SADDLE

The transition state in a simple chemical reaction is to be optimized. Extra data are required. After the first geometry, specifying the reactants, the second geometry, specifying the products, is defined, using the same format as that of the first geometry.

A SADDLE calculation works entirely in Cartesian coordinates, regardless of the coordinate system used in the data set. The output, however, is in internal coordinates. All coordinates are optimized, so if SYMMETRY is present, it will only be used in setting up the starting geometry, after that it will be ignored. See also BAR=n.nn.

Example

*
*
* Locate the approximate transition state for the reaction
*
* 2CH4 -> H2 + C2H6
*
* For this calculation to work, the C2H6 system had to be put into the
* D2h geometry.
*
T=4H bar=0.04 geo-ok SADDLE mndo


C 0.000000 0   0.000000 0    0.000000 0 0 0 0  0.0704
C 2.521764 1   0.000000 0    0.000000 0 1 0 0  0.0704
H 1.104340 1 144.595726 1    0.000000 0 1 2 0 -0.0178
H 1.104167 1  90.121334 1 -125.283960 1 1 2 3 -0.0175
H 1.104166 1  90.116566 1  125.235484 1 1 2 3 -0.0176
H 1.103818 1  76.622615 1    0.975117 1 2 1 3 -0.0173
H 1.104407 1 122.428917 1  120.148896 1 2 1 3 -0.0177
H 1.104394 1 122.390158 1  240.200743 1 2 1 3 -0.0178
H 1.103576 1  35.239589 1   -0.019577 1 1 2 3 -0.0175
H 1.103620 1  33.058707 1    0.000000 0 2 1 3 -0.0176

C 0.000000 0   0.000000 0    0.000000 0 0 0 0  0.0164
C 1.521764 1   0.000000 0    0.000000 0 1 0 0  0.0164
H 1.208966 1 111.093608 1    0.000000 0 1 2 0 -0.0055
H 1.108971 1 111.085646 1 -120.010199 1 1 2 3 -0.0055
H 1.108949 1 111.082331 1  119.998535 1 1 2 3 -0.0055
H 1.108982 1 111.087018 1     .998485 1 2 1 3 -0.0055
H 1.109025 1 111.084646 1  120.004584 1 2 1 3 -0.0055
H 1.109003 1 111.081348 1  240.005897 1 2 1 3 -0.0055
H 2.0      1  90        1    0        1 1 2 3
H 1.0      1  90        1   70        1 9 1 2

*
* The resulting gradient is large, because the geometry is near but not at the transition state.
* To refine the transition state, a TS calculation must be run.

When things go wrong

If, in a SADDLE calculation, you get the message,

      " BOTH SYSTEMS ARE ON THE SAME SIDE OF THE TRANSITION STATE -
        GEOMETRIES OF THE SYSTEMS ON EACH SIDE OF THE T.S. ARE AS FOLLOWS"

There are two ways to proceed.

(A) If the gradient is small (less than about 10) at the point where this message is printed, then find the geometry that has the smallest gradient near to the end of the run, and use that as the starting geometry for a transition state refinement. If the geometry refines correctly (PRECISE might be needed in order to get rid of methyl rotations, etc.), and a FORCE calculation verifies that it is a transition state, then the calculation has been successful.

(B) If the gradient is large, the examine the output to find two adjacent points that are on opposite sides of the transition state.  Geometries on one side (call it the left side) will be indicated by the message "FOR POINT  nn FIRST STRUCTURE", points on the other side will be indicated by "FOR POINT  nn SECOND STRUCTURE" 
Check that the direction cosine for both is greater than 0.9. 
Edit the output file to delete everything except the two geometries.
Reduce the value of the BAR by a factor of about 10. If BAR was not used previously, then start with BAR=0.05.
Run the new SADDLE calculation

In general, in a SADDLE calculation, avoid using extra keywords such as PRECISE.