Subject: Ionization Potentials

From: "Fernando De La Vega" <[email-masked]>
Date: Sun, 07 Apr 2002 21:40:33 -0700

Dear All,
  I want to calculate the Ionization Potential (IP) for a particular system. 
I know I can get an approximate value through Koopman's theorem but I would 
like to be more exact. I understand that I can calculate the IP from the 
energy difference between the molecule and its +1 cation. However, what I am 
not sure about is whether I should calculate the energy for the +1 cation 
using the geometry of the of the neutral molecule or whether I should 
optimize the geometry of the cation and then calculate the difference. Your 
comments on the subject are really appreciated. I will summarize if there 
are any responses. Thanks.

Fernando De La Vega

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From: Marcin Makowski <[email-masked]>
Date: Mon, 8 Apr 2002 10:24:48 +0200 (MET DST)

Computing energy of the cation at the same geometry will give you vertical
IP, while  relaxing geometry will result in adiabatic IP. I dont know
which of these you are interested in, because both of them might be
measured in experiment (but different type).

Yours sincerely,
Marcin Makowski
Ph.D. - student in Theoretical Chemistry
[email-masked]

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From: elewars <[email-masked]>
Date: Mon, 08 Apr 2002 12:19:01 -0400

Hello,

Fernando De La Vega asked about ionization energies.

Whether you use for your calculation (1) the cation at the geometry of the
neutral or (2) the optimized geometry of the cation, depends on which ionization
energy (ionization potential) you want:
(1) gives the _vertical_ ionization energy,
(2) gives the _adiabatic_  IE

The adiabatic values seem to be of more interest to chemists, since they
represent the energy difference between two "stable" species (potential energy
surface relative minima), but compilations of IE do not always state clearly
whether their listed IEs are vertical or adiabatic; an exception is R. D. levin
and S. G. Lias, "Ionization Potential and  and Appearance Potential
Measurements, 1971-1981", National Bureau of Standards, Washington DC, 1982.

Vertical IEs seem to be slightly (up to 0.2 eV) higher than adiabatic (e.g.
experimental CH3OH, 11.0 and 10.9 eV). Koopman' IEs are about 1.0 eV too high
(CH3OH 12.06 by HF/6-31G*, 12.12 by MP2(fc)/6-31G*, HOMO energy). The difference
of the calculated cation and neutral energies can give accurate IEs if post-HF
calculations are used; for CH3OH (eV):

vertical    HF/6-31G* 9.66, MP2(fc)/6-31G* 10.82

adiabatic  HF/6-31G* 9.38, MP2(fc)/6-31G* 10.6

Note that experimental and calculated vertical IEs are for electronically but
not geometrically relaxed cations; adiabatic IEs are for completely relaxed
cations; and Koopman' IEs refer to cations that are neither electronically nor
geometrically relaxed.

E. Lewars
================

Fernando De La Vega wrote:

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