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.ba...@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.ba...@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.ba...@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*
_______________________________________________
Wien mailing list
Wien@zeus.theochem.tuwien.ac.at
http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien
SEARCH the MAILING-LIST at:  
http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html

Reply via email to