Dummy atom in Gaussian

Recently, I am working on some Gaussian jobs. The error “angle Alpha is outside the valid range of 0 to 180” always makes me upset.

Learning by oneself is really low-efficient and time-consuming. : (

I got some help from internet. Dummy atom could solve this problem. Please read the article first to know dummy atom at http://www.cup.uni-muenchen.de/ch/compchem/geom/internal.html. Please pay attention to the example at the end of the paper.

The origin coordinate is

C1
C2  1  r2
H3  1  r3  2  a3
H4  1  r3  2  a3     3  120.
H5  1  r3  2  a3     3  -120.
N6  2  r6  1  180.0  3  180.0

r2=1.45
r3=1.1
r6=1.2
a3=110.

The coordinate with dummy atom is

C1
C2  1  r2
H3  1  r3  2  a3
H4  1  r3  2  a3  3  120.0
H5  1  r3  2  a3  3  -120.0
X6  2  1.0  1  90.0  3  0.0
N7  2  r7  6  90.0  1  180.0

r2=1.45
r3=1.1
r7=1.2
a3=110.

So the key to set dummy atom in your system is:

  1. Add it before the atom you want to modify. Here is N7;
  2. Divide the problem angle to half. Here is N7-C2-C1. X should have a bond with C2 and an angle with C1. Set the bond length to 1.0 and the angle to 90.0;
  3. Dihedral analysis. The original N7 has dihedral N6-C2-C1-H3. Put X in the plane of C2-C1-H3, so that you do not need to change the dihedral for the new N7. You may tell the old dihedral N6-C2-C1-H3 is the same with N7-C2-X6-C1. So X6-C2-C1-H3 is 0.
  4. Modify the N7 data. N7 still has a bond with C2. N7 form a new angle with X6, which is 90.0. And find a nearby atom, here is C1. Since the old dihedral N6-C2-C1-H3 is the same with N7-C2-X6-C1, you may set N7-C2-X6-C1 as original  N6-C2-C1-H3 with 180.0

Now, you may use it to optimize or scan operations.

 

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