Example: Ligand Field DFT: Co 2+

Download Co_LFDFT.run

Application of the Ligand Field DFT approach for a Co2+ d7 electron configuration. Here Co is surrounded by 6 Water molecules.

First an average of configuration calculation (AOC) is performed, where 7 electrons are equally distributed over the 5 orbitals that have the most dominant Co 3d character. Depending on the electron configuration this might be a non-trivial task. Symmetry NOSYM should be specified.

$ADFBIN/adf  << eor
Atoms
 Co        0.000000    0.000000    0.000000
 O         2.113495    0.000000    0.000000
 H         2.687264   -0.787336    0.000000
 H         2.687264    0.787336    0.000000
 O         0.000000   -2.016255    0.000000
 H         0.000000   -2.587300    0.791160
 H         0.000000   -2.587300   -0.791160
 O        -2.113495    0.000000    0.000000
 H        -2.687264   -0.787336    0.000000
 H        -2.687264    0.787336    0.000000
 O         0.000000    2.016255    0.000000
 H         0.000000    2.587300    0.791160
 H         0.000000    2.587300   -0.791160
 O         0.000000    0.000000    2.124849
 H         0.785128    0.000000    2.701737
 H        -0.785128    0.000000    2.701737
 O         0.000000    0.000000   -2.124849
 H         0.785128    0.000000   -2.701737
 H        -0.785128    0.000000   -2.701737
End

Charge 2

Occupations
  A 56 1.4 1.4 1.4 1.4 1.4
End

Symmetry nosym

Basis
type TZP
core small
End
relativistic scalar zora

XC
gga bp86
End
End input
eor

When the AOC calculation is ready, you need to make sure that indeed the partially occupied orbitals are dominantly d orbitals. In the ADF output you can find the character of the MOs in the list of all MOs, ordered by energy, with the most significant SFO gross populations.

Next two LFDFT calculations are performed, first without spin-orbit coupling (soc 0), next including spin-orbit coupling (soc 1). In this case there is 1 shell, and the nlval for 3d is ‘3 2’. The MO indices should be the fractionally occupied levels of the AOC calculation (in this case 29 30 31 32 33).

$ADFBIN/lfdft <<eor
nshell 1
nlval1 3 2
MOind1 29 30 31 32 33
soc    0
eor

$ADFBIN/lfdft <<eor
nshell 1
nlval1 3 2
MOind1 29 30 31 32 33
soc    1
eor