
	O atom	H atoms
Bader’s original work	−1.16	+0.58
This work	−1.157	+0.576

Partial charge

O atom

H atoms

Bader’s original work

−1.16

+0.58

This work

−1.157

+0.576

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{=3D}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 =3D -1.157, and in the hydrogen region it's 1 = - 0.4238 =3D +0.576.

=
Here's the relevant section of the = paper:
=3D=3D=3D=3D
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

Bader=E2=80=99s = original work =E2=88=921.16 +0.58
This = work =E2=88=921.157 +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) = 354=E2=80=93360" , 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.
=

=             &n= bsp;           &nbs= p;            =             &n= bsp;           &nbs= p;          =

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

=
--=20
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
=

=
=	O = atom	H = atoms
Bader=E2=80=99s = original work	=E2=88=921.16	+0.58
This = work	=E2=88=921.157	+0.576

Christopher J. Cramer

Elmore H. Northey Professor

University of Minnesota

Department of = Chemistry

Minneapolis, MN 55455-0431

Phone: (612) 624-0859 || = FAX: (612) = 626-7541

email: cramer.**.umn.edu

jabber: cramer.**.jabber.umn.edu

http://pollux.chem.umn.edu/~cr= amer

of Computational = Chemistry: Theories and Models, = 2nd Edition")

=
=	Partial charge =
=	O = atom	H = atoms
Bader=E2=80=99s = original work	=E2=88=921.16	+0.58
This = work	=E2=88=921.157	+0.576