I am attempting to determine the transition state structure for the following reaction using the QST2 method on Gaussian 16:

$$\ce{CH3 + N -> HCN + H2 }$$

However, I am struggling to put together my input file. How do I format my z-matrices for multiple reactants and products? Here is what I have so far (though it is not running successfully):


#T UB3LYP/6-31G** opt=(qst2, redundant)

CH3 + N -> HCN + H2 reactants

0 3
X 1 1.0
H 1 r 2 90.0
H 1 r 2 90.0 3 120.0
H 1 r 2 90.0 3 -120.0

r 1.0828

CH3 + N -> HCN + H2 reactants

0 3

CH3 + N -> HCN + H2  products

0 1
X 1 1.0
H 1 rh 2 90.0
N 1 rn 2 90.0 3 180.0

rh 1.0587
rn 1.1326

0 1
H 1 r

r 0.741

I'm pretty new to this type of calculation on Gaussian so any help would be much appreciated!

  • $\begingroup$ The reverse reaction, which has the same transition state, breaks a triple bond. Even CC will likely have trouble with that. Consider computing the relaxed PES for extending the CN bond in the presence of H$_2$, since the reaction is a single atom coming from infinity. $\endgroup$ Dec 14, 2017 at 16:23

1 Answer 1


TL;DR: QST2 will not work.

For QST2 calculations you need one reactant structure and one product structure, the program will then attempt a transition state guess. The essential point is, that the supplied structures have to have identical ordering (in Cartesian coordinates), or identical z-matrices, although I am not sure that this works. For more on how to run these types of calculations, see Finding transition states in Gaussian – focus: Electrophilic Addition Reaction.

QST2 (and QST3 by extension) only make sense, if you know that you will have a single transition state - they are especially difficult to use for dissociation or association reactions because of the implied supermolecular approach. The minima $(\ce{CH3 + N})$, $(\ce{HCN + H2})$ will be very shallow and not well converged. From these structures it would be very difficult to guess the transition state, so you would at least have to supply a guess for it (QST3).

Additionally you are going from a "quintet" state (doublet + quartet) or "triplet" state (doublet + doublet, or doublet - quartet) to a singlet state (singlet + singlet). You could probably force your reactants into a broken singlet state (doublet - doublet), then it's a problem of how much sense this makes. In the other cases it is doubtful that the program will get a good guess for the electronic state of the TS.

DFT is maybe not the best methodology to investigate this kind of reaction in the first place.

In any case, here is an input file (tested with Gaussian16 rev. A.03) that will at least start running. SCF will not converge. Also reactants and products have not been optimised beforehand (which you must do). This is only for illustrative purposes.

#P BP86/def2SVP
gfinput gfoldprint iop(6/7=3)

N + CH3

0 1
C        0.000000000      0.000000000      1.891780658
N        0.000000000      0.000000000     -2.108219342
H        1.029433990      0.000000000      2.255740536
H       -0.514716995     -0.891515987      2.255740536
H       -0.514716995      0.891515987      2.255740536

HCN + H2

0 1
C        0.649013613      0.189100642      0.000000000
N       -0.817143840      0.083131316      0.000000000
H        0.964811208     -1.929636661     -0.499461795
H        0.964811208     -1.929636661      0.499461795
H        1.696363276      0.452628980      0.000000000

  • $\begingroup$ Thank you very much for your response! However, I am still confused as to how I am supposed to format the Cartesian coordinates in this way? I am relying on databases which would provide the coordinates for each species separately so how do I combine them into a single coordinate? $\endgroup$
    – pennypeat
    Nov 28, 2017 at 4:11
  • 2
    $\begingroup$ @pennypeat Use the molecular viewer and combine them in your best guess orientation. Optimise the supermolecular structure (might not converge because it's a very shallow pot. hyp. surf.), check for imaginary modes. Do same for product super molecule. Remember that the ordering must be identical. $\endgroup$ Nov 28, 2017 at 4:22

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