Experiment 8 TITLE: Binding Energy, Ionization Potential, Electron Affinity Objectives:1. To understand the ionization potential (IE), the electron affinity (EA), and the binding energy (BE). 2. To calculate the IE and the EA using both Koopman's Theorem and Direct SCF method. 3. To determine the binding energy Introduction

All GAMESS calculations that you performed until now was done using the Hartree-Fock (HF) method. HF method is a reasonable model in many problems particularly in providing reliable predictions for geometries of molecules. However, the limitation is that the theory does not include electron correlation. For many molecular systems such effect is very critical for determining the energetics and properties of molecular systems, particularly in calculating the energetics of reactions and bond dissociations. In this experiment, we will study the ionization potential, the electron affinity, the binding energy, and the Koopman's theorem. Ionization energies (IE) measure the tendency of a neutral atom to resist to the loss of electrons. Thus, it takes a considerable amount of energy. While the electron affinity (EA) is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion.

As you already know frontier orbitals are a collective term for the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO). We use the approach known as "Koopman's theorem". Based on the "Koopman's theorem" IE is equal to the Highest Occupied Molecular Orbital (HOMO) energy (this method is exact for only hydrogenic atoms) and the EA is the energy of the Lowest Unoccupied Molecular Orbital (LUMO) approximately. Thus, the Koopman's Theorem takes the minus of the energy of the HOMO for the ionization energy and the eigenvalue of the LUMO for the EA (no sign change).

We can also use the "Direct SCF method" or the "subtraction method" in which the IE is equals to the energy of the ion minus the energy of the neutral form of the molecule and the EA is the energy of the neutral form and the anionic form of the molecule. Therefore, the binding energy is defined as;

Binding Energy (BE) = [ E(Monomer 1) + ZPE(Monomer 1) ] + [ E(Monomer 2) + ZPE(Monomer 2) ] - [ E(Dimer) + ZPE(Dimer) ]

Precedure1. Calculating the properties of PH2, PH2+, and PH2-

2. Binding Energy in HF Result1. Calculating the properties of PH2, PH2+, and PH2- PH2PH2-PH2+

coordinatecartesiancartesiancartesian

Charge0-1+1

Multiplicity213

wavefunctionROHFRHFROHF

Basis set6-31G(2d,p)6-31G++(2d,p)6-31G(2d,p)

Vibrational scaling factor0.8560.9040.856

Table 1: Ionization Energy (Reference experimental IE of PH2 = 9.82 eV or 226.45 kcal/mol)Molecule

Koopmans` theorem IESCF IE

-HOMO(opt)LUMO(opt)E(Hartree)(opt)ZPE(Hartree)(Hessian)

PH2-0.15630.1592 0.014379

PH2+----0.012384

PH2-----0.012434

Table 2: Electron Affinity (Reference experimental EA of PH2 = 1.26 eV or 29.06 kcal/mol)

2. Binding Energy in HF E(H) (Hartree) E(F) Hartree E(HF) (Hartree) ZPE(HF) (Hartree) B. Energy (Hartree) B. Energy (Kcal/mol)

HF/STO-3G -0.4665818504-97.9865050330-98.55462715090.0101950.0913452675

HF/631G -0.4982329092-99.3602181659-99.96150596320.0094210.0936338849

HF/631G(d) -0.4982329092-99.3617921172-99.97622049050.0099280.0936338849

HF/631G(2d,p) -0.4982329092-99.3622294747

-100.01762884500.0102390.1674054611

HF/6311+G (2d,p) -0.4998098153-99.3976089299-100.02421991020.0102050.116597065

HF/6311+G (2df,p) -0.4998098153-99.3983077787

-100.02480462000.0103090.116378014

MP2/6311G(2d,p) -0.4998098153-99.3949092786-100.01762884500.0096310.113278714

MP2/6311+G (2d,p) -0.4998098153-99.3976089299-100.02421991020.0095090.117292165

MP2/6311+G(2df,pd) -0.4998098153-99.3983077787-100.02480462000.0095520.117135026

Question ( Part 1)

i. Compare the calculated results with the experimental values. What is your conclusion? ii. Compare the results of these two different methods. Which method provides a better prediction? iii. What are the fundamental differences between these two methods?

Question (Part 2)

i. What basis set provides the most accurate results? Why? ii. Compare your calculated results with the experimental value. Determine the error (in kcal/mol) with respect to the experimental value for each calculation and fill out the table below; HF/STO-3G

HF/631G

HF/631G(d)

HF/631G(2d,p)

HF/6311+G (2d,p)

HF/6311+G (2df,pd)

MP2/6311G(2d,p)

MP2/6311+G (2d,p)

MP2/6311+G (2df,p)

MP2/6311+G(2df,pd)

Discussion

Conclusion

References1. Atkins, P. & de Paulo, J. (2009). Physical Chemistry. 9th edition. Oxford: Oxford University Press. 2. http://www.sparknotes.com/chemistry/organic2/conformations/section1.rhtml3. http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes4. https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/rotconf1.htm