1

6. Exploring the potential energy surface

So far we have focused on the equilibrium structures - minima on PES. But other stationary points

- in particular the transition states - are equally interesting and important in chemistry. Transition

states are often difficult to observe experimentally because they are not populated at equilibrium:

they are short lived, transient species.

Knowledge of transitions state structures provides insights into the mechanism of the reaction. The

energy (and thermochemistry) of the transition state allows calculating the reaction barriers and

from them the rates of reactions.

In addition, we can also study the paths along the PES, e.g. from one minimum (=reactants)

through the saddle point (transition state) to another local minimum (=products). We can scan the

PES along one or more reaction coordinates (bond lengths, angles, torsions) and follow reaction

paths through transition states.

6.1. Optimizing transition structures.

There are three ways to optimize transition state structures:

1. Using Opt=TS. The geometry specification now has to be a geometry for the transition

structure. To converge, this requires really good initial guess geometry, otherwise it may have

trouble converging.

Common errors you may encounter:

Eigenvalue test fails. You will see a line like this in your output (not at the very end so

make sure you scroll up!):

Optimization stopped.

-- Wrong number of Negative eigenvalues: Desired= 1 Actual= 0

-- Flag reset to prevent archiving.

To fix that you may try NoEigenvalueTest option, i.e. Opt=(TS,NoEigenvalueTest).

The optimization runs out of steps or otherwise fails to converge. For example, you may

see this:

Optimization stopped.

-- Number of steps exceeded, NStep= 22

-- Flag reset to prevent archiving.

Then proceed as you would for a difficult optimization case from geometry optimization:

o Examine the optimization steps and find the best structure. Note that this may not be one

with the lowest energy, because you are not looking for a minimum, but for a saddle

point on PES.

o Use this structure, best by reading it from the checkpoint file, which you should always

save: Geom =(AllCheck,Step=n), and calculate vibrational frequencies.

2

o Start the optimization again using the calculated frequencies (from the checkpoint file).

Opt=(TS,ReadFC) Geom =AllCheck. Note that here you do not use Step=n

anymore.

When you have a good guess and frequencies, it rarely fails the eigenvalue test. But if it does,

you know what to do (hint: number one).

2. Finding the good guess for the right transition state structure may be difficult. Gaussian can help

you out and try to generate the guess for you, using a so-called STQN method. It is called by

Opt=QST2 and the input file must include two title and molecule specification sections, like this:

#HF/6-31G(d) Opt=QST2 Test

TS optimization: reactants

0 2

coordinates of reactant molecule(s)

TS optimization: products

0 2

coordinates of reactant molecule(s)

It is important that the corresponding atoms appear in the same order in both structures, i.e. that

the numbering of atoms matches. On the other hand, the orientations of the molecules do not

matter.

3. For even more difficult cases, you can supply both the reactants and products AND your guess

for the transition structure. This is requested by Opt=QST3 and, you guessed it, the file will have

three title and molecular specification sections: reactants, products and the transition state structure

in that order.

6.2. Characterizing stationary points on PES

Geometry optimizations converge to a structure on the potential energy surface where the forces

are essentially zero. The final structure may correspond to a minimum on the potential energy

surface, or it may represent a saddle point, which is a minimum with respect to some directions on

the surface and a maximum in one or more others. First order saddle points which are a maximum

in exactly one direction and a minimum in all other orthogonal directions correspond to transition

state structures linking two minima.

There are two pieces of information from the output which are critical to characterizing a

stationary point:

• The number of imaginary frequencies.

• The normal mode corresponding to the imaginary frequency.

Imaginary frequencies are listed in the output of a frequency calculation as negative numbers. By

definition, a structure which has n imaginary frequencies is an nth order saddle point. Ordinary

3

transition structures are 1st order saddle points and therefore characterized by single imaginary

frequency.

Log files may be searched for this line as a quick check for imaginary frequencies:

% grep imagin job.log

****** 1 imaginary frequencies (negative signs) *******

It is important to keep in mind that finding exactly one imaginary frequency does not guarantee

that you have found the transition structure in which you are interested. Saddle points always

connect two minima on the potential energy surface, but these minima may not be the reactants

and products of interest. Whenever a structure yields an imaginary frequency, it means that there

is some geometric distortion for which the energy of the system is lower than it is at the current

structure (indicating a more stable structure). In order to fully understand the nature of a saddle

point, you must determine the nature of this deformation.

One way to do so is to look at the normal mode corresponding to the imaginary frequency and

determine whether the displacements that compose it tend to lead in the directions of the structures

that you think the transition structure connects. The symmetry of the normal mode is also relevant

in some cases.

The following table summarizes the most important cases you will encounter when attempting to

characterize stationary points.

4

Example: Optimize transition state for the proton shift in formic acid and calculate the vibrational

frequencies.

1. First we will try the Opt=TS method. Using Gabedit, this is a reasomable guess for the

transition structure:

I made it by dragging the proton from the -OH group closer to the =O. In the transition structure

it should be right in the middle. This is my input file:

%chk=formicts.chk

# HF/6-31G(d) Opt=(TS,ModRedundant)

Formic TS

0 1

C -0.1096490000 0.4954820000 0.0814140000

O 1.0793510000 0.5124820000 0.3964130000

O -0.7066490000 -0.6775180000 -0.2605860000

H -0.6746490000 1.4044820000 0.0764140000

H 0.5492690000 -0.8135300000 0.0181350000

Running this input file resulted in a crash due to the failed eigentest.

Inlcuding NoEigenTest as another option to Opt ran out of steps. Examination of the

structures during the optimization steps in Gabedit suggested that the last one is actually OK:

O

C H

H O

⇌

O

C

H

H

O

5

In the next step I used this structure to calculate frequencies and then used the frequencies to

finish the optimization:

%chk=formicts.chk

#HF/6-31G(d) Freq Geom=AllCheck Guess=Read Test

--Link1--

%chk=formicts.chk

#HF/6-31G(d) Opt=(TS,ModRedundant,ReadFC) Geom=AllCheck Guess=Read Test

2 5

3 5

I also added the distances between the hydrogen (5) and two oxygens as redundant coordinates,

so that I can read their distances. (note: this may not work with the new Gaussian). This

optimization finished in three steps. This is the part of the output:

R5 R(2,5) 1.287 -DE/DX = 0.0

R6 R(3,5) 1.2873 -DE/DX = 0.0002

The H-O distances are very nearly the same for both oxygens, as expected. We could do better

with Tight or VeryTight optimization.

Finally, the optimized transition structure can be used to calculate vibrational frequencies for it:

%chk=formicts.chk

#HF/6-31G(d) Freq Geom=AllCheck Guess=Read Test

You will notice there is one large imaginary frequency (reported as negative: 2453 cm-1) which

corresponds to the motion of the proton in the direction of the two oxygens. This is the mode that

6

takes the proton through the transition state: from one oxygen to the other (never mind the crazy

bonds drawn by Gabedit):

It is therefore the right transition state.

2. To try the STQN method, it is necessary to generate the two end states. We should probably

optimize them first, we will try to get away with the Gabedit structures. The carboxylic acid

fragment from the fragment library looks like this:

7

This will be one of our end states, e.g. the ‘reactant’. Note that atom numbers (you can turn them

on under Labels in the Menu - as you of course well know). To get the other state, we just need

to switch the numbers of the two oxygens. That is a lot easier that moving the proton (although

that can be done too). The input file is: chk=formicqst2.chk

#HF/6-31G(d)

# Opt=(QST2,ModRedundant) Test

Reactant formic

0 1

C 0 -0.0142000000 0.4266000000 0.0636000000

O 0 1.1748000000 0.4436000000 0.3786000000

O 0 -0.6112000000 -0.7464000000 -0.2784000000

H 0 -0.5791990000 1.3356000000 0.0586000000

H 0 0.0298000000 -1.4594000000 -0.2224000000

2 5

3 5

Product formic

0 1

C 0 -0.0142000000 0.4266000000 0.0636000000

O 0 -0.6112000000 -0.7464000000 -0.2784000000

O 0 1.1748000000 0.4436000000 0.3786000000

H 0 -0.5791990000 1.3356000000 0.0586000000

H 0 0.0298000000 -1.4594000000 -0.2224000000

2 5

3 5

The second geometry is just copied and pasted first one with the two oxygens exchanged by

cutting and pasting.

I also added ModRedundant to make sure I get the distances. The redundant coordinate input

is the same as before

If I needed to physically generate a second geometry and then insert it after the first on in the

input file, that can be done in Gabedit with Insert/Gaussian/Geometry in the main

Menu.

Running the job finished quickly - in 8 steps. The optimized bonds lengths are perfectly equal:

R4 R(2,5) 1.2875 2.3008 0.9604 -DE/DX = 0.0

R5 R(3,5) 1.2875 0.9604 2.3008 -DE/DX = 0.0

and pretty much the same as from the TS optimization.

8

6.3. Reaction path following

As we have just seen, one way to verify that the transition state is the right one is to examine the

imaginary frequency normal mode. Another, more accurate way to determine what reactants and

products the transition structure connects is to perform an IRC calculation to follow the reaction

path and thereby determine the reactants and products explicitly.

An IRC calculation examines the reaction path leading down from a transition structure on a

potential energy surface. It starts at the saddle point and follows the path in both directions from

the transition state, optimizing the geometry of the molecular system at each point along the path.

In this way, the calculation definitively connects two minima on the potential energy surface by a

path which passes through the transition state between them. Once this is confirmed, you can then

go on to compute an activation energy for the reaction by comparing the (zero-point corrected)

energies of the reactants and the transition state.

In Gaussian, a reaction path calculation is requested with the IRC keyword in the route section.

Before you can run one, however, certain requirements must be met. An IRC calculation begins at

a transition structure and steps along the reaction path a fixed number of times (the default is 10)

in each direction, toward the two minima that it connects. However, in most cases, it will not step

all the way to the minimum on either side of the path.

IRC calculations absolutely require initial force constants to proceed! They can be provided:

by saving the checkpoint file from the preceding frequency calculation (used to verify that

the optimized geometry to be used in the IRC calculation is in fact a transition state), and

then specify the RCFC option in the route section. Note that it has to be RCFC and not

ReadFC !!!

by computing them at the beginning of the IRC calculation (CalcFC).

Note that one of RCFC and CalcFC must be specified!

Here is the procedure for running an IRC calculation:

Optimize the starting transition structure.

Run a frequency calculation on the optimized transition structure. This is done for several

reasons:

o To verify that the first job did in fact find a transition structure.

o To determine the zero-point energy for the transition structure.

o To generate force constant data needed in the IRC calculation.

Perform the IRC calculation (requested with the IRC keyword). This job will help you to

verify that you have the correct transition state for the reaction when you examine the

structures that are downhill from the saddle point.

9

In some cases, however, you will need to increase the number of steps taken in the IRC in order to

get closer to the minimum; the MaxPoints option specifies the number of steps to take in each

direction as its argument.

You can also continue an IRC calculation by using the IRC=(Restart,MaxPoints=n)

keyword, setting n to some appropriate value (provided, of course, that you have saved the

checkpoint file).

By default the IRC calculation reports only the energies and reaction coordinate at each point on

the path. If geometrical parameters along the path are desired, these should be defined as redundant

internal coordinates via Geom=ModRedundant or as input to the IRC code via IRC(Report=Read).

The various components of an IRC study are often run as a single, multi-step job.

Example:

To follow the reaction path of the proton transfer for the formic acid, with the frequencies

calculated at the transition state saved in the checkpoint file formicts.chk, the IRC input file

would look like the following:

%chk=formicts.chk

#HF/6-31G(d) IRC=(RCFC,MaxPoints=6,Report=Read) Geom=AllCheck

Guess=Read Test

2 5

3 5

The last two lines specify the two bond lengths we like to monitor.

The summary output of the IRC calculation looks like this:

Maximum number of steps reached.

Calculation of REVERSE path complete.

Reaction path calculation complete.

----------------------------

! IRC Parameters !

! (Angstroms and Degrees) !

-------------------------- -----------------------

! Name Definition TS Reactant Product

-----------------------------------------------------------------------------

! R1 R(2,5) 1.2875 1.0461 1.5601

! R2 R(2,5) 1.2875 1.0461 1.5601

----------------------------------------------------------------------------

Energies reported relative to the TS energy of -188.681644

----------------------------------------------------------------------------

10

--------------------------------------------------------------------------

Summary of reaction path following

--------------------------------------------------------------------------

Energy Rx Coord R1 R2

1 -0.03262 -0.62615 1.04606 1.04606

2 -0.02492 -0.52181 1.08103 1.08103

3 -0.01728 -0.41745 1.11933 1.11933

4 -0.01036 -0.31310 1.15974 1.15974

5 -0.00482 -0.20874 1.20152 1.20152

6 -0.00124 -0.10439 1.24417 1.24417

7 0.00000 0.00000 1.28750 1.28750

8 -0.00124 0.10439 1.33130 1.33130

9 -0.00482 0.20874 1.37557 1.37557

10 -0.01036 0.31310 1.42037 1.42037

11 -0.01728 0.41745 1.46586 1.46586

12 -0.02492 0.52181 1.51230 1.51230

13 -0.03262 0.62615 1.56006 1.56006

--------------------------------------------------------------------------

Total number of points: 12

What this means is that in maximum number of steps (6 in each direction) it did not reach the two

minima. In the above table the transition state is in the middle (step 7 with energy 0.00000) and

the energies of each step in both directions from the TS are listed. The reaction coordinate (Rx

Coord) is some complex combination of coordinates, but, as requested, the two bond lengths

between the proton and oxygens are listed.

Increasing the number of steps to 50 is more successful. We can find this in the output:

… PES minimum detected on this side of the pathway. Magnitude of the gradient = 0.0000554

Calculation of FORWARD path complete.

Begining calculation of the REVERSE path.

followed by the optimized structure:

Input orientation:

---------------------------------------------------------------------

Center Atomic Atomic Coordinates (Angstroms)

Number Number Type X Y Z

---------------------------------------------------------------------

1 6 0 -0.259506 0.427776 0.008746

2 8 0 0.959505 0.325760 0.255031

3 8 0 -0.748551 -0.680341 -0.263185

4 1 0 -0.808239 1.354071 0.030566

5 1 0 0.550147 -0.820072 -0.005294

---------------------------------------------------------------------

and for the reverse direction:

PES minimum detected on this side of the pathway.

Magnitude of the gradient = 0.0000831

Calculation of REVERSE path complete.

Reaction path calculation complete.

…..

11

Input orientation:

---------------------------------------------------------------------

Center Atomic Atomic Coordinates (Angstroms)

Number Number Type X Y Z

---------------------------------------------------------------------

1 6 0 -0.267912 0.321548 -0.009068

2 8 0 1.023010 0.371549 0.275573

3 8 0 -0.862430 -0.662031 -0.284866

4 1 0 -0.704984 1.311067 0.046186

5 1 0 1.383305 -0.509205 0.220224

---------------------------------------------------------------------

That means it converged both in forward and backward directions and we have the optimized

structures of the reactants and products.

The summary shows that the end points are indeed stationary points:

Energies reported relative to the TS energy of -188.681644

-------------------------------------------------------------------------- Summary of reaction path following

--------------------------------------------------------------------------

Energy Rx Coord R1 R2

1 -0.08067 -2.50074 0.95324 0.95324

2 -0.08067 -2.49719 0.95338 0.95338

3 -0.08066 -2.48384 0.95386 0.95386

…. 51 -0.08066 2.48386 2.30117 2.30117

52 -0.08067 2.49734 2.30473 2.30473

53 -0.08067 2.50004 2.30541 2.30541

--------------------------------------------------------------------------

Total number of points: 52

The calculations took 52 points, which mean 26 in each direction.

You can read the last geometry to Gabedit and convince yourself that it is the optimized formic

acid structure. Unfortunately, Gabedit does not seem to allow you to view the step by step progress

of the reaction path following. What you can do is to cut (or copy) the summary table from the

output and make a nice plot of the PES along the reaction path in Matlab, SigmaPlot, Origin, Excel

etc.

Reaction coordinate

-2 -1 0 1 2

Energ

y (A

.U.)

-0.08

-0.06

-0.04

-0.02

0.00

12

You can see that the TS energy is set to zero, but the reactant and product (since they are the same)

are about 0.08 Hartree below it. That means that the energy barrier is 0.08 Hartree. However, this

is only the electronic energy, not the total energy ! To calculate thermodynamic barrier and the

rate constant, we will need the thermochemistry calculations.

6.4. Calculating reaction barriers and rate constants

Optimization of the transition state and calculating its vibrational parameters also gives us the

thermochemistry, ie. the ZPE, thermal corrections etc. That allows us to calculate the

thermodynamic barriers to reactions and from them the reaction rates.

According to transition state theory (TST) the rate of a gas phase reaction at a given temperature

T is given by:

Tk

Gh

TkTk

B

B‡

exp)( (5.1)

where the G‡ is the free energy barrier, i.e .the difference between the energy of the transition

state and the reactants:

reactGGG ‡‡

(5.2)

Calculation of the reaction rates is therefore straightforward:

1. Optimize the reactanct(s), calculate the vibrational frequencies and thermochemistry.

2. Optimize the transition state (TS), calculate its vibrational frequencies and thermochemistry.

Obviously, this has to be done at the same level of theory as 1.

3.(optional) Run the IRC calculation from the TS to make sure you found the right transition state.

4. From the free energies of the reactant(s) and the TS, calculate the barrier height using formula

(5.2).

5. From the barrier height calculate the rate using formula (5.1).

For calculations of rates for different isotopes and temperatures, you can use freqchk, just be

careful with calculating the total free energies (remember you have to add the electronic energy).

6.5. Potential surface scan

A potential energy surface scan allows you to explore a region of a potential energy surface. A

normal scan calculation performs a series of single point energy calculations at various structures,

thereby sampling points on the potential energy surface.

There are two kinds of potential surface scans:

rigid. Rigid PES scan consists of single point energy evaluations over a rectangular grid

involving selected internal coordinates. In other words, the energy is calculated only for a

specified set of geometries, which are held constant.

13

relaxed. Relaxed PES scan samples points on the potential energy surface and performs a

geometry optimization of the remaining non-scanned coordinates at each one.

1. Rigid potential energy surface scan.

The rigid PES scan can be requested by the keyword Scan in the route section. The molecular

structure must be defined using Z-matrix coordinates.

The number of steps and step size for each variable are specified on the variable definition lines,

following the variable’s initial value in the following format:

name initial-value number-of-points increment-size

For example:

#HF/6-311+G(d,p) Scan Test

C-H PES Scan

0 2

C

H 1 R

R 0.7 35 0.05

When only one parameter follows a variable name, that variable is held fixed throughout the entire

scan. Note that this is a bit different from freezing coordinates during Z-matrix optimizations!

When all three parameters are included, that variable will be allowed to vary during the scan. Its

initial value will be set to initial-value; this value will increase by increment-size at each of

number-of-points subsequent points.

When more than one variable is allowed to vary, then all possible combinations of their values

will be included. For example, the following variable definitions will result in a total of 20 scan

points:

R 1.0 4 0.1

A 60.0 3 1.0

There are five values of R and four values of A, and the program will compute energies at all 20

structures corresponding to the different combinations of them. The results of a potential energy

surface scan appear following this heading within Gaussian output:

Scan completed.

Summary of the potential surface scan:

N R SCF

---- --------- -----------

1 0.7000 -38.04119

2 0.7500 -38.11738

3 0.8000 -38.17272….

14

34 2.3500 -38.19043

35 2.4000 -38.19023

36 2.4500 -38.19007

---- --------- -----------

You can plot the results of the scan to get a picture of the region of the potential energy surface

that you've explored. By doing so, you may be able to determine the approximate location of the

minimum energy structure. However, remember that rigid potential energy surface scans do not

include a geometry optimization.

C-H bond length (Angstrom)

0.5 1.0 1.5 2.0 2.5

En

erg

y (

Ha

rtre

e)

-38.25

-38.20

-38.15

-38.10

-38.05

2. Relaxed potential energy surface scan

A relaxed PES scan is requested by using the Opt=ModRedundant keyword and including the

S code on one or more variables in the Add Redundant input section

atoms(s) S number-of-steps step-size

All rules concerning ModRedundant apply, i.e. you can scan a bond length, angle, torsion or a

single Cartesian coordinate. It will always start from the value of the cooreindate in your input

geometry (molecule specification)

Again, if you specify more than one coordinate, muti-dimensional scan will be performed over the

corresponding grid of points. For example:

RHF/6-31G(d) Opt=ModRedundant Test

Water relaxed PES scan

0 1

O -0.464 0.177 0.0

H -0.464 1.137 0.0

H 0.441 -0.143 0.0

1 2 S 20 0.05

15

will run a relaxed PES scan for water, incrementing the O-H bond length by 0.05 Angstroms ten

times and optimizing the structure at each point.

In the output, Gaussian will tell you which coordinate it will scan and which are the other

coordinates (note that initially the bond length is set to 0.75 A): ----------------------------

! Initial Parameters !

! (Angstroms and Degrees) !

-------------------------- ----------------------

! Name Definition Value Derivative Info.

!

----------------------------------------------------------------------------

! R1 R(1,2) 0.75 Scan

!

! R2 R(1,3) 0.9599 estimate D2E/DX2

!

! A1 A(2,1,3) 109.4731 estimate D2E/DX2

!

----------------------------------------------------------------------------

Number of optimizations in scan= 21

and after the scan is done:

Summary of Optimized Potential Surface Scan

1 2 3 4 5

Eigenvalues -- -75.94099 -75.97678 -75.99772 -76.00803 -76.01074

R1 0.75000 0.80000 0.85000 0.90000 0.95000

R2 0.94258 0.94430 0.94532 0.94642 0.94739

A1 108.55551 107.78770 107.04742 106.26206 105.45388

6 7 8 9 10

Eigenvalues -- -76.00807 -76.00162 -75.99257 -75.98175 -75.96981

R1 1.00000 1.05000 1.10000 1.15000 1.20000

R2 0.94824 0.94898 0.94963 0.95019 0.95068

A1 104.62974 103.79781 102.96623 102.14224 101.33169

11 12 13 14 15

Eigenvalues -- -75.95718 -75.94420 -75.93112 -75.91808 -75.90522

R1 1.25000 1.30000 1.35000 1.40000 1.45000

R2 0.95110 0.95147 0.95180 0.95209 0.95234

A1 100.53892 99.76682 99.01706 98.29028 97.58639

16 17 18 19 20

Eigenvalues -- -75.89260 -75.88029 -75.86829 -75.85665 -75.84537

R1 1.50000 1.55000 1.60000 1.65000 1.70000

R2 0.95257 0.95278 0.95297 0.95314 0.95330

A1 96.90477 96.24454 95.60471 94.98435 94.35747

21

Eigenvalues -- -75.83444

R1 1.75000

R2 0.95345

A1 93.77380

where the eigenvalues are energies (in Hartree) and R1, R2 and A are the geometry

parameters. Again, you can plot the energy versus the scanned coordinate (though it is not as

easy to extract):

16

O-H bond length (Angstrom)

0.8 1.0 1.2 1.4 1.6 1.8

Energ

y (

Hart

ree)

-76.00

-75.95

-75.90

-75.85

You already know how to define and freeze coordinates with Opt=ModRedundant. You can

use Opt=ModRedundant to perform rigid surface scans (by freezing all other coordinates) and

therefore avoid the Z-matrix. Or by mixing and matching frozen and scanned coordinates you can

perform semi-relaxed PES scans.

For example, the following input would scan water O-H bond, relax the other O-H bond but keep

the H-O-H angle frozen:

RHF/6-31G(d) Opt=ModRedundant Test

Water relaxed PES scan

0 1

O -0.464 0.177 0.0

H -0.464 1.137 0.0

H 0.441 -0.143 0.0

1 2 S 20 0.05

2 1 3 F

This would be a two-dimensional scan over both O-H bond lengths, keeping the H-O-H angle

fixed. 0 1

O -0.464 0.177 0.0

H -0.464 1.137 0.0

H 0.441 -0.143 0.0

1 2 S 20 0.05

1 3 S 20 0.05

2 1 3 F