MolecularIEs - Maple Help

Koopman's Theorem and Molecular IEs

 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.



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.

 >
 $\left[\begin{array}{c}1388.89000000\\ 1050.20000000\\ 1215.61000000\\ 1525.41000000\end{array}\right]$
 $\left[\begin{array}{c}1433.52000000\\ 1123.71000000\\ 1331.13000000\\ 1706.57000000\end{array}\right]$
 $\left[\begin{array}{r}0\\ 0\\ 0\\ 0\end{array}\right]$ (2.1)



Koopman's Theorem and IEs 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

 >
 ${\mathrm{AObasis}}{≔}{"dzp"}$
 ${\mathrm{AOmethod}}{≔}{\mathrm{HartreeFock}}$ (3.1)

Step 1: Specify Parameters

 > $m≔1:\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{molecule}≔"methane":$

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

 >
 ${\mathrm{molec}}{≔}\left[\left[{"C"}{,}{0}{,}{0}{,}{0}\right]{,}\left[{"H"}{,}{0.55410000}{,}{0.79960000}{,}{0.49650000}\right]{,}\left[{"H"}{,}{0.68330000}{,}{-0.81340000}{,}{-0.25360000}\right]{,}\left[{"H"}{,}{-0.77820000}{,}{-0.37350000}{,}{0.66920000}\right]{,}\left[{"H"}{,}{-0.45930000}{,}{0.38740000}{,}{-0.91210000}\right]\right]$ (3.2)
 >
 ${\mathrm{table}}{}\left({\mathrm{%id}}{=}{18446744966272555454}\right)$
 ${\mathrm{table}}{}\left({\mathrm{%id}}{=}{18446744966267056670}\right)$
 ${\mathrm{HOMO}}{≔}{5}$ (3.3)

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

 >
 $\left[\begin{array}{c}1424.85241559\\ 0\\ 0\\ 0\end{array}\right]$
 $\left[\begin{array}{c}1301.05550066\\ 0\\ 0\\ 0\end{array}\right]$ (3.4)



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

 >



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?



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).