# Why does a symmetric stretch mode not have an imaginary frequency?

Currently studying potential energy systems and our professor asked a question at the end of a lecture and I can't wrap my head around it.

He said, "Picture a 2D PES for the simple reaction of H+H2 -> H2 + H. The direction of the symmetric stretch mode would be to the top right corner where both distances between the hydrogens in the transition state are large. This mode does not have an imaginary frequency. Why is that?"

I can't think of a reason that it would be an imaginary (negative) frequency since the TS must already have one for it to be a TS. So my only thought is that the TS already has one and it cannot have two.

Is my thought process too basic? What am I missing here?

• Does the symmetric mode contribute to the transition state? Isn't the dissociation path equivalent to a high amplitude assymmetric stretch? Jan 12, 2023 at 20:02

The path over the transition state is like an inverted parabola so has a negative frequency, i.e. its an upside down harmonic oscillator if you like. But orthogonal to this is a second potential like a Morse potential and this has a normal (positive) frequency, just as is the case of a saddle point. Buy a pack of Pringles if you want to see 3D saddle points :)

You can see this in the energy contour plot calculated for H+H2 below (using the LEPS potentials). The hydrogen molecule is vibrating as it approaches the H atom and the product is similarly vibrating. The diagonal is the symmetric stretch at the saddle point ( circle) and the asym. stretch is along the reaction path, shown as a short line at right angles to the other. The transition state is located at the highest point between the two valleys on approach and retreat and this has the shape of a saddle as you can see from the potential.

(I just noticed that the pic has been clipped on the right shouid read H+H2.)

Below is a sketch of the atomic arrangements when considering a linear attack.

The blue diagonal represents the symmetric stretch. The reaction coordinate (dotted line) takes the lowest possible path from reactant (top left) to product (bottom right). If I stretch the atomic arrangement, I make it worse, and if I scrunch the arrangement, I make it (much) worse. So this is like a vibration in a molecule, with a minimum somewhere in the middle.

On the other hand, starting from the transition state ("‡"), the asymmetric stretch makes one bond longer (the two atoms that will no longer be bound) and one bond shorter (the two atoms that will have a bond in dihydrogen). So there is a maximum in the middle, and both directions have lower energy.

Another way of saying it is: The "middle" hydrogen "wants" to make a single bond, not two or zero. The asymmetric stretch gets the transition state there, the symmetric stretch doesn't.

For an actual potential energy surface, see e.g. https://people.uleth.ca/~roussel/C4000foundations/slides/07PES.pdf

• So is the reason that the symmetric stretch mode doesn't have an imaginary frequency due to the minimum it has? Because imaginary frequencies signify a maximum somewhere on the PES? Jan 13, 2023 at 21:32