\tb$?

This has to do with nuclear spin statistics. Because oxygen-16 is a boson (spin 0), the wavefunction of the entire molecule (which comprises translational, rotational, vibrational, electronic, and nuclear spin components) must be symmetric upon exchange of the two oxygen-16 atoms.

The translational component is symmetric, as usual. The electronic ground state is ${}^1\Sigma_\mathrm{g}$, which is totally symmetric. The nuclear spin component must be symmetric by virtue of the fact that both oxygen atoms are spin 0; this means that there is only one possible spin state that they can be in (cf. hydrogen-1 has a spin of 1/2 and two possible spin states). Consequently, exchange of the oxygen atoms leads to no change in the nuclear spin wavefunction.

As a result, this means that the vibrational component times the rotational component must be symmetric; or in other words, the vibrational and rotational components must have the same parity.

Since the peak structure at $\pu{2349 cm-1}$ arises due to the antisymmetric stretch, the vibrational ground state $v = 0$ is symmetric with respect to exchange of oxygen, and must therefore be paired with rotational levels with even $J$, which are likewise symmetric. In a similar manner, the vibrational excited state $v = 1$ is antisymmetric, and must be paired with odd $J$ levels.

In $\ce{CO}$ where there are no nuclear spin statistics at work, the following transitions are all allowed in the $R$-branch ($\Delta J = +1$): \begin{align} (v,J) = (0,0) &\to (1,1) & \text{line 1}\\ (0,1) &\to (1,2) & \text{line 2}\\ (0,2) &\to (1,3) & \text{line 3}\\ (0,3) &\to (1,4) & \text{line 4}\\ \end{align}

and so on. However, in $\ce{CO2}$ the state $(v,J) = (0,1)$ is not permissible, and hence line 2 does not appear in the spectrum. Similarly, neither does line 4, or line 6, and so on; the separation between adjacent lines is therefore doubled.

Why are the peak separations in the $P$-branch larger than in the $R$-branch?

This is because the rotational constant $\tb_0$ in the vibrational ground state is different from the rotational constant $\tb_1$ in the first vibrational excited state. The rotational energy levels (in wavenumber units) are therefore by:

$$\tilde{G}(v,J) = v\omega_0 + \tb_vJ(J+1)$$

where $\omega_0$ is the wavenumber of the pure vibrational transition (i.e. the band origin). For the transition $(v,J) = (0,J) \to (1,J-1)$ in the $P$-branch, the associated energy is therefore

$$\tilde{P}(J) = \omega_0 + \tb_1J(J-1) - \tb_0J(J+1)$$

The separation between adjacent lines in the $\ce{CO2}$ spectrum is then

\begin{align} \tilde{P}(J) - \tilde{P}(J+2) &= \tb_1J(J-1) - \tb_0J(J+1) - \tb_1(J+1)(J+2) + \tb_0(J+2)(J+3) \\ &= 4J(\tb_0 - \tb_1) + 6\tb_0 - 2\tb_1 \end{align}

You can easily verify that if $\tb_0 = \tb_1 = \tb$, then this simplifies directly to $4\tb$, as expected.

However, because $\tb_0 \neq \tb_1$, the gap between adjacent lines acquires a $J$-dependency, because the term $(\tb_0 - \tb_1)$ is nonzero. In the case of the asymmetric stretch, it turns out that $\tb_0 > \tb_1$ (can you figure out why? It's best explained using the moment of inertia $I$, since $\tb = \hbar/8\pi cI$). Therefore, as $J$ increases (i.e. as we move farther left along the spectrum), the separation between lines increases.

An exactly analogous calculation will explain why the lines in the $R$-branch tend to "bunch":

$$\tilde{R}(J+2) - \tilde{R}(J) = 4J(\tb_1 - \tb_0) + 6\tb_1 - 2\tb_0$$

$\endgroup$