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.)
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 $\ce{CO2}$ in its excited $\Pi$ state). When the molecule bents, the degeneracy is lifted as it can bend in the $x$ or $y$ direction. Because of 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
$$ V^0=ar^2+br^4+\cdots, $$ 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 $r$.
$$ \Delta V=\alpha r^2 + \beta r^4 + \cdots $$
If the interaction is so large that $a<\alpha/2$, 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.
