molecular modeling notes 8

8/13/2019 Molecular Modeling Notes 8

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o Use this structure, best by reading it from the checkpoint file, which you shouldalways save: Geom =( Al l Check, St ep=n), and calculate vibrationalfrequencies.

o Start the optimization again using the calculated frequencies (from the checkpoint

file). Opt =( TS, ReadFC) Geom =Al l Check. Note that here you do notuse St ep=n anymore.

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

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

TS opt i mi zat i on: r eact ants

0 2

coordinates of reactant molecule(s)

TS opt i mi zat i on: product s

0 2

coordinates of reactant molecule(s)

It is important that the corresponding atoms appear in the same order in both structures, i.e. thatthe 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 willhave three title and molecular specification sections: reactants, products and the transition state

structure in that order .

5.2. 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 energysurface, 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 amaximum 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:


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• 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

transition structures are 1

st

order saddle points and therefore characterized by single imaginaryfrequency.

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

% gr ep i magi n j ob. l og*** *** 1 i magi nar y f r equenci es ( negat i ve si gns) *** ****

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 reactantsand 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 currentstructure (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 thestructures 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.


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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=f ormi ct s. chk# HF/ 6- 31G( d) Opt =( TS, ModRedundant )

For mi c TS

0 1C - 0. 1096490000 0. 4954820000 0. 0814140000O 1. 0793510000 0. 5124820000 0. 3964130000O - 0. 7066490000 - 0. 6775180000 - 0. 2605860000H - 0. 6746490000 1. 4044820000 0. 0764140000H 0. 5492690000 - 0. 8135300000 0. 0181350000

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

Inlcuding NoEi genTest as another option to Opt ran out of steps. Examination of thestructures during the optimization steps in Gabedit suggested that the last one is actually OK:

O

CH

HO

⇌ O

C

H

H

O


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

%chk=f ormi ct s. chk#HF/ 6- 31G( d) Fr eq Geom=Al l Check Guess=Read Test

- - Li nk1- -%chk=f ormi ct s. chk#HF/ 6- 31G( d) Opt =( TS, ModRedundant , ReadFC) Geom=Al l Check Guess=Read Test

2 5

2 5

I also added the distances between the hydrogen (5) and two oxygens as redundant coordinates,so that I can read their distances. This optimization finished in three steps. This is the part of the

output:

R5 R( 2, 5) 1. 287 - DE/ DX = 0. 0R6 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 Ti ght or VeryTi ght optimization.

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

%chk=f ormi ct s. chk#HF/ 6- 31G( d) Fr eq Geom=Al l Check Guess=Read Test

You will notice there is one large imaginary frequency (reported as negative: 2453 cm-1) whichcorresponds to the motion of the proton in the direction of the two oxygens. This is the mode

that takes the proton through the transition state: from one oxygen to the other (never mind thecrazy bonds drawn by Gabedit):


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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:

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

on under Label s in the Menu - as you of course well know). To get the other state, we justneed 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:


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chk=f ormi cqst 2. chk#HF/ 6- 31G( d)# Opt =( QST2, ModRedundant ) Test

React ant f ormi c

0 1C 0 - 0. 0142000000 0. 4266000000 0. 0636000000O 0 1. 1748000000 0. 4436000000 0. 3786000000O 0 - 0. 6112000000 - 0. 7464000000 - 0. 2784000000H 0 - 0. 5791990000 1. 3356000000 0. 0586000000H 0 0. 0298000000 - 1. 4594000000 - 0. 2224000000

2 52 5

Product f ormi c

0 1C 0 - 0. 0142000000 0. 4266000000 0. 0636000000

O 0 - 0. 6112000000 - 0. 7464000000 - 0. 2784000000O 0 1. 1748000000 0. 4436000000 0. 3786000000H 0 - 0. 5791990000 1. 3356000000 0. 0586000000H 0 0. 0298000000 - 1. 4594000000 - 0. 2224000000

2 53 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 I nsert / Gaussi an/ Geomet r y in the mainMenu.

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. 0R5 R( 3, 5) 1. 2875 0. 9604 2. 3008 - DE/ DX = 0. 0

and pretty much the same as from the TS optimization.


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5.3. Reaction path following

As we have just seen, one way to verify that the transition state is the right one is to examine theimaginary frequency normal mode. Another, more accurate way to determine what reactants and

products the transition structure connects is to perform an I RC calculation to follow the reaction path and thereby determine the reactants and products explicitly.

An I RC 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 fromthe 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 I RC keyword in the route section.

Before you can run one, however, certain requirements must be met. An IRC calculation beginsat 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 notstep 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 verifythat 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 (Cal cFC).

Note that one of RCFC and Cal cFC 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 I RC keyword). This job will help you toverify that you have the correct transition state for the reaction when you examine the

structures that are downhill from the saddle point.


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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 MaxPoi nt s option specifies the number of steps to take ineach direction as its argument.

You can also continue an IRC calculation by using the I RC=( Rest ar t , MaxPoi nt s=n)

keyword, setting n to some appropriate value (provided, of course, that you have saved thecheckpoint 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 viaI RC( Repor t =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 thefrequencies calculated at the transition state saved in the checkpoint file f or mi ct s. chk, theIRC input file would look like the following:

%chk=f ormi ct s. chk#HF/ 6- 31G( d) I RC=( RCFC, MaxPoi nt s=6, Repor t =Read) Geom=Al l CheckGuess=Read Test

2 53 5

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

The summary output of the IRC calculation looks like this:

Maxi mum number of st eps r eached.

Cal cul at i on of REVERSE path compl ete.React i on path cal cul at i on compl et e.

- - - - - - - - - - - - - - - - - - - - - - - - - - - -! I RC Par ameters !! ( Angst r oms and Degr ees) !

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

! Name Def i ni t i on TS React ant Product- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -! R1 R( 2, 5) 1. 2875 1. 0461 1. 5601! R2 R( 2, 5) 1. 2875 1. 0461 1. 5601- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Ener gi es r eport ed r el at i ve t o t he TS energy of - 188. 681644

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


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- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Summary of r eact i on path f ol l owi ng

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Ener gy Rx Coor d R1 R2

1 - 0. 03262 - 0. 62615 1. 04606 1. 046062 - 0. 02492 - 0. 52181 1. 08103 1. 081033 - 0. 01728 - 0. 41745 1. 11933 1. 119334 - 0. 01036 - 0. 31310 1. 15974 1. 159745 - 0. 00482 - 0. 20874 1. 20152 1. 201526 - 0. 00124 - 0. 10439 1. 24417 1. 244177 0. 00000 0. 00000 1. 28750 1. 287508 - 0. 00124 0. 10439 1. 33130 1. 331309 - 0. 00482 0. 20874 1. 37557 1. 37557

10 - 0. 01036 0. 31310 1. 42037 1. 4203711 - 0. 01728 0. 41745 1. 46586 1. 4658612 - 0. 02492 0. 52181 1. 51230 1. 5123013 - 0. 03262 0. 62615 1. 56006 1. 56006

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Tot al number of poi nt s: 12

What this means is that in maximum number of steps (6 in each direction) it did not reach thetwo 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 mi ni mum detected on t hi s si de of t he pathway.

Magni t ude of t he gradi ent = 0. 0000554Cal cul at i on of FORWARD pat h compl et e.Begi ni ng cal cul at i on of t he REVERSE pat h.

followed by the optimized structure:

I nput or i ent at i on:- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Cent er At omi c At omi c Coordi nat es ( Angst r oms)Number Number Type X Y Z- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

1 6 0 - 0. 259506 0. 427776 0. 0087462 8 0 0. 959505 0. 325760 0. 2550313 8 0 - 0. 748551 - 0. 680341 - 0. 2631854 1 0 - 0. 808239 1. 354071 0. 0305665 1 0 0. 550147 - 0. 820072 - 0. 005294

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

and for the reverse direction:

PES mi ni mum detected on t hi s si de of t he pathway.Magni t ude of t he gradi ent = 0. 0000831

Cal cul at i on of REVERSE path compl ete.React i on path cal cul at i on compl et e.

…..


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I nput or i ent at i on:- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Cent er At omi c At omi c Coordi nat es ( Angst r oms)Number Number Type X Y Z- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

1 6 0 - 0. 267912 0. 321548 - 0. 0090682 8 0 1. 023010 0. 371549 0. 2755733 8 0 - 0. 862430 - 0. 662031 - 0. 2848664 1 0 - 0. 704984 1. 311067 0. 0461865 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:

Ener gi es r eport ed r el at i ve t o t he TS ener gy of - 188. 681644- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Summary of r eact i on path f ol l owi ng- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Energy Rx Coor d R1 R21 - 0. 08067 - 2. 50074 0. 95324 0. 953242 - 0. 08067 - 2. 49719 0. 95338 0. 953383 - 0. 08066 - 2. 48384 0. 95386 0. 95386

….51 - 0. 08066 2. 48386 2. 30117 2. 3011752 - 0. 08067 2. 49734 2. 30473 2. 3047353 - 0. 08067 2. 50004 2. 30541 2. 30541

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Tot al number of poi nt s: 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.


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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.

5.4. Calculating reaction barriers and rate constants

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

T k G

h

T k T k

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:

react GGG ‡‡ (5.2)

Calculation of the reaction rates is therefore straightforward:

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

Reaction coordinate

-2 -1 0 1 2

E n e r g y ( A . U

. )

-0.08

-0.06

-0.04

-0.02

0.00


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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 f r eqchk, just becareful with calculating the total free energies (remember you have to add the electronic energy).

5.5. Potential surface scan

A potential energy surface scan allows you to explore a region of a potential energy surface. Anormal 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 gridinvolving selected internal coordinates. In other words, the energy is calculated only for a

specified set of geometries, which are held constant.

relaxed. Relaxed PES scan samples points on the potential energy surface and performs ageometry 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 molecularstructure 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 2CH 1 R

R 0. 7 35 0. 05


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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-matrixoptimizations! 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. 1A 60. 0 3 1. 0

There are five values of R and four values of A, and the program will compute energies at all 20structures corresponding to the different combinations of them. The results of a potential energy

surface scan appear following this heading within Gaussian output:

Scan compl et ed.

Summary of t he potent i al sur f ace scan:N R SCF

- - - - - - - - - - - - - - - - - - - - - - - -1 0. 7000 - 38. 041192 0. 7500 - 38. 11738

3 0. 8000 - 38. 17272….

34 2. 3500 - 38. 1904335 2. 4000 - 38. 1902336 2. 4500 - 38. 19007

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

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


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C-H bond length (Angstrom)

0.5 1.0 1.5 2.0 2.5

E n e r g y ( H a r t r e 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) initial-value S number-of-steps step-size

All rules concerning ModRedundant apply, i.e. you can scan a bond length, angle, torsion or asingle Cartesian coordinate. The initial-value does not have to be the value in your inputgeometry (molecule specification)

Again, if you specify more than one coordinate, muti-dimensional scan will be performed overthe corresponding grid of points.

For example:

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

Wat er r el axed PES scan

0 1O - 0. 464 0. 177 0. 0

H - 0. 464 1. 137 0. 0H 0. 441 - 0. 143 0. 0

1 2 0. 75 S 20 0. 05

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.


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

coordinates:- - - - - - - - - - - - - - - - - - - - - - - - - - - -! I ni t i al Par amet er s !! ( Angst r oms and Degr ees) !

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

! Name Def i ni t i on Val ue Der i vat i ve I nf o.!- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

! R1 R( 1, 2) 0. 75 Scan!! R2 R( 1, 3) 0. 9599 est i mate D2E/ DX2!! A1 A( 2, 1, 3) 109. 4731 est i mate D2E/ DX2!- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Number of opt i mi zat i ons i n scan= 21

and after the scan is done:

Summary of Opt i mi zed Potent i al Sur f ace Scan1 2 3 4 5

Ei genval ues - - - 75. 94099 - 75. 97678 - 75. 99772 - 76. 00803 - 76. 01074R1 0. 75000 0. 80000 0. 85000 0. 90000 0. 95000R2 0. 94258 0. 94430 0. 94532 0. 94642 0. 94739A1 108. 55551 107. 78770 107. 04742 106. 26206 105. 45388

6 7 8 9 10Ei genval ues - - - 76. 00807 - 76. 00162 - 75. 99257 - 75. 98175 - 75. 96981

R1 1. 00000 1. 05000 1. 10000 1. 15000 1. 20000R2 0. 94824 0. 94898 0. 94963 0. 95019 0. 95068A1 104. 62974 103. 79781 102. 96623 102. 14224 101. 33169

11 12 13 14 15Ei genval ues - - - 75. 95718 - 75. 94420 - 75. 93112 - 75. 91808 - 75. 90522

R1 1. 25000 1. 30000 1. 35000 1. 40000 1. 45000R2 0. 95110 0. 95147 0. 95180 0. 95209 0. 95234A1 100. 53892 99. 76682 99. 01706 98. 29028 97. 58639

16 17 18 19 20Ei genval ues - - - 75. 89260 - 75. 88029 - 75. 86829 - 75. 85665 - 75. 84537

R1 1. 50000 1. 55000 1. 60000 1. 65000 1. 70000R2 0. 95257 0. 95278 0. 95297 0. 95314 0. 95330A1 96. 90477 96. 24454 95. 60471 94. 98435 94. 35747

21Ei genval ues - - - 75. 83444

R1 1. 75000R2 0. 95345A1 93. 77380

where the ei genval ues 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):


17/17

O-H bond length (Angstrom)

0.8 1.0 1.2 1.4 1.6 1.8

E n e r g y ( H a r t r e e )

-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 coordinatesyou 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

Wat er r el axed PES scan

0 1O - 0. 464 0. 177 0. 0H - 0. 464 1. 137 0. 0H 0. 441 - 0. 143 0. 0

1 2 0. 75 S 20 0. 052 1 3 F

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

fixed at 110 degrees:

0 1O - 0. 464 0. 177 0. 0H - 0. 464 1. 137 0. 0H 0. 441 - 0. 143 0. 0

1 2 0. 75 S 20 0. 051 3 0. 75 S 20 0. 052 1 3 110. F

molecular modeling notes 8

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