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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