From NWChem
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4:35:27 AM PDT  Thu, May 7th 2015 

I'm uncertain how to set up the brokensymmetry input for a molecule with multiple unpaired electrons,
i.e. a Fe2S2(SCH3)4 cluster which has got a high spin configuration on both irons.
The input for the high spin state looks like follows:
# Fe2(III) Fe1(III) high spin (s=5/2 alpha) (s=5/2 beta), S=1
charge 2
dft
xc pbe0
odft
mult 1
convergence nolevelshifting
direct
grid fine
iterations 300
# atoms 112 and 1324 comprise a FeS(SCH3)2 subsystem each
cdft 1 12 spin 5.0
cdft 13 24 spin 5.0
vectors input atomic output "highspin.mos"
end
# the HOMO is orbital #93 for alpha and beta spins.
Do I have to "shift" all of the highest occupied 5 beta MOs up by one orbital and constrain the
charge of this beta subsystem to +1?
Thanks in advance!




Edoapra Forum:Admin, Forum:Mod, bureaucrat, sysop


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11:32:54 AM PDT  Thu, May 7th 2015 

I am not quite sure how to do this with CDFT
However, you can obtain a brokensymmetry solution using the regulat DFT by means of the fragment guess,
by first converging two separate Fecontaining fragment, one fragment with Fe spinup (a.k.a. excess of alpha electrons), the second with Fe spindown (a.k.a. excess of beta electrons).
Here is an example that might be close of what you are looking for.
start fe2s2sch34
#geometry from http://dx.doi.org/10.1038/nchem.2041
geometry big
Fe 5.22 1.05 7.95
S 5.00 0.95 5.66
S 4.77 3.18 8.74
C 6.00 4.34 8.17
H 6.46 4.81 9.01
H 5.53 5.08 7.55
H 6.74 3.82 7.60
C 3.33 1.31 5.18
H 2.71 0.46 5.37
H 3.30 1.54 4.13
H 2.97 2.15 5.73
Fe 5.88 1.05 9.49
S 6.10 0.95 11.79
S 6.33 3.18 8.71
C 5.10 4.34 9.28
H 5.56 5.05 9.93
H 4.67 4.84 8.44
H 4.34 3.81 9.81
C 7.77 1.31 12.27
H 7.84 1.35 13.34
H 8.42 0.54 11.90
H 8.06 2.25 11.86
S 3.86 0.28 9.06
S 7.23 0.28 8.38
end
geometry fe1
Fe 5.22 1.05 7.95
S 5.00 0.95 5.66
S 4.77 3.18 8.74
C 6.00 4.34 8.17
H 6.46 4.81 9.01
H 5.53 5.08 7.55
H 6.74 3.82 7.60
C 3.33 1.31 5.18
H 2.71 0.46 5.37
H 3.30 1.54 4.13
H 2.97 2.15 5.73
end
geometry fe2
Fe 5.88 1.05 9.49
S 6.10 0.95 11.79
S 6.33 3.18 8.71
C 5.10 4.34 9.28
H 5.56 5.05 9.93
H 4.67 4.84 8.44
H 4.34 3.81 9.81
C 7.77 1.31 12.27
H 7.84 1.35 13.34
H 8.42 0.54 11.90
H 8.06 2.25 11.86
end
geometry s2
S 3.86 0.28 9.06
S 7.23 0.28 8.38
end
basis spherical
* library 631g*
end
title "first Fe fragment spin up"
charge 1
set geometry fe1
dft
mulliken
mult 6
xc pbe0
vectors input atomic output fe1.mos
end
task dft ignore
title "second Fe fragment spin down"
charge 1
set geometry fe2
dft
mult 6
vectors input atomic output fe2.mos
end
task dft ignore
title "neutral s2 fragment"
charge 0
set geometry s2
dft
odft
mult 1
vectors input atomic output s2.mos
end
task dft ignore
title " fe2s2sch34"
charge 2
set geometry big
dft
mult 1
vectors input fragment fe1.mos fe2.mos s2.mos output big
maxiter 99
end
task dft

Edited On 10:30:04 AM PDT  Fri, May 8th 2015 by Edoapra




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12:37:55 AM PDT  Fri, May 8th 2015 

Thanks! Seems this kind of [2Fe2S] cluster is a popular problem.
As for the mixed valence state Fe(III)Fe(II) with a total charge of 3:
would the Fe(II) subsystem be defined as charge 0 and mult 5?




Edoapra Forum:Admin, Forum:Mod, bureaucrat, sysop


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10:28:06 AM PDT  Fri, May 8th 2015 

The Fe(II) fragment should have charge 2 and mult 5.
The sum of the charge of the fragments must be equal to the total charge of the aggregate molecule

Edited On 10:28:38 AM PDT  Fri, May 8th 2015 by Edoapra



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