From NWChem
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| 4:35:27 AM PDT - Thu, May 7th 2015 |
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I'm uncertain how to set up the broken-symmetry 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 1-12 and 13-24 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!
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Edoapra Forum:Admin, Forum:Mod, bureaucrat, sysop
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Forum Vet
Threads 3
Posts 844
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| 11:32:54 AM PDT - Thu, May 7th 2015 |
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I am not quite sure how to do this with CDFT
However, you can obtain a broken-symmetry solution using the regulat DFT by means of the fragment guess,
by first converging two separate Fe-containing fragment, one fragment with Fe spin-up (a.k.a. excess of alpha electrons), the second with Fe spin-down (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 6-31g*
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
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Edited On 10:30:04 AM PDT - Fri, May 8th 2015 by Edoapra
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Just Got Here
Threads 1
Posts 3
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| 12:37:55 AM PDT - Fri, May 8th 2015 |
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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?
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Edoapra Forum:Admin, Forum:Mod, bureaucrat, sysop
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Forum Vet
Threads 3
Posts 844
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| 10:28:06 AM PDT - Fri, May 8th 2015 |
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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
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Edited On 10:28:38 AM PDT - Fri, May 8th 2015 by Edoapra
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