Modes of vibration of triatomic molecules

Question

$$\omega=0,\sqrt{\frac Km},\sqrt{\frac{K(2m+M)}{Mm}}$$

There are three modes here but actually triatomic linear molecule have 4 vibrational modes (e.g. CO2).

1. So where does that remaining one mode? Or is it four frequencies itself?(including plus and minus, since it is square root, but negative frequency doesn't make sense)
2. How can I know which frequency corresponds to which mode of vibration (e.g.: symmetric stretching frequency, anti symmetric stretching frequency etc.)
3. Or is there any problem with my understanding?

Answer 1

A linear triatomic molecule (like $CO_2$) in 3-dimensional space should have 4 normal modes of vibration:

• symmetric stretch mode
• asymmetric stretch mode
• bending mode (horizontal)
• bending mode (vertical)

However, the two bending modes (horizontal and vertical) have the same frequency because of symmetry.

_{(image from Tec-Science.com - Equipartition theorem - Triatomic linear molecules)}

Answer 2

This is toy model: they are only considering motion in the $x$ direction. In 3-d a triatomic molecule has 9 modes of which 6 are zero frequency (3 rotations, and 3 translations) and the other 4 are genuine "vibrations." Of the three frequecies they display above, the zero frequency one is the $x$ translation mode.

Answer 3

I think that the basic problem here is the difference between normal modes and their frequencies.

Each normal mode corresponds to a well definite and independent displacement field of the atoms. There is the possibility of degenerate modes, i.e., independent modes with the same frequency. Therefore, the information that only three frequencies are present does not automatically imply only three normal modes.

However, before any conclusion from the analogy with the real CO$_2$ molecule can be drawn, we should know what exactly the hypotheses underlying the question are.

Here we have three different possible scenarios.

1. the motion is constrained in one dimension (the three atoms can only move along a straight line). This case is different from the case of a linear molecule in three dimensions (see next points). In this completely one-dimensional case, the system has only three independent vibrational modes. Remember that each normal mode corresponds to a linear combination of the original displacements. With three independent positions, we may have only three independent displacements and three independent modes. In such a case, there is no degenerate mode. The zero-frequency mode corresponds to the displacement such that no relative distance is changed (so no restoring force: all the atoms have the same displacement in the same direction). The two remaining modes (symmetric and asymmetric stretching) have different frequencies. Without explicit calculation of the eigenvectors of the dynamical matrix, the correct assignment of the two non-zero frequencies to the corresponding modes may be puzzling.

2. the motion is not constrained to one dimension (a full three-dimensional motion), and there are only two elastic nearest-neighbor interactions. In such a case, we have nine degrees of freedom, therefore, nine normal nodes. Being the molecule linear, five modes have zero frequency due to the global translational and rotational symmetry. Still, two additional zero-frequency modes exist, corresponding to the bending of the molecule around the central atom. The remaining two normal modes are like in the one-dimensional case.

3. finally, realistic models of the interactions introduce one additional elastic constant for angular deviations from the straight geometry, inducing two degenerate non-zero frequencies. These modes are absent in the model as shown in the figure but are present in a real diatomic molecule like CO$_2$.