A fundamental problem that
has not yet been pointed out in this thread is that Bader (or, perhaps more
informatively, AIM) charges computed using a pseudopotential basis set are
unlikely to match the same charges computed with an all-electron basis set even
if the pseudopotential is "perfect" (where "perfect" means that the density of
the valence electrons is precisely the same with the pseudopotential as in an
all-electron calculation at the same level with an infinitely flexible treatment
of the core).
By using a pseudopotential, the atomic basins
will be defined by the zero-flux surfaces of those electrons included in the
calculation. There is certainly no reason to suspect that the ignored core
electron density will (i) have the same zero-flux surfaces or (ii) integrate to
the correct integer number of electrons within the valence zero-flux
surfaces.
As such, unless one wants to define a new model
chemistry for AIM with missing core electrons (which doesn't seem particularly
profitable to me, personally), one should avoid AIM analysis in the absence of
an all-electron basis set. If the core is REALLY small, then perhaps one can be
confident that the density of the missing electrons will be vanishingly small by
the time one reaches the zero-flux surface about any atom, but since these
surfaces may be as near as half a bond-length to another atom, that's a pretty
steep requirement (so to speak).
Chris
Cramer On Aug 22, 2010, at 5:38 PM, Alan Shusterman alan
. reed.edu wrote: I think the Bader charge
you have is ok. The lanl2dz replaces core electrons on Au with a model
potential. Just figure out how many core electrons on Au were replaced (-N)
and add that to the Bader charge you have right now (+Q) to get the "correct"
Bader charge. Alan On 8/20/2010 10:27 AM, neranjan
perera neranjan007],[gmail.com wrote: Hi,
Thank you very much for the help given to
understand how it is calculated,
I do have a another
question concerning the bader charges,
I used to create a cube file
using gaussian, and converted it to bader charges. I used b3lyp theory
and lanl2dz basis set for Au atoms and 6-31g** for the other atoms.
To calculate the partial chargers using , subtracting
the Bader charge > from the Valance number of electrons gives me a good
result for the atoms which basis 6-31g** was used. But not for the Au
(gold) atoms.
The bader charge for the gold atoms are around 18.8~18.9
How can I calculate the partial charge for the Au atoms when
lanl2dz basis is used?
Thank you very much.
Regards,
Neranjan Perera
The numbers have simply been substracted from their atomic
numbers, as is done in population analysis schemes
PKI
On Wed, Aug 18, 2010 at 2:33 AM, Robert
McGibbon rmcgibbo{=} princeton.edu <owner-chemistry*-*ccl.net> wrote:
Seems very
simple. They just subtract the number of electrons in the Bader
region from the number of protons. Thus, the partial charge on
oxygen is 8 - 9.1566 = -1.157, and in the hydrogen region
it's 1 - 0.4238 = +0.576.
Here's the relevant section of the paper: ==== Three Bader
regions were found, each containing one atom. The total charge in
each one of the regions around the hydrogen atoms contained
0.4238 electrons and the oxygen region contained 9.1566
electrons, which gives a sum of 10.0041 electrons. The atomic
partial charges are the same as found by Bader [2] (see Table 1), showing that these two different
algorithms yield the same results.
Table 1.
Partial charge of oxygen and hydrogen in an isolated water
molecule
|
|
|
| Partial charge
|
| O atom | H atoms |
| Baderâs
original work | â1.16 |
+0.58 |
| This work |
â1.157 | +0.576 |
---- Robert McGibbon
Princeton University Undergraduate Class of
2011
On
Tue, Aug 17, 2010 at 11:32 PM, neranjan perera neranjan007]*[ gmail.com <owner-chemistry()ccl.net>
wrote: Hi,
In the
paper " G. Henkelman et al. / Computational Materials Science
36 (2006) 354â360" , they have got bader chargers for
water (H2O) as,
For Hydrogen
0.4238 e
Oxygen 9.1566 e
and then converted these Bader chargers into atomic partial
charges;
Hydrogen +0.58
Oxygen -1.16
So I want to know how to do the
above conversion ?
Thanks
Neranjan Perera.
On
Tue, Aug 17, 2010 at 9:21 PM, Radoslaw Kaminski
rkaminski.rk[#] gmail.com <owner-chemistry|ccl.net>
wrote: What do you understand by
these 'atomic partial charges'?
Radek
2010/8/17 neranjan perera neranjan007!A! gmail.com
<owner-chemistry|,|ccl.net>
Hi,
how
can i convert "bader chargers" to "atomic
partial chargers"?
Thanks.
Neranjan Perera.
neranjan007 *-* gmail.com
--
Graduate Student
Department of Chemistry
University of
Connecticut
--
Radoslaw
Kaminski, M.Sc. Eng.
Ph.D. Student
Crystallochemistry Laboratory
Department of
Chemistry
University of Warsaw
Pasteura 1, 02-093 Warszawa, Poland
http://acid.ch.pw.edu.pl/~rkaminski/
--
Graduate Student
Department of Chemistry
University of
Connecticut
--
Alan Shusterman
Chemistry Department
Reed College
Portland, OR 97202-8199
503-517-7699
http://blogs.reed.edu/alan/
"Nature doesn't make long speeches." Lao Tzu 23 --
Christopher J. Cramer Elmore H. Northey
Professor University of Minnesota Department of Chemistry 207 Pleasant St. SE Minneapolis, MN 55455-0431 -------------------------- Phone: (612) 624-0859 || FAX: (612) 626-7541 Mobile: (952) 297-2575 email: cramer.**.umn.edu jabber: cramer.**.jabber.umn.edu http://pollux.chem.umn.edu/~cramer (website includes information about the textbook
"Essentials of Computational
Chemistry: Theories and Models, 2nd
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