[Pw_forum] Oxidation state for dopants in TiO2
myoussef at mit.edu
Wed Jan 20 05:03:00 CET 2016
If the supercell is charge neutral which is the default, then one aims at simulating a neutral substitutional defect with respect to Ti. In this case V in 4+ oxidation state which corresponds in the naive fully ionic picture to V losing 4 of its valence electrons to oxygen and retaining the last one as you said. The question arises whether this electron is localized on the V site and it is really in 4+ oxidation state. The other alternative is that the electron delocalizes and the supercell contains V in 5+ oxidation state and an extra "free" electron. In my opinion the best way to analyze this problem is to start from the most positive oxidation state and systematically reduce it. In the case of V you can do 6 separate relaxation simulations starting from tot_charge=+1 which corresponds to V on 5+ oxidation state, all the way to tot_charge=-4 which corresponds to V in 0 oxidation state. After these 6 simulations are done you can track the changes in the charge density and spin density when you go from the oxidation state q to the oxidation state q-1. In my experience V in oxides such as TiO2 can take oxidation states from 5+ to 2+. If you compare 1+ and 2+ you will notice that the extra electron you add to 2+ to achieve 1+ never localizes on V and as such 2+ is likely the lowest oxidation state for V in these oxides.
If you do not want to do this lengthy analysis and you just want to check whether you have 4+ or not , check the spin density. V4+ will likely have a net magnetic moment close to 1 Bohr Mag.
Also there are many cases in literature where achieving certain oxidation state never happens because of electron (hole) delocalization. For example, if you try to model neutral hydrogen interstitial in ZnO or ZrO2, you will get interstitial proton and a free delocalized electron. See for example:
A final word of caution, analyzing the oxidation states of transition metal dopants and their changes by adding or removing electrons requires very dense and accurate grids for representing the charge (and spin) density. The reason is that the change of the localized charge (if any) on the transition metal defect while going from formal oxidation state q to q-1 is usually very low (in my experience in the order of 0.1 e). This observation was discussed in this article: http://www.nature.com/nature/journal/v453/n7196/full/nature07009.html
(It is also fun and instructive to follow the debate that this paper raised in literature!)
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