When calculating the number of vibrational modes for a molecule, the formulas differ for linear $(n = 3N – 5)$ and non-linear $(n = 3N – 6)$ molecules, where $n$ is number of modes and $N$ is number of atoms in a molecule. I have hard time to conceptualize this strictly in terms of subtracting the combination of harmonic motions for respective atomic coordinates, which lead to to a) translation of the whole system b) rotation of the system, even though it was rather fun to prove the aforementioned relations. Yet, it was only an abstract exercise.
I would like know what extra type of mode – longitudinal or transversal – is being present in the linear molecule?
Perhaps this question is rather incomprehensible, so maybe it will be demonstrative to ask an additional, but specific question. Regarding modes of the linear 3 atom molecule – can be the $\delta_{xz}$ mode be superposed from $\delta_{xy}$ and "antisymmetric stretch" modes, reducing the total number of orthogonal modes to 3? (I think the answer should be no because it would violate the aforementioned relations for number of modes, but I cannot see why.)
I am adding a picture of non-linear molecule modes for quick visual comparison.
EDIT
This picture perhaps helped me to understand a bit more why there is an additional mode in the linear molecule. This is related to two orthogonal directions perpendicular to linear axis in which the atoms in transverse mode can move/vibrate. Can you confirm that the extra mode is indeed always transversal?
Best Answer
I am not too certain of what you are asking. If you are asking why do we not look at a molecule and then just identify the number of independent modes of vibration the answer is that it would be too hard. Real oscillations are a linear superposition of these normal modes, even undergrad spectra are often complicated. Only very simple diatomic give nice Raman and IR spectra, or very symmetrical molecules.
In general we can expand a potential function about an equilibrium configuration and retain second order terms. The first term is the absolute value so will be set to zero. The second term vanishes by assumption that we are at an equilibrium point. \begin{equation} U(\boldsymbol q)=U(q^i_0)+\sum _i\frac{\partial U}{\partial q^i}\bigg|_{0}\eta _i+\sum_{i,j}\frac 12 \bigg(\frac{\partial ^2U}{\partial q^i\partial q^j}\bigg)\bigg|_0\eta _i\eta _j+\sum _{i,j,k}\frac 1 6\bigg(\frac{\partial ^3U}{\partial q^i\partial q^j\partial q^k}\bigg)\bigg|_0\eta _i\eta _j\eta _k+\dots \end{equation} Where $\eta_i = q^i-q^i_0$ and $\dot \eta _i=\dot q^i-\dot q^i_0=\dot q^i$ are variations to the equilibrium configuration. This analysis leads to the following Lagrangian, \begin{equation} \mathscr L=\frac 12\sum _{i,j}(M_{ij}\dot \eta _i\dot \eta _j-V_{ij}\eta_i\eta_j) \end{equation} And a series of coupled harmonic oscillator ordinary differential equations. \begin{equation} \sum _{j}(M_{ij}\ddot \eta_j+V_{ij}\eta_j)=0 \end{equation} A special type of solution is a normal mode, \begin{equation} \eta _j=a_j\cos (\omega t-\varphi) \end{equation} The general solution to the small oscillations problem is then a linear superposition of normal modes, \begin{equation} \eta _r=\sum _r\boldsymbol a_r\cos (\omega _rt-\varphi _r) \end{equation} So far we have not counted modes per say. Although mechanically the aim is to compute $\eta$ in chemistry we normally only bother with the frequency of the mode. Performing this analysis with molecules is a little tricky except for linear symmetric molecules (the symmetry simplifies the problem).
If we consider a linear symmetric triatomic molecule like CO2 for instance then we expect $3(3)-5=4$. Performing the above analysis will give two modes ($\omega_1$ is discounted as it corresponds to translation in the internuclear axis so was already counted). \begin{equation} \omega _1=0,\ \ \ \ \ \ \ \omega_2=\sqrt{\frac{k}{m_1}},\ \ \ \ \ \ \omega _3=\sqrt{\frac{km_T}{m_2m_1}} \end{equation} Where $m_T=2m_1+m_2$ (total mass). (follow the procedure in Goldstein and interpretation of their oscillations). You may ask why this does't agree with the number of modes we expect by the simple counting above?
We can split the total number of vibrations into transverse and longitudinal modes. For an $N$ atom system, we look down the internuclear axis, one is free translation leaving $N-1$ longitudinal oscillations. Therefore the total number of transverse oscillations is $2N-4$. Which solves that problem.
So what about the transverse modes? These are generally complicated and depend on the phase of the oscillations that we choose. This area can lead to ro-vibrational coupling and orbital angular momentum. The quantisation of this is important to quantum chemical spectroscopy.
So you can see that in general real modes are complicated and their interpretation and form depend on phase too. Could we use modes as a basis for other modes? Yes, that is what we do. Could we use modes of different molecules as a basis to predict other molecules modes? Yes ... PROVIDED they are of the same symmetry. In this way we can use the same modes of CO2 as for Carbon disulphide CS2, or silicon tetrachlorides SCl4 modes for methane CH4.
Loss of symmetry changes the problem. How "okay" this is depends really. It is really the symmetry of the molecule that dictates this. This is seen when computing the modes of a non-linear triatomic molecule, they are not too far removed from the linear case as shown here but the number of them is not the same and they are slightly different. Therefore using vibrations of one molecule as a basis for another is flawed. The frequencies we get are not the same!
Does this answer your question?
EDIT TO ANSWER
These two degenerate modes of the CO2 are the complicated transverse modes I was referring to in my answer. If they are degenerate then they have the same roots of the characteristic equation and hence the same expression of the frequencies. However they are complex because of phase and rotational effects and they need NOT be the same. See Goldstein and Lissajous figures which should answer your question :)
EDIT TO EDIT
Perhaps it would be more intuitive to say that the absence of rotational angular momentum about the internuclear axis is a result of there being no degenerate modes down this axis, or perhaps that there is a $C_{\infty}$ symmetry axis? Or perhaps the fact that there is rotational angular momentum about the transverse axes means there is ro-vibrational coupling, due to the degeneracies? Asking are there any degeneracies along the internuclear axis of polyatomic molecules is a harder question to answer. Certainly we would attempt to construct the analysis from a point where this was not the case.