If I have a diatomic molecule, so a linear molecule, $3N-5=6-5=1$ holds, since this vibrational degree of freedom corresponds to the stretching coordinate beetwen the two nuclei.

If I consider water for example, I have a nonlinear triatomic molecule, so the formula says $3N-6=9-6=3$ vibrational degrees of freedom. Also in this case it makes sense since they correspond to the bending angle and to the $2$ stretching lenghts.

While, If I consider $\ce{CO2}$, in this case I have a triatomic linear molecule, so $3N-5=9-5=4$ vibrational degrees of freedom. Of course $2$ of these correspond to the two stretching lenghts, but I really don't understand what the other two correspond to. Since the molecule is linear, there shouldn't be angle to consider, otherwise it would not be linear anymore.

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    $\begingroup$ The molecule is linear in the ground state. There are two bending motions in addition to the symmetric stretch and anti-symmetric stretch. $\endgroup$ – MaxW Dec 17 '18 at 20:53

Molecular vibrations can certainly occur by changing the O-C-O bond angle away from 180º. When we call the molecule linear we are saying that the geometric minimum is a linear configuration; however, vibrational modes can (and do) break that symmetry. What's important is that the average nuclear position is a linear molecule (or if you're doing a geometry calculation you would say that it's the 0 K positions).

So the two vibrational modes you're missing are the two O-C-O bends at 90º from each other.

Also as a note: you seem to be suggesting that the two stretches correspond with the two bonds stretching. That's not quite true. One of them corresponds to both C-O bonds stretching outward at the same time (in phase) and the other will correspond one compressing while the other stretches (out of phase).

For a full derivation of the normal modes of CO$_2$ there are many sites, but this one looked good: http://www.colby.edu/chemistry/PChem/notes/NormalModesText.pdf

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