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Koopman's Theorem with Applications in Medicine

Copyright (c) RDMCHEM LLC 2019

 

Overview

Koopman's Theorem and Periodic Trends

Application of Koopman's Theorem: Ionization Energies and NSAIDs

Appendix

Overview

 

Very often, chemical reactivity can be thought of in terms of ionization energies.  For example, the lower an atom or molecule's ionization energy, the higher its reactivity.  For an atom or molecule, X, its ionization energy, IE, can be defined as

 

X(g) -->  X+(g) + 1e-         ΔE = IE

 

Of course, ionization energies will depend on from which molecular orbital an electron is removed.   The lowest ionization energy would correspond to removing an electron from the highest occupied molecular orbital (HOMO).   So, given its importance, it is often useful to calculate an atom or molecule's ionization energy to predict or explain its chemical reactivity. In this activity, we explore two approaches. One approach is to calculate explicitly the energy of X, and X+ and take energy differences:

 

IE = E(X+) − E(X)          (1)

 

For smaller molecules, this is a straightforward approach. However, as the size of the molecule increases, this approach becomes increasingly difficult.  Another more approximate method, known as Koopman's Theorem, requires the calculation of the electronic structure of X only:

 

Koopman's Theorem: The lowest (vertical) ionization energy of a molecule can be approximated as the negative of the highest occupied molecular orbital (HOMO) energy:  

IE_Koopman = − e(HOMO)          (2)

                                        

 

More generally, ionization from the ith molecular orbital (MO) is approximately the negative of the MO energy.  In this exercise, we will explore to some degree the quality, or lack thereof, of Koopman's theorem for approximating ionization energies.  We will then consider an application in which the ionization energies of a series of non-steroidal anti-inflammatory drugs (NSAIDs) can be related to their medicinal activity!

 

Koopman's Theorem and Periodic Trends

 

In this section, we explore Koopman's theorem for a series of atoms (and molecules) and compare calculated ionization energies to their experimentally observed values. In particular, we consider periodic trends for atoms as one proceeds down a group or across a period.  We consider  Group 1A (H - Cs), Group 7A (F - I), and  Period 2 (Li - Ne).  For those that are interested, we also consider a series of isoelectronic binary compounds: methane, ammonia, water, and hydrogen fluoride.

Initialize

Here we initialize the Maple packages we wish to use as well as define the experimental values of the quantities we wish to calculate for comparison.

 

withplots:withScientificConstants:withQuantumChemistry:digits15: 

Atomic Ionization Energies and Periodic Trends

 

In this section, we calculate the energies of the neutral and cation species for Group 1A, Group 7A, and Period 2 elements and use equations (1) and (2) to determine IEs.  We then plot the results along with experimental values to assess Koopman's theorem.    We will use Hartree Fock theory with a DZP basis, which allows us to calculate energies of Group 1A elements up to cesium and Group 7A elements up to iodine.

 

The following Maple input is divided into steps.  The first 3 steps are used to calculate the IE of the H.  Step 4 requires you to repeat Steps 1 - 3 to calculate IEs for the remaining elements in Group 1A.  Once you have completed the series, you can plot the results and then repeat the process for Group7A and Period 2.  For logistical reasons, we consider the following variables to be specified before each calculation:

 

Group1A series has n = 6 elements.  The index m labels which atom:  

   H (m = 1),  

   Li (m = 2),

   Na (m = 3),

   K (m = 4),

   Rb (m = 5), and

   Cs (m = 6).  

   Num_neutral = 1, Num_cation = 0.

 

Group7A series has n = 4 elements.  The index m labels which atom:  

   F (m = 1),  

   Cl (m = 2),

   Br (m = 3), and

   I (m = 4).  

   Num_neutral = 1,

   Num_cation = 2.

 

Period 2 series has n = 8 elements.  The index m labels which atom:  

   Li (m = 1),

   Be (m = 2),

   B (m = 3),

   C (m = 4),

   N (m = 5),

   O (m = 6),

   F (m = 7), and

   Ne (m = 8).  

   Num_neutral = varies according to ground electron configuration and Hund's Rule,

   Num_cation = varies according to ground electron configuration and Hund's Rule.

 

 

Step 0:  Initialize vectors (Run once for each series: Group 1A, Group 7A, and Period 2)

 

AObasisdzp:     KoopmanIEVector8,0:IEVector8,0:IE_Group1AVector1312.04, 520.22, 495.84, 418.81, 403.03, 375.70, 392.96:IE_Group7AVector1681.03, 1251.18,1139.85, 1008.39:IE_Period2Vector520.22, 899.50, 800.63, 1086.44, 1402.32, 1313.93, 1681.03, 2080.64:

 

 

Step 1: Specify parameters

 

n6:      # for Group 1A, n = 6, for Group7A, n = 4, for Period2, n = 8

m1:     # Start with m = 1 and increase as you proceed down or across a group or period, respectively.

elementH:      # Provide the chemical symbol

Num_neutral1:  # Number of unpaired electrons in neutral

Num_cation0:   # number of unpaired electrons in cation ground electron configuration

 

 

Step 2: Calculate energies of neutral atom and cation. Determine HOMO.

 

Atomelement,0,0,0:NeutralHartreeFockAtom,charge=0, spin=Num_neutral, basis=AObasis:CationHartreeFockAtom,charge=1, spin=Num_cation, basis=AObasis:for i from 1 by 1 while Neutralmo_occi>0 do end do:HOMOi1; 

HOMO1

(2.2.1)

 

 

Step 3: Calculate IE (in kJ/mol) using Equations (1) and (2) 

 

KoopmanIEm2625.50 Neutralmo_energyHOMO;IEm2625.50Catione_totNeutrale_tot;

_rtable18446744829320603638

_rtable18446744829320603758

(2.2.2)

 

 

Step 4: Repeat steps 1 - 3 for each element in group or period

 

 

Step 5: Plot Ionization Energies and Compare

 

XVectorn,ii :AplotX,IE_Group1A,color=Blue:BplotX,KoopmanIE,color=Red:CplotX,IE,color=Green:displayA,B,C

 

How do ionization energies compare with experimental values using Equations (1) and (2)? How does the *trend* in ionization energies compare?

Answer

We see that Equation (1) using Hartree-Fock and the DZP basis compares well with experiment.  Approximate IEs using Koopman's theorem (Equation (2)) are qualitatively correct with the same trend as experimental values.

 

Repeat Steps 0 - 5 for Group 7A and Period 2.

 

Ionization Energies of Small, Isoelectronic Binary Hydrides

 

In this section, we consider Koopman's theorem for an isoelectronic series of binary hydrides: methane (CH_4), ammonia (NH_3), water (H_2O), and hydrogen fluoride (HF).  Similar to what we did above, we let

   m = 1 (methane),

   m = 2 (ammonia),

   m = 3 (water), or

   m = 4 (hydrogen fluoride).

 

Step 0: Initialize vectors

 

AObasistzp:IE_molsVector1388.89,1050.20,1215.61,1525.41:IE_HFlimVector1433.52, 1123.71, 1331.13, 1706.57: KoopmanIEVector4,0:IEVector4,0:

 

 

Step 1: Specify Parameters

 

m3:moleculewater:

 

 

Step 2: Calculate energies of neutral molecule and cation. Determine HOMO.

 

molecMolecularData molecule,geometry3d:

NeutralHartreeFockmolec,charge=0, spin=0, basis=AObasis:CationHartreeFockmolec,charge=1, spin=1, basis=AObasis:for i from 1 by 1 while Neutralmo_occi>0 do end do:HOMOi1; 

HOMO5

(2.3.1)

 

 

Step 3: Calculate IE (in kJ/mol) using Equations (1) and (2) 

 

KoopmanIEm 2625.50 Neutralmo_energyHOMO;IEm2625.50Catione_totNeutrale_tot;

001330.652320260

001066.414776300

(2.3.2)

 

Step 4: Repeat steps 1 - 3 for ammonia (m=2), water (m=3), and hydrogenfluoride (m=4)

 

 

Step 5: Plot Ionization Energies and Compare

 

Note: Equation (1) = Blue,   Equation (2) (Koopman's Thm) = Red, Experiment = Green

 

 

XVector4,ii :AplotX,IE_mols,color=Blue:BplotX,KoopmanIE,color=Red:CplotX,IE,color=Green:displayA,B,C

We find that Koopman's theorem is slightly larger than the experimental ionization for each molecule, but the trend is correct.  

 

Optional

It is important to note that ionization energies calculated using Koopman's theorem using Equation (2) or explicitly using Equation (1)  depend on the level of theory and basis.  Using higher levels of theory than Hartree Fock can improve calculated ionization energies. Repeat the calculations above for methane - hydrogen fluoride with the tzp basis.  Did the calculations improve?

 

 

Application of Koopman's Theorem: Ionization Energies and NSAIDs

 

Now that we are familiar with Koopman's theorem, let's use it to predict activities of non-steroidal anti-inflammatory drugs (NSAIDs). The compounds we consider are based on substituted salicylic acids (SAA), benzoic acids (BZA), and phenols [1,2].  Figure 1 depicts the structures considered.  R-groups include 3-OH, 4-OH, 5-OH, 3-F, 5-F, 3-Cl, 4-Cl, 5-Cl, 3-IPR (isopropyl), 4-IPR, and 5-IPR.  When X=H, we have SAA, and when S=OH, we have BZA.

 

Figure 1: Skeletal structure represents substituted salicylic acids (SAA) and benzoic acids (BZA) as NSAIDs. Substitution of the R group occurs at the numbered positions.   The skeletal structure was generated with the SkeletalStructure command.

 

Salicylic acid derivative   (SA_index)

Coordinate Filename

Activity

H (salicylic acid)   (1)

H.xyz

3.33

3-Cl    (2)

3Cl.xyz

3.89

3-F    (3)

3F.xyz

3.82

3-OH    (4)

3OH.xyz

4.43

3-IPR    (5)

3IPR.xyz

3.92

4-Cl    (6)

4Cl.xyz

3.31

4-OH    (7)

4OH.xyz

3.02

4-IPR    (8)

4IPR.xyz

3.29

5-Cl    (9)

5Cl.xyz

4.06

5-F    (10)

5F.xyz

3.82

5-OH    (11)

5OH.xyz

4.61

5-IPR    (12)

5IPR.xyz

4.12


                   
Use the Maple input below to estimate the ionization energy of each salicylic acid derivative.          

 

Step 1: Initialize and read in activities of salicylic acid derivatives.   You will not have to reinitialize again.

 

withQuantumChemistry:Digits15:withplots:ActivitiesVector12, ReadXYZactivities.dat,1:IonizationEnergiesVector12:

 

 

Step 2: Specify salicylic acid derivative to be calculated.  The "SA_index" and coordinate filename can be found in Table 1:

 

SA_index2;        # Comment: 1 = H (salicylic acid), 2 = 3-Cl, 3 = 3-F, etc. See Table 1

SA_index2

(3.1)

molec ReadXYZ3Cl.xyz;

molecO,−0.25234200,−1.82785300,−0.00028600,O,3.34848900,0.38099100,0.00014600,O,2.35622900,−1.66916500,−0.00007400,C,−0.23597400,−0.46877200,−0.00006500,C,0.96542400,0.29463800,0.00005900,C,−1.45886000,0.22292000,−0.00002900,C,0.90804900,1.70344800,0.00022600,C,−1.50685900,1.60886400,0.00014300,C,−0.31660900,2.35789300,0.00027100,C,2.23723300,−0.41825800,0.00001200,Cl,−3.00103200,−0.72522200,−0.00015800,H,1.83746100,2.25851900,0.00032100,H,−2.47031400,2.10464600,0.00018100,H,−0.36247200,3.44041100,0.00040400,H,4.15935500,−0.17151400,0.00013100,H,0.68008900,−2.17947200,−0.00034200

(3.2)

PlotMoleculemolec;

 

 

Step 3: Calculate energy using DFT with 6-31g basis.  For improved results, you may try a larger basis set at a computational (time) cost.

 

dataDensityFunctionalmolec,basis=6-31g;

table%id=18446744829309859870

(3.3)

for i from 1 by 1 while datamo_occi=2 do end do: HOMOi1; 

HOMO44

(3.4)

IonizationEnergiesSA_index1 datamo_energyHOMO27.2116;

IonizationEnergies26.55694334

(3.5)

Now repeat Steps 2 - 3 for SA_index = 2 to 12!  

 

 

Once you have calculated the ionization energy for each salicylic acid derivative, you can plot the ionization energies verses activity and use the interactive CurveFitting function to calculate a linear fit.

 

plotspointplotIonizationEnergies,Activities;

#CurveFittinginteractiveIonizationEnergies,Activities, x;      # Choose "Plot" for linear fit then "Done"

fIE9.352957349768440.879064241187929IE

fIE9.35295735+−10.87906424IE

(3.6)

 

Now you can use the function f(x) to estimate the activity of new derivatives!  Consider a derivative with a fluorine at the 3-position.   The coordinate file is "4F.xyz".  

 

 

molec ReadXYZ4F.xyz;

molecO,−0.57423900,2.20316900,−0.00033600,O,−2.61515200,−1.50531600,0.00009700,O,−2.76883700,0.76845100,−0.00021900,C,0.08316400,1.00840100,−0.00013300,C,−0.59083300,−0.24584600,0.00001100,C,1.48479400,1.04612000,−0.00008400,C,0.16589100,−1.43790100,0.00021100,C,2.17656000,−0.15135800,0.00011300,C,1.55297100,−1.40487300,0.00026600,C,−2.04454000,−0.25944200,−0.00005300,H,1.99737200,1.99781200,−0.00019900,H,−0.36039200,−2.38396500,0.00032000,F,3.56252000,−0.11046100,0.00016300,H,2.14977600,−2.30713700,0.00041800,H,−3.59295500,−1.42528600,0.00004700,H,−1.55869500,2.05168100,−0.00036800

(3.7)

PlotMoleculemolec;

 

#GeometryOptimizationmolec, DensityFunctional, basis=6-31g;

dataDensityFunctionalmolec,basis=6-31g;

table%id=18446744829479293342

(3.8)

for i from 1 by 1 while datamo_occi=2 do end do:HOMOi1; 

HOMO40

(3.9)

IE1 datamo_energyHOMO27.2116;

IE6.69549138

(3.10)

fIE

3.46719030

(3.11)

 

 

Do you expect the 4-F derivative to be more or less active then the average derivative?

Answer

Less... The average activity is 3.80. The predicted activity of the 4-F derivative is only 3.5.  

 

How could the prediction of activity be improved, if possible?

Answer

Our estimation of activity is based on two approximations:  1) Koopman's theory for calculating IE, and 2) estimating activity as a linear function of IE.   From the point plot, it is clear that there is only a trend relating IE and activity and that the relationship is not exactly linear.  We can only improve how we estimate IE, either using a higher level of electronic theory, a larger basis, or calculating IE using Eq. (1) from the Overview instead of Koopman's theorem.   We should point out, however, that improving our estimation of IE does not necessarily mean the linear fit will be any better. See Ref. 2 for more details on this matter!
   
One possible way to improve the prediction of activity would be to include more derivatives in the linear fit. Here we chose only a subset of the activity data presented in Ref. 2.

 

Appendix

References

1. Mehler, et al.,  Int. J. of Quant. Chem., 35, 205 (1989).

2. Shakman and Mazziotti, J. Phys. Chem. A, 111, 7223 (2007).

3. A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory (Dover Books, New York, 1996).