S Ling lingsa... at gmail.com
Fri Jul 3 23:24:57 UTC 2015


For V(II), you can try the same values which I gave you. If it doesn't
work, then please try to reduce +7/-7 to something like +5/-5. The general
idea is to the break the symmetry by adding or substracting electrons from
the ALPHA and BETA spin channels, and by adding/substracting differnet
number of electrons from different atoms, hopefully you can get different
oxidation states on different atoms after the SCF calculation. The total
number of electrons should be corrected after the rescaling, but in certain
cases, you might end up with more than 5 d electrons in the ALPHA or BETA
spin channel even after the scaling, and therefore you see the problems you
mentioned earlier. I am not very clear about how the rescaling was done.
Maybe some other people on this mailing list can give you more
information. I would suggest you to play with some small systems to get
more feeling about it. Keep in mind that the &BS subsection just break the
symmetry of your initial guess. You may or may not get the exact magnetic
ordering or oxidation states which you expect after the SCF calculation,
and therefore, you should always check the spin moments on different atoms
after the SCF calculation.


On 3 July 2015 at 13:34, Rizwan Nabi <rizwan... at gmail.com> wrote:

> Hi S Ling
> Thank you for your valuable inputs
> The new NEL values you gave me are working fine, but I just have one query
> about V(II) STATE  do I have to use same values of NEL you have given me
> for V(II) or is there any change required. And would you please give a
> general idea of how to fix these values for any metal such that I may use
> that idea for future work on other metals with different multiplicity.
> Here I am mentioning your comment.
> The electronic configuration of V2+ is [Ar]3d34s0. Therefore, instead of
> +7 and -7, you should set NEL of the 3d orbital to +3 and -3 for ALPHA and
> BETA spin channels, respectively. "+3" is performed on the electronic
> configuration of atomic V which is [Ar]3d34s2, and you end up with
> (3+3)/2=3 for the ALPHA spin channel. Similarly, for the BETA spin channel,
> you have (3-3)/2=0.
> Your help is cordially appreciated!
> Regards
> Rizwan
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