IOp(3/110=…)

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Supplementary User’s Guide
for the modified version of GAUSSIAN03 package
(rev. D.01 and D.02)
This version of GAUSSIAN03
has been modified and tested by
The Simons Group
Henry Eyring Center for Theoretical Chemistry
Department of Chemistry, University of Utah
Salt Lake City, UT 84112, U.S.A.
Selecting the centers for modifications
IOp(3/110=…) and IOp(7/110=…)
These IOps are used to tell Gaussian the range of atoms (centers in the geometry
input list in Gaussian .com file) in which there are the nuclei whose charges are to
be modified.
Example #1:
To modify only the charge on center number 1, one needs to specify
IOp(3/110=1) and IOp(7/110=1).
Example #2:
To modify two nuclei whose numbers (in the geometry input specification) are 1
and 2, one needs to specify IOp(3/110=2) and IOp(7/110=2).
Example #3:
To modify four nuclei whose numbers (in the geometry input specification) are
13, 26, 27, and 67, one needs to specify IOp(3/110=67) and IOp(7/110=67).
Briefly, the values of option 110 in overlay 3 and option 110 in overlay 7 have to be
set to the highest number of center that is to be modified, no matter how many
centers will actually be modified. Note that option 110 should have the same
value in both overlays (i.e., IOp(3/110=X1) and IOp(7/110=X2), where X1=X2).
Warning: the largest possible value for this option is 90 in both overlays (see
“Limitations and known problems” section at the end of this manual).
Modifying the nuclei charges on selected centers
IOp(3/111-…-200) and IOp(7/111-…-200)
These IOps are used to both pick the center (for modification) and set the
additional charge that is to be added to it (or subtracted from it).
To modify nuclear charge on center #N the IOptions 3/(110+N) and 7/(110+N)
should be set to a proper value. The additional charge which is to be added to (or
subtracted from) a given nucleus needs to be multiplied by 105 (100000) and
such a value should be set as a value of both IOptions 3/(110+N) and 7/(110+N).
Example #1:
To add the charge of 0.1 a.u. to the center number 1, one needs to specify
IOp(3/111=10000) and IOp(7/111=10000) because:
 the center number is 1 which corresponds to options 111 (110+1) in
overlays 3 and 7 , and
 the charge to be added is 0.1 a.u. which corresponds to values of 10000 for
options 111 in overlays 3 and 7 (because 0.1×105 = 10000).
Example #2:
To modify two nuclei whose numbers in the geometry input specification are 4
and 17, by adding an extra positive charge of 0.03 a.u. to nucleus on center 4
and charge of 0.58 a.u. to the nucleus on center 17, one needs to specify
IOp(3/114=3000,3/127=58000) IOp(7/114=3000,7/127=58000)
because
 the number of the first center (that is to be modified) is 4 which
corresponds to options 114 (110+4) in overlays 3 and 7 , and
 the number of the second center (that is to be modified) is 17 which
corresponds to options 127 (110+17) in overlays 3 and 7 , and
 the charge to be added to the first modified center is 0.03 a.u. which
corresponds to values of 3000 for options 114 in overlays 3 and 7 (because
0.03×105 = 3000) for this center, and
 the charge to be added to the second modified center is 0.58 a.u. which
corresponds to values of 58000 for options 127 in overlays 3 and 7
(because 0.58×105 = 58000) for this center.
Recall, that even though you are choosing the center numbers here, you still need
to set the proper range by using IOp(3/110=…) and IOp(7/110=…)
(see the preceding section).
Input examples
Here we present a few examples of Gaussian03 input files showing the single
point energy calculations and geometry optimization jobs that involve modified
nuclear charges.
Example #1:
%chk=testSP
#p RHF 6-31G(d) SCF=(Tight,NoVarAcc,Maxcycle=512,IntRep)
GFInput IOp(6/7=3) IOp(3/32=1)
IOp(3/110=3,3/111=33000,3/112=34000,3/113=33000)
IOp(7/110=3,7/111=33000,7/112=34000,7/113=33000)
Test job #1
0
1
O
C
N
H
H
H
1
2
2
3
3
co2
nc3
nco3
hc4
hco4
dih4
hn5
hnc5
dih5
hn6
hnc6
dih6
co2
nc3
hc4
hn5
hn6
1
1
2
2
nco3
hco4
hnc5
hnc6
3 dih4
1 dih5
1 dih6
1.277811
1.496812
116.154
1.002000
135.119
172.107
1.004634
107.935
-119.695
1.002000
112.357
0.000
In this job the first three atoms are modified: charges of 0.33, 0.34, and 0.33 a.u.
are added to nuclei of the oxygen atom, carbon atom, and nitrogen atom,
respectively. So the resulting charges for atoms O, C, N are 8.33, 6.34, and 7.33
a.u., respectively, while those for hydrogen atoms remain unchanged. This is
achieved in the 4th and 5th line of the input file. The Hartree-Fock single point
energy for such a system will be calculated, with the 6-31G(d) basis sets, for the
given geometry.
Example #2:
%chk=testOPT
#p UHF 6-31G(d) SCF=(Tight,NoVarAcc,Maxcycle=512,IntRep)
GFInput IOp(6/7=3) IOp(3/32=1) OPT GUESS=READ
IOp(3/110=3,3/111=33000,3/112=34000,3/113=33000)
IOp(7/110=3,7/111=33000,7/112=34000,7/113=33000)
Test job #2
-1
O
C
N
H
H
H
2
co2
nc3
nco3
hc4
hco4
dih4
hn5
hnc5
dih5
hn6
hnc6
dih6
1
2
2
3
3
co2
nc3
hc4
hn5
hn6
1
1
2
2
nco3
hco4
hnc5
hnc6
3 dih4
1 dih5
1 dih6
1.277811
1.496812
116.154
1.002000
135.119
172.107
1.004634
107.935
-119.695
1.002000
112.357
0.000
In this job the first three atoms are modified: charges of 0.33, 0.34, and 0.33 a.u.
are added to nuclei of the oxygen atom, carbon atom, and nitrogen atom,
respectively. So the resulting charges for atoms O, C, N are 8.33, 6.34, and 7.33
a.u., respectively, while those for hydrogen atoms remain unchanged. This is
achieved in the 4th and 5th line of the input file. The Hartree-Fock geometry
optimization for such an anion will be performed, with the 6-31G(d) basis sets.
Example #3:
%chk=testOPT2
#p UHF aug-cc-pVDZ SCF=(Tight,NoVarAcc,Maxcycle=512)
GFInput IOp(6/7=3) IOp(3/32=1) OPT GUESS=READ
IOp(3/110=5,3/111=33000,3/112=34000,3/115=33000)
IOp(7/110=5,7/111=33000,7/112=34000,7/115=33000)
Test job #3
-1 2
O 0.000000
C 0.000000
H -0.700350
H 1.412538
N 1.343557
H 2.094138
0.000000
0.000000
0.097094
-0.830293
0.000000
0.000000
0.000000
1.277811
1.987802
2.498962
1.937584
1.273783
In this job there are also three modified atoms: charges of 0.33, 0.34, and 0.33
a.u. are added to nuclei of the oxygen atom, carbon atom, and nitrogen atom,
respectively. So the resulting charges for atoms O, C, N are 8.33, 6.34, and 7.33
a.u., respectively, while those for hydrogen atoms remain unchanged. This is
achieved again in the 4th and 5th line of the input file. The Hartree-Fock geometry
optimization for such an anion will be performed, with the 6-31G(d) basis sets.
(Please note how the centers for modifications are specified via IOps).
Limitations and known problems
The maximum number of modified centers in one job is limited to 90
which is related to the array size (used for storing IOps).
The maximum number of atoms in a molecule is not limited and it is the
subject to the same limitations as in any regular Gaussian03 input files.
If the number of atoms in a molecule is larger than 90 then the highest
possible number for a modified center is 90. This means that centers
numbered with 91 and larger numbers can be present but cannot be
modified.
The modified version of Gaussian03 has been tested for single point
energy and geometry optimization jobs. Calculating harmonic vibrational
frequencies is also possible although one should keep in mind that
changing the nuclear charges does not affect the atomic masses in any case
so the molecular quantities that depend on atomic masses (such as
vibrational frequencies) are calculated in a standard G03 manner.
The value of nuclear repulsion energy is calculated with the use of
modified nuclear charges.
Some specific modifications of the nuclear charges showed to produce
unexpected errors during geometry optimization jobs. In particular,
changing any nuclear charge by 1 a.u. (or more) might lead to such
behavior since, de facto, it changes the element used. In such cases the
errors reported by MinBas appear (due to the comparison made by
Gaussian03 in which the atomic number is compared with its nuclear
charge). Such an error, if happens, can be overcome by running a
preliminary single point energy job and using its checkpoint file in a
geometry optimization run (together with specifying both guess=read
and IOp(4/5=1)).
If a version where the modified charges affect nuclear repulsion energy is
used one can expect the unusually long (or short) chemical bonds after
geometry optimization.
In some cases both the SCF procedure and the geometry optimization
process can be more difficult to converge it is recommended to use
MaxCycle keyword to increase the number of maximum SCF iterations
and optimization steps to assure successful convergence achievement.
The molecular properties, such as population analysis (e.g., Mulliken
population analysis, ESP-fitted charges according to the Merz-SinghKollman scheme, etc.), multipole moments, polarizability tensor, and
other quantities that depend on the nuclear charges will be evaluated with
the modified nuclear charges introduced by the user.
Please report any problems to:
Jack Simons (simons@chemistry.utah.edu) or
Piotr Skurski (piotr@hydrogen.hec.utah.edu).
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