# Why is carbon dioxide bent in its first excited state?

R. H. Crabtree writes in The Organometallic Chemistry of the Transition Metals (p 145, 6th ed.) that

$\ce{CO2}$ is linear in the free state but bent [...] in the first excited state...

Why is this so? (The context refers to an electronically excited state.)

• If you think about how linear molecules would deform (vibrational modes) as they become excited you can think of the carbon atom and the oxygen atoms shifting up and down w.r.t each other making a bent shape (i.e. a "v" and inverted "v" with the tip of the "v" being the carbon atom). Unfortunately I can't quote you anything, I'm digging back to what I learned in University in one class. – J. Ari Apr 3 '17 at 22:43

This effect is the so-called Renner-Teller effect and is a consequence of a coupling between vibrational and electronic motion and thus a breakdown of the Born-Oppenheimer Approximation. You might be familiar with the Jahn-Teller effect which is related to the Renner-Teller effect. In degenerate states of linear molecules with three or more atoms, the total energy of the molecule can be reduced by bending the molecule. Consider a three atomic linear molecule in a degenerate state (such as CO$_2$ in its excited $\Pi$ state). When the molecule bents, the degeneracy is lifted as it can bent in the $x$ or $y$ direction. Because symmetry requirements, the potential has to be an even function of the bending coordinate. Ignoring the lifting of the degeneracy we may approximate the zero-order potential with a Taylor expansion as

$$V^0=ar^2+br^4+\ldots,$$ where $r$ is the bending coordinate and $a$ and $b$ are constants describing the potential ignoring the vibronic interaction. When we do consider the vibronic interaction, the degenerate potential splits in two, but the splitting must still be an even function of $r$.

$$\Delta V=\alpha r^2 + \beta r^4 + \ldots$$

If the interaction is so large that $a<\tfrac{1}{2}\alpha$, the potential will not have its minimum at $r=0$, but will have two minima at $\pm r_e$ and the molecule has a lower energy when it is bent.

See also Herzberg Vol 3, p 26.

• Okay, the J-T effect applies to non-linear molecules while the R-T effect applies to linear. – RBW Apr 4 '17 at 21:13
• That is correct :) – Paul Apr 5 '17 at 10:00
• Just for future reference: $\ce{CO2}$ is preferable to CO$_2$ as the latter might break in applications other than your own browser. (Even CO<sub>2</sub> would be better.) – Martin - マーチン Jun 20 '17 at 9:58