sun at magnet.fsu.edu
Tue Dec 10 16:31:13 CET 2013
Thank you for your explanations. I think I would cc this email to the forum. Thanks again!
Graduate Research Assistant
National High Magnetic Field Laboratory
Condensed Matter Science
Chemical & Biomedical Engineering
FAMU-FSU College of Engineering
Florida State University
sun at magnet.fsu.edu
----- Original Message -----
From: "Peter Blaha" <pblaha at theochem.tuwien.ac.at>
To: "Jifeng Sun" <sun at magnet.fsu.edu>
Sent: Tuesday, December 10, 2013 6:56:42 AM
Subject: Re: Questions
>> f) Have you checked the forces ?? Are the O in equillibrium ???
> I checked the forces and found out they are not equal to 0, since I didn't relax the structure.
> I just used the experimental lattice constants. Is that ok?
I don't know. Usually I do not trust powder x-ray data for light
elements too much.
>> g) from the charges you listed below, it seems in fact that Mo1 and 3 have LESS (spin-dn and
> total) electrons, so they should be the "6+ ions" ???
> To be more clear, those values I listed are the integrated values directly from case.outputt files,
> which I think are the number of electrons. And you can see a difference between spin-dn and spin-up,
> which we think that's where the magnetic stuff comes from. So, which means Mo1 and 3 are not 6+
> ions. We need 1 electron in d-orbital, so those Mo1 and 3 are a mixture of 5+ and 4+ (even with 6+),
> That's why we want to know how many 5+, 4+ and 6+ we have in our system.
>> h) why has Mo 6+ (5s^14d^5) ?????
> Mo has one 5s electron and five 4d electrons in its valence. So I am thinking a Mo6+ meaning it will
> lose both the 5s and 4d orbitals electrons.
As I explained to you before: A Mo6+ ion DOES NOT loose all its
electrons. This is a "formal" valence, which is a very useful concept in
chemistry, but does NOT mean that Mo does not have any d-electrons.
If you add-up spin-up + dn, the Mo 1 and 3 will have a littel less
electronic charge, thus these atoms I'd interpret as "Mo6+", while the
ones with a bit larger charge are Mo 4+ or so.
> Two more questions, how accurate would that "Bader" analysis be? And is that helpful to use -eece hybrid
> functional to calculate the 3d charge density?
What means "accurate" ??? You cannot expect, that ANY method would lead
to 5s0 4d0 configuration. This is unphysical.
Bader charges give you larger differences than just charges within an
PS: Please use in future the wien2k-mailing list. This discussion would
have been informative for the wien2k community (and I don't have to
repeat my arguments all the time...)
> Sorry for so many questions and thank you for your patience and help!
> Jifeng Sun
> Graduate Research Assistant
> National High Magnetic Field Laboratory
> Condensed Matter Science
> Chemical & Biomedical Engineering
> FAMU-FSU College of Engineering
> Florida State University
> sun at magnet.fsu.edu
> ----- Original Message -----
> From: "Peter Blaha" <pblaha at theochem.tuwien.ac.at>
> To: "Jifeng Sun" <sun at magnet.fsu.edu>
> Sent: Monday, December 9, 2013 1:15:08 PM
> Subject: Re: Questions
> a) You will NEVER find these formal valences like 6+. This is a very useful "concept"
> in chemistry, but it does NOT mean, that a Mo6+ ion has all d-states empty !!!!
> b) Typically, "real" charges within an atomic sphere will differ ONLY by a very small number
> (often 0.02 - 0.1 e.).
> c) You don't need to integrate the (partial) DOS numerically. This is done analytically and listed in
> case.outputt. Furthermore, the total partial charges are also listed in case.scf (:qtlXXX)
> d) A better estimate of a charge state is given by the "Bader"-charges. Use the aim program.
> (But don't expect a 6+ ion).
> e) this material is magnetic ??? Ferromagnetic ?? well converged (:MMT) , k-mesh ???
> f) Have you checked the forces ?? Are the O in equillibrium ???
> g) from the charges you listed below, it seems in fact that Mo1 and 3 have LESS (spin-dn and total)
> electrons, so they should be the "6+ ions" ???
> h) why has Mo 6+ (5s^14d^5) ?????
> Am 08.12.2013 23:15, schrieb Jifeng Sun:
>> Dear Dr. Blaha,
>> I am sorry to bother you but I just have a question about WIEN2K. I am doing
>> a calculation on the material LaMo16O44. There are four inequivalent Mo. From
>> our bond valence sum (BVS) calculation, the Mo2 and Mo4 should be 6+(5s^14d^5)
>> and Mo1 and Mo3 should be 4.625+(on average). My question is whether we can calculate
>> the valence of Mo1 and Mo3. Can we get the exact value?
>> In fact, I have tried to calculate the DOS and tried to integrate the d-orbital up to EF.
>> And I found out that for spin-up: Mo1=0.9387, Mo2=0.9279, Mo3=0.9419, Mo4=0.9244; for
>> spin-down: Mo1=0.87, Mo2=0.9265, Mo3=0.8678, Mo4=0.9236. But I don't know how to interpret
>> the results and these integrated d electrons are only inside the sphere,right? Then how to
>> get the interstitial d-electrons?
>> Any insight is helpful. I also attach the .struct file. Thanks!
>> Jifeng Sun
>> Graduate Research Assistant
>> National High Magnetic Field Laboratory
>> Condensed Matter Science
>> Chemical & Biomedical Engineering
>> FAMU-FSU College of Engineering
>> Florida State University
>> sun at magnet.fsu.edu
Peter BLAHA, Inst.f. Materials Chemistry, TU Vienna, A-1060 Vienna
Phone: +43-1-58801-165300 FAX: +43-1-58801-165982
Email: blaha at theochem.tuwien.ac.at WWW:
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