[Wien] spin configuration for charge state
Komal Bapna
komal.bapna at gmail.com
Thu May 12 11:44:48 CEST 2016
Dear Prof. Peter and Laurence,
I am really thankful for your valuable suggestions.
*Prof. Laurence::*
These are the bond-valence sums, I am getting
Atom 1 equiv 1 Sr Bond-Valence Sum 2.40 2.58
Atom 2 equiv 1 Co Bond-Valence Sum 3.03 3.20
Atom 3 equiv 1 O O1 Bond-Valence Sum 2.25 2.39
Atom 4 equiv 1 O O2 Bond-Valence Sum 1.67 1.78
Please tell the two values in each line. Can I now say that these are the
valence state for respective atom. If it is so, why for Co it is not 4!!!!
*Prof. Peter:: *
I tried to stabilize a "high-spin" state for Co by changing the 0.8 on
spin-dn to 0.2, and the 0.59 to 0.9 in spin-up.
For Co up spin
0.89437 0.00000 0.00000 0.00000 -0.04204
0.00000 0.93434 0.00000 0.00000 0.00000
0.00000 0.00000 0.91726 0.00000 0.00000
0.00000 0.00000 0.00000 0.93434 0.00000
-0.04204 0.00000 0.00000 0.00000 0.89437
Total spin moment: 4.57467
For Co dn spin
0.34200 0.00000 0.00000 0.00000 -0.05202
0.00000 0.33947 0.00000 0.00000 0.00000
0.00000 0.00000 0.34654 0.00000 0.00000
0.00000 0.00000 0.00000 0.33947 0.00000
-0.05202 0.00000 0.00000 0.00000 0.34200
Total spin moment: 1.70947
Total energy is 516 meV greater than the previous case.
SPIN MAGNETIC MOMENTS OF MIXED CHARGE DENSITY
INTERSTITIAL = 0.14578
Sr = 0.00800
Co = 2.89880
O1 = 0.23087
O2 = 0.17575
total moment in the CELL = 3.87383
Looking at the moments, can I say Co is in high spin state, as the moment
is still low than its actual theoretical high spin moment (~5 muB).
Thanks and Regards
On Tue, May 10, 2016 at 5:24 PM, Komal Bapna <komal.bapna at gmail.com> wrote:
> Dear Prof. Peter
> Thank you for your kind reply..
>
> The Density matrix for Co up spin is..
> 0.84607 0.00000 0.00000 0.00000 -0.08661
> 0.00000 0.92206 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.59673 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.92206 0.00000
> -0.08661 0.00000 0.00000 0.00000 0.84607
> Total electrons: 4.13298
>
> and Density matrix for Co dn spin is:
> 0.20583 0.00000 0.00000 0.00000 0.06263
> 0.00000 0.80064 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.25923 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.80064 0.00000
> 0.06263 0.00000 0.00000 0.00000 0.20583
> Total electrons: 2.27217
>
> This gives me the net spin moment of 1.86 muB on Co.
>
> I understood by changing the occupancies we can change the spin state.
> Still I am confused whether I can use these occupation numbers of electrons
> in the respective orbitals to find the spin state of Co.Here, total number
> of electrons is 6.4, 1.4 electrons higher than the expected value for Co4+
> ion. I may consider the extra electrons due to the bonding of Co-O bonds.
> But, I am really unable to calculate the spin state from the present
> configuration. Please suggest how to calculate spin state for the given
> configuration, so that I can modify further.
>
> Thanks and Regards
>
> On Mon, May 9, 2016 at 3:59 PM, Komal Bapna <komal.bapna at gmail.com> wrote:
>
>> Sir,
>>
>> Its true that we can not generate "ionic" electron density with lstart/dstart. When I tried to modify .inst file according to the Co4+ ionic state, it showed error.
>>
>> My query is that how can we generate spin state configuration for such an ionic state, it spin states for Co and Co4+ are different and accordingly the magnetic moments.
>>
>>
>> Please suggest.
>>
>>
>> Thanks
>>
>>
>> On Fri, May 6, 2016 at 3:18 PM, Komal Bapna <komal.bapna at gmail.com>
>> wrote:
>>
>>> Dear Wien users
>>>
>>> I am working on Sr2CoO4. Here I wanted to study the system with different spin state configuration of Co4+, which is known to be valence state of Co in this system. I could understand how to create:
>>> (a) High-spin configuration
>>> (b) Intermediate spin configuration
>>> (c) Low-spin configuration
>>> for the given Co atoms in the .inst file as
>>>
>>> Co
>>> Ar 3
>>> 3, 2,2.0 N
>>> 3, 2,2.0 N
>>> 3,-3,3.0 N
>>> 3,-3,0.0 N
>>> 4,-1,1.0 N
>>> 4,-1,1.0 N (for HS state)
>>> and
>>> Co
>>> Ar 3
>>> 3, 2,2.0 N
>>> 3, 2,2.0 N
>>> 3,-3,2.0 N
>>> 3,-3,1.0 N
>>> 4,-1,1.0 N
>>> 4,-1,1.0 N (for IS state)
>>>
>>>
>>> But my query is that .inst file takes Co as neutral atom (9
>>> electrons:3d74s2) and accordingly its spin state. As if Co were in 4+ state, I would have 5 electrons in 3d state (3d54s0) rather 9 electrons as is revealed from .inst file now.
>>>
>>> Please suggest me how to give spin state for Co4+ for this system.
>>>
>>>
>>> Thanks
>>>
>>>
>>>
>>> --
>>> *Komal*
>>>
>>
>>
>>
>> --
>> *Komal*
>>
>
>
>
> --
> *Komal*
>
--
*Komal*
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