Subject: RE: Ionization Potentials

From: "Shobe, Dave" <[email-masked]>
Date: Mon, 08 Apr 2002 14:00:26 -0400

E. Lewars wrote:
>Vertical IEs seem to be slightly (up to 0.2 eV) higher than adiabatic (e.g.
>experimental CH3OH, 11.0 and 10.9 eV). 

There's a reason for IE-vert being > IE-adia.  Actually it's implicit in the
rest of E. Lewars' message, but it's worth stating explicitly:  
1. The IE-vert is the energy required to go from neutral at its geometric
energy minimum to cation at the *neutral's* geometric energy minimum; 
2. whereas the IE-adia is the energy required to go from neutral at its
geometric energy minimum to the *cation's* geometric energy minimum.  Note
that the starting point is in both cases the same: the neutral at the
neutral's geometric minimum.
3. And by definition, the cation's energy will be lower at *its* geometric
minimum than at any other geometry, including the neutral's geometric energy
minimum. QED.

--David Shobe
Süd-Chemie Inc.
phone (502) 634-7409
fax     (502) 634-7724
email  [email-masked]

Don't bother flaming me: I'm behind a firewall.

-----Original Message-----
From: elewars [mailto:[email-masked]]
Sent: Monday, April 08, 2002 12:19 PM
To: [email-masked]
Cc: [email-masked]
Subject: CCL:Ionization Potentials

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:

> 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
>
> _________________________________________________________________
> Send and receive Hotmail on your mobile device: http://mobile.msn.com
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