Symmetry in CCSD(T) frequency calculation

Questions
#1

Hello,

I have run an optimisation and frequency calculation of triazine using CCSD(T) and the cc-pVDZ basis set:

$molecule
0 1
N1
C1 N1 cn
N2 C1 cn N1 ncn
C2 N2 cn C1 cnc N1 dih1
N3 C2 cn N2 ncn C1 dih1
C3 N3 cn C2 cnc N2 dih1
H1 C1 hc N2 hcn C2 dih2
H2 C3 hc N3 hcn C2 dih2
H3 C2 hc N2 hcn C1 dih2

cn      =       1.384568
hc      =       1.080504
ncn     =       120.0000
cnc     =       120.0000
hcn     =       120.0000
dih1    =         0.0000
dih2    =       180.0000
$end

$rem
JOBTYPE OPT
METHOD CCSD(T)
BASIS cc-pVDZ

MEM_TOTAL 256000
MEM_STATIC 2000
CC_BACKEND XM
CC_MEMORY 192000
$end

@@@

$molecule
READ
$end

$rem
JOBTYPE FREQ
METHOD CCSD(T)
BASIS cc-pVDZ
USE_ABELIAN_SUBGROUP TRUE

MEM_TOTAL 256000
MEM_STATIC 2000
CC_BACKEND XM
CC_MEMORY 192000
$end

The point group of the final geometry in the optimisation is correctly assigned to D3h:

        ******************************
        **  OPTIMIZATION CONVERGED  **
        ******************************
 ----------------------------------------------------------------
             Standard Nuclear Orientation (Angstroms)
    I     Atom           X                Y                Z
 ----------------------------------------------------------------
    1      N       1.3928984145     0.0000038461    -0.0000003419
    2      C       0.6505360415     1.1267655209    -0.0000000992
    3      N      -0.6964537783     1.2062832386     0.0000371829
    4      C      -1.3010819569    -0.0000016872     0.0000001787
    5      N      -0.6964477302    -1.2062904582    -0.0000375666
    6      C       0.6505442463    -1.1267701484     0.0000004095
    7      H       1.1999583287     2.0783847903    -0.0000213112
    8      H       1.1999729949    -2.0783940543     0.0000215348
    9      H      -2.3999265605     0.0000189523     0.0000000129
 ----------------------------------------------------------------
 Molecular Point Group                 D3h   NOp =  1
 Largest Abelian Subgroup              C2v   NOp =  1
 Nuclear Repulsion Energy =         209.66858501 hartrees
 There are       21 alpha and       21 beta electrons

However, once it is loaded onto the frequency calculation, the C1 point group is assigned:

--------------------------------------------------------------
User input:
--------------------------------------------------------------

$molecule
READ
$end

$rem
JOBTYPE FREQ
METHOD CCSD(T)
BASIS cc-pVDZ
USE_ABELIAN_SUBGROUP TRUE

MEM_TOTAL 256000
MEM_STATIC 2000
CC_BACKEND XM
CC_MEMORY 192000
$end
--------------------------------------------------------------
 ----------------------------------------------------------------
             Standard Nuclear Orientation (Angstroms)
    I     Atom           X                Y                Z
 ----------------------------------------------------------------
    1      N       1.1596715452     0.7715761046    -0.0000002521
    2      C      -0.0825406748     1.2984561336    -0.0000137513
    3      N      -1.2480402024     0.6185182737     0.0000225198
    4      C      -1.0832301781    -0.7207114352     0.0000001827
    5      N       0.0883675001    -1.3900947548    -0.0000228569
    6      C       1.1657725113    -0.5777467343     0.0000141939
    7      H      -0.1522454572     2.3950806483    -0.0000465316
    8      H       2.1503411117    -1.0656859501     0.0000469227
    9      H      -1.9980975055    -1.3293798476    -0.0000000184
 ----------------------------------------------------------------
 Molecular Point Group                 C1    NOp =  1
 Largest Abelian Subgroup              C1    NOp =  1
 Nuclear Repulsion Energy =         209.66858501 hartrees
 There are       21 alpha and       21 beta electrons

I have run similar calculations using CCSD, and this did not occur; it correctly assigned C2v (Abelian subgroup) as expected. In my previous calculations, the highest three frequencies correspond to an E representation and the totally symmetric CH stretching. However, the corresponding modes from CCSD(T) do not include the totally symmetric stretch and look rather mixed. Also, frequencies that should be degenerate show significant differences compared to previous methods, which did coincide up to at most .05 cm-1.

I was wondering if these discrepancies are due to the method not correctly identifying the symmetry of the molecule, and if so, how could I fix it?

Thanks in advance for any advice!

#2

My version of Q-Chem recognizes that molecule as C1. You can use IQmol to symmetrize, the structure, which I did:

  N    1.3928998   -0.0000000    0.0000000
  C    0.6505402    1.1267687    0.0000000
  N   -0.6964499    1.2062866   -0.0000000
  C   -1.3010804    0.0000000   -0.0000000
  N   -0.6964499   -1.2062866   -0.0000000
  C    0.6505402   -1.1267687    0.0000000
  H    1.1999614    2.0783941   -0.0000000
  H    1.1999614   -2.0783941    0.0000000
  H   -2.3999228    0.0000000   -0.0000000

This geometry is recognized as D3h. The behavior you’re seeing as odd but might be related to different default thresholds for different job types, as there’s a finite tolerance for recognizing point-group symmetry.

1 Like
#3

Then, could an increase in the symmetry tolerance through SYM_TOL provide a workaround? Or is it better to symmetrise the optimised geometry and then run the frequency calculation?

#4

Adjusting SYM_TOL can be a way to get Q-Chem to recognize point-group symmetry, so that you can then optimize the molecule into a properly symmetric geometry. (Symmetrizing with IQmol skips that step but I would still optimize.) In general, I don’t think you should play with SYM_TOL for your final, production calculations; i.e., use it to get to the high-symmetry geometry but then run your production single point energies with the default setting of SYM_TOL.

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