Why does NRT not respect symmetry?

Why does NRT not respect symmetry?

Postby martin » Wed Jan 31, 2018 9:44 am

I have calculated BF3 in D3h symmetry and therefore expected a mixture of the following structures:
    - X% (maybe below threshold) all single bonds, empty p-orbital a boron
    - (1/3)(100 - X - rem)% one double bond between one of the ligands; three times because of symmetry

Instead I get one structure with double bond, weight 80%, the ones that should be symmetry equivalent to that have 7%.
What is the reason for this discrepancy? In other words, why don't I get symmetry equivalent resonance structures?
Or am I doing something wrong here?

Input:
Code: Select all
%chk=borontrifluoride.df-b97d3tzvpp.chk
#P B97D3/def2tzvpp/W06 DenFit  scf(xqc,MaxConventionalCycle=500)
int(ultrafinegrid) gfinput gfoldprint iop(6/7=3) symmetry(loose)
pop(nbo6read)

 Borontrifluoride DF-B97D3/def2-SVP

0   1
B          0.00000        0.00000        0.00000
F          0.00000        1.32333        0.00000
F          1.14604       -0.66166        0.00000
F         -1.14604       -0.66166        0.00000

$NBO
  archive file=borontrifluoride.df-b97d3tzvpp
  NRT
$END



(Partial) Output
Code: Select all
                    Distance matrix (angstroms):
                    1          2          3          4
     1  B    0.000000
     2  F    1.323330   0.000000
     3  F    1.323330   2.292072   0.000000
     4  F    1.323330   2.292072   2.292080   0.000000
 Stoichiometry    BF3
 Framework group  D3H[O(B),3C2(F)]
 Deg. of freedom     1
 Full point group                 D3H     NOp  12
 Largest Abelian subgroup         C2V     NOp   4
 Largest concise Abelian subgroup C2      NOp   2
                         Standard orientation:
 ---------------------------------------------------------------------
 Center     Atomic      Atomic             Coordinates (Angstroms)
 Number     Number       Type             X           Y           Z
 ---------------------------------------------------------------------
      1          5           0        0.000000   -0.000000    0.000000
      2          9           0       -0.000000    1.323327    0.000000
      3          9           0        1.146035   -0.661664    0.000000
      4          9           0       -1.146035   -0.661664    0.000000
 ---------------------------------------------------------------------


NBO (partial):
Code: Select all
 Summary of Natural Population Analysis:

                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    B  1    1.42580      1.99998     1.52700    0.04722     3.57420
    F  2   -0.47527      2.00000     7.46411    0.01116     9.47527
    F  3   -0.47527      2.00000     7.46411    0.01116     9.47527
    F  4   -0.47527      2.00000     7.46411    0.01116     9.47527
 ====================================================================
 * Total *  0.00000      7.99997    23.91934    0.08069    32.00000

                                 Natural Population
 ---------------------------------------------------------
   Core                       7.99997 ( 99.9996% of    8)
   Valence                   23.91934 ( 99.6639% of   24)
   Natural Minimal Basis     31.91931 ( 99.7478% of   32)
   Natural Rydberg Basis      0.08069 (  0.2522% of   32)
 ---------------------------------------------------------

    Atom No         Natural Electron Configuration
 ----------------------------------------------------------------------------
      B  1      [core]2s( 0.39)2p( 1.14)3p( 0.03)3d( 0.01)
      F  2      [core]2s( 1.86)2p( 5.60)3d( 0.01)
      F  3      [core]2s( 1.86)2p( 5.60)3d( 0.01)
      F  4      [core]2s( 1.86)2p( 5.60)3d( 0.01)

[...]

 3-Center, 4-Electron A:-B-:C Hyperbonds (A-B :C <=> A: B-C)
              [threshold for detection: 33.3%]

                                                  NBOs       3-center hybrids
                                              -------------  ----------------
       Hyperbond A:-B-:C   %A-B/%B-C   occ    BD(A-B) LP(C)  h(A)  h(B)  h(C)
      -------------------  ---------  ------  ------- -----  ----  ----  ----
   1.  F  4:- B  1-: F  2  50.7/49.3  4.1070      15      7    17    18     7
   2.  F  4:- B  1-: F  3  50.7/49.3  4.1070      15     10    17    18    10

[...]

 NATURAL RESONANCE THEORY ANALYSIS:

 Maximum reference structures : 20
 Maximum resonance structures : 300
 Memory requirements : 1390490 words of 99941692 available

 4 candidate reference structure(s) calculated by SR LEWIS
 Additional candidate reference structure taken from NBO search
 0 candidate reference structure(s) added by SR HBRES
 Initial loops searched 24 bonding pattern(s); 5 were retained
 Delocalization list threshold set to 1.00 kcal/mol for reference 1
 Delocalization list threshold set to 3.55 kcal/mol for reference 2
 Delocalization list threshold set to 3.55 kcal/mol for reference 3
 Delocalization list threshold set to 3.55 kcal/mol for reference 4
 Delocalization list threshold set to 1.00 kcal/mol for reference 5
 Reference   1:  rho*=0.56010, f(w)=0.92671 converged after   7 iterations
 Reference   2:  rho*=0.89854, f(w)=0.95653 converged after  23 iterations
 Reference   3:  rho*=0.89854, f(w)=0.95653 converged after  23 iterations
 Reference   4:  rho*=0.89854, f(w)=0.95653 converged after  23 iterations
 Reference   5:  rho*=0.44085, f(w)=0.89443 converged after   6 iterations
 Multi-ref( 5):  D(W)=0.07896, F(W)=0.00109 converged after 242 iterations
 4 reference structures have low weight (<35.0% of 74.6%); discarded

                                                fractional accuracy f(w)
                  non-Lewis             -------------------------------------
  Ref     Wgt      density      d(0)      all NBOs     val+core     valence
 ----------------------------------------------------------------------------
   5    1.00000    0.44085    0.02815     0.89443      0.92833      0.92833


 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4
     ---- --- --- --- ---
   1.  B   0   1   1   2
   2.  F   1   3   0   0
   3.  F   1   0   3   0
   4.  F   2   0   0   2

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*     79.90
    2 (2)   7.34    B  1- F  2, ( B  1- F  4), ( F  2),  F  4
    3 (2)   7.34    B  1- F  3, ( B  1- F  4), ( F  3),  F  4
    4       1.35    B  1- F  2, ( B  1- F  3), ( F  2),  F  3
    5       1.35   ( B  1- F  2),  B  1- F  3,  F  2, ( F  3)
    6       1.35   ( B  1- F  2),  B  1- F  4,  F  2, ( F  4)
    7       1.35   ( B  1- F  3),  B  1- F  4,  F  3, ( F  4)
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]

 Natural Bond Order:  (total/covalent/ionic)

     Atom       1      2      3      4
     ----   ------ ------ ------ ------
   1.  B  t 0.0000 1.0598 1.0598 1.8803
          c   ---  0.3439 0.3439 0.4579
          i   ---  0.7159 0.7159 1.4225

   2.  F  t 1.0598 2.9402 0.0000 0.0000
          c 0.3439   ---  0.0000 0.0000
          i 0.7159   ---  0.0000 0.0000

   3.  F  t 1.0598 0.0000 2.9402 0.0000
          c 0.3439 0.0000   ---  0.0000
          i 0.7159 0.0000   ---  0.0000

   4.  F  t 1.8803 0.0000 0.0000 2.1197
          c 0.4579 0.0000 0.0000   ---
          i 1.4225 0.0000 0.0000   ---


 Natural Atomic Valencies:

                      Co-    Electro-
     Atom  Valency  Valency  Valency
     ----  -------  -------  -------
   1.  B    4.0000   1.1457   2.8543
   2.  F    1.0598   0.3439   0.7159
   3.  F    1.0598   0.3439   0.7159
   4.  F    1.8803   0.4579   1.4225


 $NRTSTR
   STR        ! Wgt = 79.90%
     LONE 2 3 3 3 4 2 END
     BOND S 1 2 S 1 3 D 1 4 END
   END
 $END


Tested with G16.a03; G09.e01 and some newer version of NBO. I have no idea where to find this information, it's not in the output (or it is cryptically hidden).
martin
 
Posts: 5
Joined: Thu Mar 24, 2016 3:22 am

Re: Why does NRT not respect symmetry?

Postby ericg » Wed Jan 31, 2018 10:57 am

Martin,

You're certainly not doing anything wrong here. The problem is that NRT starts with an asymmetrical guess for important resonance structures (this occurs, at times, for three-fold symmetric species) and never recovers the symmetry.

You can obtain a symmetric solution as follows...

Your calculation yielded just one important resonance form (one of the doubly-bonded structures) that is reported in the $NRTSTR list at the end of the NRT output.

Code: Select all
 $NRTSTR
   STR        ! Wgt = 79.90%
     LONE 2 3 3 3 4 2 END
     BOND S 1 2 S 1 3 D 1 4 END
   END
 $END

I added the two other doubly-bonded structures and the singly-bonded one to $NRTSTR,

Code: Select all
 $NRTSTR
   STR
     LONE 2 3 3 3 4 2 END
     BOND S 1 2 S 1 3 D 1 4 END
   END
   STR
     LONE 2 3 3 2 4 3 END
     BOND S 1 2 D 1 3 S 1 4 END
   END
   STR
     LONE 2 2 3 3 4 3 END
     BOND D 1 2 S 1 3 S 1 4 END
   END
   STR
     LONE 2 3 3 3 4 3 END
     BOND S 1 2 S 1 3 S 1 4 END
   END
 $END

inserted this list into your input file immediately after the $NBO keylist, and reran the calculation. NRT now gives

Code: Select all
 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4
     ---- --- --- --- ---
   1.  B   0   1   1   2
   2.  F   1   3   0   0
   3.  F   1   0   3   0
   4.  F   2   0   0   2

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*(2)  31.53
    2*(2)  31.53    B  1- F  2, ( B  1- F  4), ( F  2),  F  4
    3*(2)  31.53    B  1- F  3, ( B  1- F  4), ( F  3),  F  4
    4       0.90    B  1- F  2, ( B  1- F  3), ( F  2),  F  3
    5       0.90   ( B  1- F  2),  B  1- F  3,  F  2, ( F  3)
    6       0.90    B  1- F  2,  B  1- F  3, ( B  1- F  4), ( B  1- F  4),
                   ( F  2), ( F  3),  F  4,  F  4
    7       0.45   ( B  1- F  2),  B  1- F  4,  F  2, ( F  4)
    8       0.45   ( B  1- F  3),  B  1- F  4,  F  3, ( F  4)
    9       0.45   ( B  1- F  2),  B  1- F  3,  B  1- F  3, ( B  1- F  4),
                    F  2, ( F  3), ( F  3),  F  4
   10       0.45    B  1- F  3,  B  1- F  3, ( B  1- F  4), ( B  1- F  4),
                   ( F  3), ( F  3),  F  4,  F  4
   11       0.45    B  1- F  2,  B  1- F  2, ( B  1- F  3), ( B  1- F  4),
                   ( F  2), ( F  2),  F  3,  F  4
   12       0.45    B  1- F  2,  B  1- F  2, ( B  1- F  4), ( B  1- F  4),
                   ( F  2), ( F  2),  F  4,  F  4
   13-19    0.00
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]

and symmetric bond orders and valencies.

Eric
ericg
 
Posts: 296
Joined: Sat Dec 29, 2012 9:31 am

Re: Why does NRT not respect symmetry?

Postby martin » Thu Feb 01, 2018 1:07 am

Eric,

thank you very much for your reply and explanation. Your support is much appreciated.
I have tried that before, but now realise, that I had the input for NTRSTR wrong.
I'll keep this in mind for future endeavours.

Best, Martin
martin
 
Posts: 5
Joined: Thu Mar 24, 2016 3:22 am


Return to General NBO Discussion

Who is online

Users browsing this forum: No registered users and 2 guests

cron