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Bader charge vs. oxidation state

Posted: Sun Apr 12, 2015 9:52 pm
by jzaffran
Hello

I am currently working on a research project aiming at determining oxidation state of metallic atoms in a metal oxide material.

To reach this goal, I performed a Bader analysis on my system as described in the website http://theory.cm.utexas.edu/henkelman/code/bader/

However I observed that various metallic atoms, that are expected to have different oxidation states, have very close Bader charge.

For example, all the Ni oxidation states (0, I, II, III or IV) have Bader charges in the range 1.2-1.3

Thus, it seems that Bader analysis is not an efficient tool to discriminate between several oxidation state.

What is your advice?

I would like to know in particular:

1- What is the difference between Bader charge and oxidation state? Are those 2 concepts similar?
2- How to deduce oxidation state for Bader analysis?
3- What is the correct procedure to compute oxidation state of an atom? Shall one also consider other parameters such as Magnetization of Mulliken charge?

Thanks for your help
Jeremie

Re: Bader charge vs. oxidation state

Posted: Tue Apr 14, 2015 1:56 pm
by graeme
The Bader charges should be correlated to the formal oxidation states. Typically the bader charges are significantly reduced however, because electrons do not sit entirely on the metal centers.

What you describe, however, with only tenths of charge differences over many oxidation states, does not sound right. Double-check that you are treating the core charge properly (see the "Note for VASP users" on http://theory.cm.utexas.edu/henkelman/code/bader/ ) Also make sure that your Ni atoms really are in the oxidation state that you think they are. You can do this, for example, using the DFT+U method and looking at the on-site occupancies, as well as the spin states. If you are using pure DFT the metal states will be artificially delocalized and (I expect) will have reduced charges at the metal centers.

Re: Bader charge vs. oxidation state

Posted: Tue Apr 14, 2015 5:38 pm
by jzaffran
Thanks a lot for your help
I would like to have one precision
What should be the expected difference between the Bader charges of two different oxidation state? Let us say for example that we want to distinguish Ni(II) and Ni(III). What should be the difference between their Bader charges (~0.3? more?)

Re: Bader charge vs. oxidation state

Posted: Tue Apr 14, 2015 9:43 pm
by xph
I have seen that V(III) and V(IV) only differ in Bader charge by 0.08 to 0.15 depending on the local environment. See Table 2. on page 3093 of the following paper:
http://theory.cm.utexas.edu/henkelman/p ... 4_3089.pdf
The difference between O- and O2- is more obvious.

The on-site occupancy or magnetization from DFT+U is better for distinguishing different oxidation states of 3d metals.

Re: Bader charge vs. oxidation state

Posted: Thu Jun 04, 2015 3:03 pm
by Neutrino
Hi Jeremie,

The small difference you found between different oxidation states has a fundamental reason. It is the so-called charge-self regulation mechanism. Essentially if you go from Ni3+ to Ni2+ anticipating a change of 1 in the bader charge, then this does not happen and most of the extra one electron is distributed among the surrounding ligands. This was suggested in this paper:
http://www.nature.com/nature/journal/v4 ... 07009.html

Having said , make sure first that the grid NGX,NGY,NGZ is well converged and that ADDGRID=.FALSE.

From my previous work on 3d transition metals in insulators, going from one oxidation state to the other can lead to bader change as small as 0.06e. If the change is much lower than this , then probably it is not possible to achieve that particular oxidation state at all at the level of theory you are using (GGA/LDA/...). As suggested above , you can also use DFT+U (with U=0 if you really do not want a Hubbard U on particular element) as this forces VASP to print out the occupations.

Mostafa