I'm working on adsorption of hydrogen on carbon and I want to optimize the entire structure, except for one cartesian coordinate (z of H atom).
I've tried with the input file:
%chk=adsorption.chk
#p HF/sto-3g IOp(5/13=1) opt=ModRedundant
Title
0 2
C 0 0.66323384 -0.61906673 1.33878472
C 0 1.42188029 0.61932069 1.33857966
C 0 1.42220944 -0.61917221 -1.33858523
C 0 0.66368585 0.61918818 -1.33864467
C 0 -0.66323277 0.61906367 -1.33876173
C 0 -1.42188021 -0.61932476 -1.33855567
C 0 -1.42220836 0.61916915 1.33861122
C 0 -0.66368578 -0.61919125 1.33866966
C 0 -0.76945414 1.86712664 0.80414218
C 0 0.76947856 1.86713781 0.80451949
C 0 0.76945422 -1.86712972 -0.80411519
C 0 -0.76947849 -1.86714088 -0.80449350
H 0 xh yh zh
Tv 0 4.17082252 0.00000000 0.00000000
Tv 0 2.08566380 3.73433495 0.00000000
xh 1.42188 A
yh 0.619322 A
zh 2.838567 F
The program works for a time but then throws an error:
Error in old z-matrix.
Error termination via Lnk1e in /usr/local/gaussian-2009-D.01_intel_sse4.2/g09/l9999.exe
Does anybody have any solutions to this problem, or another way to fix only one cartesian coordinate?
Dear Adam,
You can easily fix any coordinate along optimization.For example, consider the methane and assume one of C-H bond distances needs to be fixed. It is worth to note that geometry should be saved as (gjf) format. The initial (input) file can be as follows:
0 1
C
H 1 B1
H 1 B2 2 A1
H 1 B3 3 A2 2 D1 0
H 1 B4 3 A3 2 D2 0
B1 1.07000000
B2 1.07000000
B3 1.07000000
B4 1.07000000
A1 109.47120255
A2 109.47125080
A3 109.47121829
D1 -119.99998525
D2 120.00000060
Now, assume B1 needs to be fixed. You should remove B1 among variables and paste it at the end of variable after a blank line:
0 1
C
H 1 B1
H 1 B2 2 A1
H 1 B3 3 A2 2 D1 0
H 1 B4 3 A3 2 D2 0
B2 1.07000000
B3 1.07000000
B4 1.07000000
A1 109.47120255
A2 109.47125080
A3 109.47121829
D1 -119.99998525
D2 120.00000060
B1 1.07000000
Now you can save and run the calculation. It must be noted that inclusion of z-matrix in the route section is mandatory:
#hf/3-21g opt=z-matrix
In the output file :
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
Berny optimization.
Initialization pass.
----------------------------
! Initial Parameters !
! (Angstroms and Degrees) !
---------------------- ----------------------
! Name Value Derivative information (Atomic Units) !
------------------------------------------------------------------------
! B2 1.07 estimate D2E/DX2 !
! B3 1.07 estimate D2E/DX2 !
! B4 1.07 estimate D2E/DX2 !
! A1 109.4712 estimate D2E/DX2 !
! A2 109.4712 estimate D2E/DX2 !
! A3 109.4712 estimate D2E/DX2 !
! D1 -120.0 estimate D2E/DX2 !
! D2 120.0 estimate D2E/DX2 !
! B1 1.07 Frozen !
------------------------------------------------------------------------
as can bee seen, B1 is frozen during the optimization. I hope you find it helpful.
Sincerely,
Saeed
Why do you not use relative coordinate? I think there is no meaning for your purpose because the whole moiety except H moves freely in this calculation. It means the H atom moves freely compared with other atoms.
Dear Adam,
You can easily fix any coordinate along optimization.For example, consider the methane and assume one of C-H bond distances needs to be fixed. It is worth to note that geometry should be saved as (gjf) format. The initial (input) file can be as follows:
0 1
C
H 1 B1
H 1 B2 2 A1
H 1 B3 3 A2 2 D1 0
H 1 B4 3 A3 2 D2 0
B1 1.07000000
B2 1.07000000
B3 1.07000000
B4 1.07000000
A1 109.47120255
A2 109.47125080
A3 109.47121829
D1 -119.99998525
D2 120.00000060
Now, assume B1 needs to be fixed. You should remove B1 among variables and paste it at the end of variable after a blank line:
0 1
C
H 1 B1
H 1 B2 2 A1
H 1 B3 3 A2 2 D1 0
H 1 B4 3 A3 2 D2 0
B2 1.07000000
B3 1.07000000
B4 1.07000000
A1 109.47120255
A2 109.47125080
A3 109.47121829
D1 -119.99998525
D2 120.00000060
B1 1.07000000
Now you can save and run the calculation. It must be noted that inclusion of z-matrix in the route section is mandatory:
#hf/3-21g opt=z-matrix
In the output file :
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
Berny optimization.
Initialization pass.
----------------------------
! Initial Parameters !
! (Angstroms and Degrees) !
---------------------- ----------------------
! Name Value Derivative information (Atomic Units) !
------------------------------------------------------------------------
! B2 1.07 estimate D2E/DX2 !
! B3 1.07 estimate D2E/DX2 !
! B4 1.07 estimate D2E/DX2 !
! A1 109.4712 estimate D2E/DX2 !
! A2 109.4712 estimate D2E/DX2 !
! A3 109.4712 estimate D2E/DX2 !
! D1 -120.0 estimate D2E/DX2 !
! D2 120.0 estimate D2E/DX2 !
! B1 1.07 Frozen !
------------------------------------------------------------------------
as can bee seen, B1 is frozen during the optimization. I hope you find it helpful.
Sincerely,
Saeed
It is impossible to freeze cartesian coordinates as told by Mauro.
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