How many coordinates should be used for the potential energy surface of a non-rotating linear molecule? (3N-6) or (3N-5)?

Question

Take, for example, $\ce{CO2}$. It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.

So if I wanted to build a potential energy surface for this molecule that was valid near and far from equilibrium, it should depend on these coordinates. But how do I rectify this with the fact that $\ce{CO2}$ has four vibrational modes? As we learn in undergraduate chemistry, it has two stretches and two degenerate bends.

If I build a Hamiltonian in the three valence coordinates, could it possibly describe the ground state energy correctly? That is, would it contain the zero-point energy for all four modes? Would it be able to find the right low-lying excited states of the bending mode(s)?

EDIT: Wildcat has brought up the point of symmetry and point groups below, and perhaps that can help get to the heart of my question. When we say that $\ce{CO2}$ is a linear molecule, that means that the lowest energy configuration has a bond angle of $\pi$. But in no way do I want to restrict the symmetry of the molecule. I want to work with a potential surface that is valid for symmetric and asymmetric geometries.

But here is where I get confused. Let's say I did restrict it to be linear. I would then have a two-dimensional potential, a function of the two bond lengths. Of the seven normal modes, I only keep two: the symmetric stretch and anti-symmetric stretch. The ground state energy would be the sum of the zero-point energy of these two modes. So therefore the number of modes is equal to the number of coordinates I used to write my Hamiltonian.

If I further constrict the symmetry, and impose the $D_{\infty h}$ point group, then the potential is now a function of one parameter. I will now only have one vibrational mode, the symmetric stretch, and the ground state energy is the zero-point energy of this mode. So, again, the number of modes is equal to the number of coordinates I use to write my Hamiltonian.

But what if the symmetry were not restricted in any way? I can write the potential as a function three internal coordinates, be they bond lengths and angle, othogonormal rectilinear coordinates, jacobi coordinates, etc. This potential, and the Hamiltonian I construct with it, should find the same modes described above, as well as the bending modes. But will it find both bending modes? I don't see why it should. Based on the two reduced-dimensional examples above, I would expect the number of modes to equal the number of coordinates. I would expect therefore, not to find the right ground-state energy, etc.

Specifically, I am concerned about the accuracy of dynamical simulations performed using this three-dimensional potential.

Likewise, what about a tetratomic liner molecule like $\ce{C2H2}$? I would use three bond lengths, two bond angles, and one dihedral angle, for a total of 6 coordinates - but it has 7 vibrational modes. Would I model the dynamics properly?

Answer 1

The Cartesian Hessian ($\mathbf{H}_C$) can be obtained by from the Hessian computed in internal coordinates ($\mathbf{H}_i$) through a non-linear transformation $$\mathbf{H}_C = \mathbf{B}^T\mathbf{H}_i\mathbf{B}$$ where $B_{ij}=\frac{\partial R_i}{x_j}$ and $R_i$ and $x_i$ is an internal and Cartesian coordinate, respectively.

In the case of the degenerate bending mode you can obtain the components of $\mathbf{H}_C$ that you need from $\frac{\partial^2 E}{\partial \theta^2}$, where $\theta$ is the bond angle, through suitable choices of $B_{ij}$'s.

So the Hessian generated in the three internal coordinates has all the information you need.

Answer 2

It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.

Yes, that is true, if you do not impose any restriction on geometry (such as, for instance, the linearity), the $\ce{CO2}$ molecule will indeed have $3N-6=3*3-6=3$ internal degrees of freedom and the same number of independent structural parameters (as which you can naturally choose two bond lengths and the angle between them). However, the point is that when you start to impose some restrictions on the geometry (symmetry), the number of internal degrees of freedom can increase by one (if you require your molecule to be linear), while the number of independent structural parameters will stay the same (though, few of them will be fixed and not varying anymore).

As a result the following three quantities which you mixed up are usually different:

• the number of internal degrees of freedom: $3N-6$ (for a non-linear molecule) or $3N-5$ (for a linear molecule);
• the number of independent structural parameters: always $3N-6$;
• the number of varying (i.e. not-fixed) independent structural parameters: depends on the point group.

Thus, for instance, if you require the $\ce{CO2}$ molecule to be linear the number of internal degrees of freedom will be $3N-5=3*3-5=4$. It does not mean though that you need 4 independent structural parameters to describe it, 3 is still enough. And besides, by requiring the $\ce{CO2}$ to be linear you make it to belong to $\mathrm{C_{\infty v}}$ point group which fixes one of the parameters, the angle between bonds. So, while you have 4 internal degrees of freedom, you still need just 3 structural parameters to describe the molecule one of which is fixed in addition to that. So, if you want to scan PES, you have to vary just two of them: two bond lengths.

You can decrease the number of varying structural parameters even further: to just 1, namely, just the bond length, by making an additional (but quite reasonable) assumption that two $\ce{C-O}$ bonds in the $\ce{CO2}$ molecule have the same length, i.e. that the $\ce{CO2}$ molecule belongs to $\mathrm{D_{\infty h}}$ point group. Note that the number of internal degrees of freedom and the number of independent structural parameters will still be 4 and 3 respectively.

Answer 3

You need to incorporate all $3N-5$ modes (both bends and both stretches) in order to accurately calculate the ZPE even if, per Jan's answer, you can neglect one of the bending modes in constructing an accurate model of the PES. Even though the bends appear identical in mind's eye, both of their (degenerate) ZPE's must be included in order for the ground state calculated by the model to be accurate.

The way I think about it is, imagine a $\ce{CO2}$ molecule is oriented along the $z$-axis, with the carbon at the origin. If you only allow one bending mode, say with both oxygens translating toward $+x$ and the carbon toward $-x$, you have to hold the orientation of that motion fixed and your system will be unable to accommodate vibrational movement of any atoms in the $y$-direction. Accurate evaluation of the ZPE requires a model that allows bending along both axes.

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The maddening thing about vibrating linear molecules in general, though, is that most linear combinations of a certain pair of these degenerate bending motions actually contain a rotational component. Using $\ce{CO2}$ as an example and ignoring anharmonicity, if the bending vibrations are $90^\circ$ out of phase with one another, the oxygen atoms gyrate around the $\ce{O=C=O}$ axis just like they would in a rotating molecule with a bent equilibrium geometry (imagine the orange dot is $\ce{C}$ and the green dot is a projection of the positions of the two $\ce{O}$'s):

^Source

Of course, the radius of the circle of projection of the motion of the oxygens will be quite small, I believe less than a degree, but it still is a rotation about the molecular axis.

Phase angles other than $90^\circ$ result in some combination of a bending motion and a rotation, where only in the $\theta=0^\circ$ case does the rotational aspect vanish:

For any linear molecule with three or more atoms, the degenerate bends involving the three atoms at each terminus of the molecule are the pair that can generate rotation. For $\ce{CO2}$ that means the entire molecule; for acetylene each vibration involves the two carbons and one of the hydrogens; etc. EDIT: The more I think about it, the more confident I am that every pair of degenerate bends in a linear molecule actually has this capability to generate rotation. It's just that any bends not centered at the central atom (if it exists) will cause the molecule to precess, as well as rotate, due to the asymmetry of the two 'arms' of the bending motion.

So, to some extent, I would actually argue your question is ill-posed: Linear molecules are always rotating! They're just rotating in a manner inconsistent with the standard rigid-rotor assumption, and (I think?) at energies much higher than those one would expect to see in conventional rotational spectroscopy.