Example: NOCV: CH2 – Cr(CO)5¶
Example for calculation of ETS-NOCV for spin-restricted fragments. ETS-NOCV: energy analysis using the Natural Orbitals for Chemical Valence. The CH2 molecule and Cr(CO)5 are the fragments, which form Cr(CO)5 CH2 molecule.
First the two fragments are calculated.
$ADFBIN/adf << eor
Title CrCO5--[CH2] run from CrCO5 and CH2 closed shell fragments,FULL electron calc.!
atoms cartesian
C -0.429104 1.732058 -0.225052
H 0.407023 2.440417 -0.352323
H -1.385325 2.281354 -0.254124
end
basis
Type DZP
Core None
end
symmetry NOSYM
xc
gga becke perdew
end
endinput
eor
mv TAPE21 t21.CH2
$ADFBIN/adf << eor
Title [CrCO5] run
atoms cartesian
Cr -0.248053 -0.169062 0.005810
C -0.072963 -2.080685 0.229583
O 0.030811 -3.223220 0.361925
C -0.182894 0.049840 1.909128
O -0.142780 0.212309 3.050403
C -0.299940 -0.409118 -1.894730
O -0.331795 -0.521589 -3.042336
C -2.138631 -0.242152 0.075713
O -3.295036 -0.249916 0.115045
C 1.624487 0.092244 -0.083118
O 2.763411 0.288575 -0.140976
end
basis
Type DZP
Core None
Cr $ADFRESOURCES/TZP/Cr
end
symmetry NOSYM
xc
gga becke perdew
end
endinput
eor
mv TAPE21 t21.Crfragment
Next these fragments are used in the calculation of the full complex. The keys ETSNOCV and ‘PRINT etslowdin’ are needed in this case to to analyze the bonding in the molecule with respect to the fragments. The symmetry must be NOSYM.
$ADFBIN/adf << eor
Title:CrCO5--[CH2], etsnocv activated by etsnocv and print etslowdin
atoms
C -0.429104 1.732058 -0.225052 f=f1
Cr -0.248053 -0.169062 0.005810 f=f2
C -0.072963 -2.080685 0.229583 f=f2
O 0.030811 -3.223220 0.361925 f=f2
C -0.182894 0.049840 1.909128 f=f2
O -0.142780 0.212309 3.050403 f=f2
C -0.299940 -0.409118 -1.894730 f=f2
O -0.331795 -0.521589 -3.042336 f=f2
C -2.138631 -0.242152 0.075713 f=f2
O -3.295036 -0.249916 0.115045 f=f2
C 1.624487 0.092244 -0.083118 f=f2
O 2.763411 0.288575 -0.140976 f=f2
H 0.407023 2.440417 -0.352323 f=f1
H -1.385325 2.281354 -0.254124 f=f1
end
fragments
f1 t21.CH2
f2 t21.Crfragment
end
symmetry NOSYM
xc
gga becke perdew
end
ETSNOCV RHOKMIN=1.e-3 EKMIN=1.5 ENOCV=0.05
print etslowdin
endinput
eor