(26 Nov 96)

Section 3 - Input Examples

The distribution of GAMESS contains a number of short examples, named EXAM*.INP. You should run all of these tests to be sure you have installed GAMESS correctly. The correct answers are shown in the comments preceeding each of the short input tests. The "correct" answers were obtained on a RS/6000 computer, other machines may differ in the last energy digit, or the last couple of gradient digits.

The examples are listed in the rest of this section, and serve as a useful tutorial about GAMESS input.

ExampleDescription
1CH2 RHF geometry optimization
2CH2 UHF + gradient
3CH2 ROHF + gradient
4CH2 GVB + gradient
5CH2 RHF + CI gradient
6CH2 MCSCF geometry optimization
7HPO RHF + gradient
8H2O RHF + MP2 gradient
9H2O MCSCF + MCQDPT energy correction
10H2O RHF + hessian
11HCN RHF IRC
12HCCH RHF geometry optimization
13H2O RHF properties
14H2O CI transition moment
15C2- GVB/ROHF on 2-pi-u state
16Si GVB/ROHF on 3-P state
17CH2 GVB/ROHF + hessian
18P2 RHF with effective core potential
19CH2 spin-orbit coupling
20I- exponent TRUDGE optimization
21CH3 OS-TCSCF hessian
22H3CN UHF + UMP2 energy
23SiH3- PM3 geometry optimization
24H2O SCRF test case
25??? internal coordinate example
26H3PO localized orbital test
27NH3 DRC example
28H2O-NH3 Morokuma decomposition
29FNH2OH surface scan
30HCONH2(H2O)3 effective fragment solvation

EXAM01.

1-A-1 CH2 RHF geometry optimization using GAMESS.

Although internal coordinates are used (COORD=ZMAT), the optimization is done in Cartesian space (NZVAR=0). This run uses a criterion (OPTTOL) on the gradient which is tighter than is normally used.

This job tests the sp integral module, the RHF module, and the geometry optimization module.

Using the default search METHOD=STANDARD,
FINAL E= -37.2322678015, 8 iters, RMS grad= .0264308
FINAL E= -37.2308175316, 7 iters, RMS grad= .0320881
FINAL E= -37.2375723414, 7 iters, RMS grad= .0056557
FINAL E= -37.2379944431, 6 iters, RMS grad= .0017901
FINAL E= -37.2380387832, 8 iters, RMS grad= .0003391
FINAL E= -37.2380397692, 6 iters, RMS grad= .0000030

           $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE COORD=ZMT NZVAR=0 $END
           $SYSTEM TIMLIM=2 MEMORY=100000 $END
           $STATPT OPTTOL=1.0E-5  $END
           $BASIS  GBASIS=STO NGAUSS=2 $END
           $GUESS  GUESS=HUCKEL $END
           $DATA
          Methylene...1-A-1 state...RHF/STO-2G
          Cnv  2
          
          C
          H  1 rCH
          H  1 rCH  2 aHCH
          
          rCH=1.09
          aHCH=110.0
           $END
 

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EXAM02.

3-B-1 CH2 UHF calculation on methylene ground state.

This test uses the default choice, COORD=UNIQUE, to enter the molecule. Only the symmetry unique atoms are given, and they must be given in the orientation which GAMESS expects.

This job tests the UHF energy and the UHF gradient. In addition, the orbitals are localized.

The initial energy is -37.228465066.
The FINAL energy is -37.2810867258 after 11 iterations.
The unrestricted wavefunction has = 2.013.
Mulliken, Lowdin charges on C are -0.020584, 0.018720.
The spin density at Hydrogen is -0.0167104.
The dipole moment is 0.016188.
The RMS gradient is 0.027589766.
FINAL localization sums are 30.57 and 25.14 Debye**2.

           $CONTRL SCFTYP=UHF MULT=3 RUNTYP=GRADIENT LOCAL=BOYS $END
           $SYSTEM TIMLIM=1 MEMORY=100000 $END
           $BASIS  GBASIS=STO NGAUSS=2 $END
           $GUESS  GUESS=HUCKEL $END
           $DATA
          Methylene...3-B-1 state...UHF/STO-2G
          Cnv  2
          
          Carbon     6.0
          Hydrogen   1.0    0.0      0.82884      0.7079
           $END
 

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EXAM03.

3-B-1 CH2 ROHF calculation on methylene ground state.

The wavefunction is a pure triplet state ( = 2), and so has a higher energy than the second example.

For COORD=CART, all atoms must be given, and as in the present case, these may be in an unoriented geometry. GAMESS deduces which atoms are unique, and orients the molecule appropriately. The geometry here is thus identical to the second example.

This job tests the ROHF wavefunction and gradient code. It also tests the direct SCF procedure.

The initial energy is -37.228465066.
The FINAL energy is -37.2778767089 after 7 iterations.
Mulliken, Lowdin charges on C are -0.020346, 0.019470.
The Hydrogen atom spin density is 0.0129735.
The dipole moment is 0.025099 Debye.
The RMS gradient is 0.027505548

           $CONTRL SCFTYP=ROHF MULT=3 RUNTYP=GRADIENT COORD=CART $END
           $SYSTEM TIMLIM=1 MEMORY=100000 $END
           $SCF    DIRSCF=.TRUE. $END
           $BASIS  GBASIS=STO NGAUSS=2 $END
           $GUESS  GUESS=HUCKEL $END
           $DATA
          Methylene...3-B-1 state...ROHF/STO-2G
          Cnv  2
          
          Hydrogen   1.0    0.82884     0.7079   0.0
          Carbon     6.0
          Hydrogen   1.0   -0.82884     0.7079   0.0
           $END
 

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EXAM04.

1-A-1 CH2 TCSCF calculation on methylene.

The wavefunction has two configurations, exciting the carbon sigma lone pair into the out of plane p.

Note that the Z-matrix used to input the molecule can include identifying integers after the element symbol, and that the connectivity can then be given using these labels rather than integers.

This job tests the GVB wavefunction and gradient.

The initial GVB-PP(1) energy is -37.187342653.
The FINAL energy is -37.2562020559 after 10 iters.
The GVB CI coefs are 0.977505 and -0.210911, giving a pair overlap of 0.64506.
Mulliken, Lowdin charges for C are 0.020810, 0.055203.
The dipole moment is 1.249835.
The RMS gradient = 0.019618475.

           $CONTRL SCFTYP=GVB  RUNTYP=GRADIENT  COORD=ZMT  $END
           $SYSTEM TIMLIM=1 MEMORY=100000 $END
           $BASIS  GBASIS=STO NGAUSS=2 $END
           $SCF    NCO=3  NSETO=0  NPAIR=1  $END
           $DATA
          Methylene...1-A-1 state...GVB...one geminal pair...STO-2G
          Cnv  2
          
          C1
          H1  C1 rCH
          H2  C1 rCH  H1 aHCH
          
          rCH=1.09
          aHCH=99.0
           $END
normally a GVB-PP calculation will use GUESS=MOREAD
           $GUESS  GUESS=HUCKEL  $END
 

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EXAM05

CH2 CI calculation.

The wavefunction is RHF + CI-SD, within the minimal basis, containing 55 configurations. Two CI roots are found, and the gradient of the higher state is then computed.

Note that CI gradients have several restrictions, which are further described in the $LAGRAN group.

FINAL energy of RHF = -38.3704885128 after 10 iters.
State 1 EIGENvalue = -37.4270674136, c(1) = 0.970224
State 2 EIGENvalue = -37.3130036824, c(29) = 0.990865
The upper state dipole moment is 0.708275 Debye.
The upper state has RMS gradient 0.032264079

           $CONTRL SCFTYP=RHF CITYP=GUGA RUNTYP=GRADIENT $END
           $SYSTEM TIMLIM=3 MEMORY=300000 $END
           $BASIS  GBASIS=STO NGAUSS=3 $END
           $GUESS  GUESS=HUCKEL $END
look at all state symmetries, by using C1 symmetry
           $CIDRT  GROUP=C1 IEXCIT=2 NFZC=1 NDOC=3 NVAL=3 $END
ground state is 1-A-1, 1st excited state is 1-B-1
           $GUGDIA NSTATE=2 $END
compute properties of the 1-B-1 state
           $GUGDM  NFLGDM(1)=1,1 IROOT=2 $END
compute gradient of the 1-B-1 state
           $GUGDM2 WSTATE(1)=0.0,1.0 $END
           $DATA
          Methylene...CI...STO-3G basis
          Cnv   2
          
          Carbon    6.0
          Hydrogen  1.0      0.0       0.82884        0.7079
           $END

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EXAM06.

1-A-1 CH2 MCSCF methylene geometry optimization.

The two configuration ansatz is the same as used in the fourth example.

The optimization is done in internal coordinates, as NZVAR is non-zero. Since a explicit $ZMAT is given, these are used for the internal coordinates, rather than those used to enter the molecule in the $DATA. (Careful examination of this trivial triatomic's input shows that $ZMAT is equivalent to $DATA in this case. You would normally give $ZMAT only if it is somehow different.)

This job tests the MCSCF wavefunction and gradient.

At the initial geometry:
The initial energy is -37.187342653,
the FINAL E= -37.2562020559 after 14 iterations,
the RMS gradient is 0.0256396.

After 4 steps,
FINAL E= -37.2581791686, RMS gradient=0.0000013,
r(CH)=1.1243359, ang(HCH)=98.8171674

           $CONTRL SCFTYP=MCSCF RUNTYP=OPTIMIZE NZVAR=3 COORD=ZMT $END
           $SYSTEM TIMLIM=5 MEMORY=100000 $END
           $BASIS  GBASIS=STO NGAUSS=2 $END
           $DATA
          Methylene...1-A-1 state...MCSCF/STO-2G
          Cnv  2
          
          C
          H 1 rCH
          H 1 rCH 2 aHOH
          
          rCH=1.09
          aHOH=99.0
           $END
           $ZMAT   IZMAT(1)=1,1,2,   1,1,3,   2,2,1,3  $END

Normally one starts a MCSCF run with converged SCF orbitals, as Huckel orbitals normally do not converge. Even if they do converge, the extra iterations are very expensive, so use MOREAD for your runs!

           $GUESS  GUESS=HUCKEL $END

two active electrons in two active orbitals.

           $DRT    GROUP=C2V FORS=.TRUE. NMCC=3 NDOC=1 NVAL=1 $END
 

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EXAM07.

1-A' HPO RHF calculation using GAMESS.

This job tests the HONDO integral and gradient package, due to the d function on phosphorus. The input also illustrates the use of a more flexible basis set than the methylene examples.
Although HUCKEL would be better, HCORE is tested.

The initial energy is -397.591192627,
the FINAL E= -414.0945320854 after 18 iterations,
The dipole moment is 2.535169.
The RMS gradient is 0.023723942.

           $CONTRL SCFTYP=RHF RUNTYP=GRADIENT $END
           $SYSTEM TIMLIM=20 MEMORY=300000 $END
           $GUESS  GUESS=HCORE $END
           $DATA
          HP=O ... 3-21G+* RHF calculation at STO-2G* geometry
          Cs
           
          Phosphorus 15.0
              N21 3
              L    1
              1    0.039             1.0              1.0
              D    1
              1     0.55         1.0
           
          Oxygen    8.0       1.439
              N21  3
           
          Hydrogen  1.0      -0.3527854           1.36412
              N21 3
           
           $END
 

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EXAM08.

1-A-1 H2O RHF + MP2 gradient calculation.

This job generates RHF orbitals which should be saved for use with EXAM9. This run, together with EXAM9, shows a much more typical MCSCF calculation, which should always be started with some sort of SCF MOs. This job also tests the 2nd order Moller-Plesset code.

The FINAL E is -75.5854099058 after 9 iterations.
E(MP2) is -75.7060362017, RMS grad=0.017449458
dipole moments are SCF=2.435688, MP2=2.329367

           $CONTRL SCFTYP=RHF MPLEVL=2 RUNTYP=GRADIENT $END
           $SYSTEM TIMLIM=2 MEMORY=100000 $END
           $BASIS  GBASIS=N21 NGAUSS=3 $END
           $GUESS  GUESS=HUCKEL $END
           $DATA
          Water...RHF/3-21G...exp.geom...R(OH)=0.95781,A(HOH)=104.4776
          Cnv      2
          
          OXYGEN     8.0
          HYDROGEN   1.0   0.0     0.7572157    0.5865358
           $END

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EXAM09.

1-A-1 H2O 2nd order MC-QDPT calculation

This job finds the Full Optimized Reaction Space MCSCF (or CAS-SCF) wavefunction for water. Its initial RHF orbitals are taken from EXAM8. The MCSCF wavefunction contains 37 configurations. The second order perturbation theory correction to the MCSCF energy is then obtained.

MCSCF:
On the 1st iteration, the energy is -75.601726235.
The FINAL E= -75.6386218833 after 13 iterations,
with c(1) = 0.988446 and dipole moment = 2.301626

MC-QDPT:
E(MCSCF)= -75.6386218833, E(MP2)= -75.7109706204

           $CONTRL SCFTYP=MCSCF MPLEVL=2 $END
           $SYSTEM TIMLIM=8 MEMORY=100000 $END
           $BASIS  GBASIS=N21 NGAUSS=3 $END
          ---- EXPERIMENTAL GEOM, R(OH)=0.95781A, HOH=104.4776 DEG.
           $DATA
          WATER...3-21G BASIS...FORS-MCSCF...EXPERIMENTAL GEOMETRY
          CNV      2
          
          OXYGEN     8.0
          HYDROGEN   1.0   0.0        0.7572157         0.5865358
           $END
           $GUESS  GUESS=MOREAD  NORB=13 $END
           $DRT    GROUP=C2V FORS=.TRUE. NMCC=1 NDOC=4 NVAL=2 $END
           $MCQDPT INORB=0 MULT=1 NMOFZC=1 NMODOC=0 NMOACT=6
                   ISTSYM=1 NSTATE=1 $END
          ---- CONVERGED 3-21G WATER VECTORS, E=-75.585409913 - - -
           $VEC
           1  1 0.98323195E+00 0.95883436E-01 0.00000000E+00 ...
           ... vectors deleted to save paper ...
          13  3 0.35961579E+00 0.28728587E+00 0.35961579E+00
           $END
 

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EXAM 10.

This run duplicates the first column of table 6 in Y.Yamaguchi, M.Frisch, J.Gaw, H.F.Schaefer, and J.S.Binkley J.Chem. Phys. 1986, 84, 2262-2278.

FINAL energy at the VIB 0 geometry is -74.9659012159.

If run with METHOD=ANALYTIC,
the FREQuencies are 2170.05, 4140.00, and 4391.07
the INTENSities are 0.17129, 1.04807, and 0.70930
the mean POLARIZABILITY is 0.40079

If run with METHOD=NUMERIC, NVIB=2,
the FREQuencies are 2170.14, 4140.18, and 4391.12
the INTENSities are 0.17169, 1.04703, and 0.70909

           $CONTRL SCFTYP=RHF RUNTYP=HESSIAN UNITS=BOHR NZVAR=3 $END
           $SYSTEM TIMLIM=4 MEMORY=100000 $END
           $FORCE  METHOD=ANALYTIC   $END
           $CPHF   POLAR=.TRUE. $END
           $BASIS  GBASIS=STO NGAUSS=3 $END
           $DATA
          Water at the RHF/STO-3G equilibrium geometry
          CNV      2
          
          OXYGEN       8.   0.0   0.0            0.0702816679
          HYDROGEN     1.   0.0   1.4325665478  -1.1312080153
           $END
           $ZMAT   IZMAT(1)=1,1,2,   1,1,3,   2,2,1,3  $END
           $GUESS  GUESS=HUCKEL   $END
 

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EXAM 11.

1A' HCN RHF Intrinsic Reaction Coordinate

This job tests the reaction path finder. The reaction is followed back to the HNC isomer. Four points on the IRC (counting the saddle point) are found,

Pt.R(N-C)R(N-H)A(HNC)Energydistance
T.S.1.221361.4376452.993-91.56485100.0
11.225331.3329658.476-91.56730970.29994
21.228021.2382764.747-91.57353460.59986
31.229741.1635072.039-91.58147750.89968
           $CONTRL SCFTYP=RHF RUNTYP=IRC NZVAR=3 $END
           $SYSTEM TIMLIM=5 MEMORY=400000 $END
           $IRC    PACE=GS2 SADDLE=.TRUE. TSENGY=.TRUE. 
                   FORWRD=.FALSE. NPOINT=3 $END
           $GUESS  GUESS=HUCKEL $END
           $ZMAT   IZMAT(1)=1,1,2  1,1,3  2,2,1,3 $END
           $BASIS  GBASIS=STO NGAUSS=3 $END
           $DATA
          HYDROGEN CYANIDE...STO-3G...INTRINSIC REACTION COORDINATE
          CS
          
          NITROGEN    7.0   -.0004620071    .0002821165    .0000000000
          CARBON      6.0   1.2208931990   -.0003427488    .0000000000
          HYDROGEN    1.0    .8654562191   1.1478852258    .0000000000
           $END
           $HESS
          ENERGY IS      -91.5648510307 E(NUC) IS       23.4154954113
           1  1 1.10665682E+00 1.58946320E-02 0.00000000E+00...
          ... 2nd derivatives deleted to save paper ...
           9  2-8.04548379E-09 0.00000000E+00 0.00000000E+00-1.42096449E-08
           $END

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EXAM 12.

This job illustrates linear bends, for acetylene. The optimal RHF/STO-2G geometry is located.

At the input geometry,
the FINAL E= -73.5036974734 after 7 iterations,
and the RMS gradient is 0.1506891.

At the final geometry, 7 steps later,
the FINAL E= -73.6046483165, RMS gradient=0.0000028,
R(CC)=1.1777007 and R(CH)=1.0749435.

           $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE NZVAR=5 $END
           $SYSTEM TIMLIM=6 MEMORY=100000 $END
           $BASIS  GBASIS=STO NGAUSS=2 $END
           $GUESS  GUESS=HUCKEL $END
           $DATA
          Acetylene geometry optimization in internal coordinates
          Dnh      4
          
          CARBON      6.0    0.0  0.0  0.70
          HYDROGEN    1.0    0.0  0.0  1.78
           $END
           $ZMAT  IZMAT(1)=1,1,2,   1,1,3,   1,2,4,
                           5,1,2,4,    5,2,1,3  $END
          ------- XZ is 1st plane for both bends -------
           $LIBE  APTS(1)=1.0,0.0,0.0,1.0,0.0,0.0 $END

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EXAM 13.

This run nearly duplicates the POLYATOM calculation of D.Neumann + J.W.Moskowitz, J.Chem.Phys. 49,2056(1968) This run differs in that the cartesian s contaminant in the d shells is retained. All properties are tested.

V(NE) = -199.1899133465
V(EE) = 37.8936602847 T = 76.0126432961
V(NN) = 9.2390200836 E(TOT)= -76.0445896821
MULL.Q(O)=-0.622844 Bond Order=0.912
DENSITY - O=286.751654 H=0.404807
MOMENTS: DZ= 2.097011
QXX=-2.369593 QYY= 2.480726 QZZ=-0.111134
OXXZ=-0.863480 OYYZ= 2.149717 OZZZ=-1.286237
ELECTRIC FIELD/GRADIENT: H(YZ)=+/-0.364926
O(Z)=-0.060754 H(Y)=+/-0.007327 H(Z)=0.001535
O(XX)=1.909584 O(YY)=-1.727606 O(ZZ)=-0.181978
H(XX)=0.301168 H(YY)=-0.258105 H(ZZ)=-0.043062
POTENTIAL - V(O)=-22.337450 V(H)=-1.006137

           $CONTRL SCFTYP=RHF RUNTYP=ENERGY UNITS=BOHR $END
           $SYSTEM TIMLIM=15 MEMORY=300000 $END
           $GUESS  GUESS=HUCKEL $END
           $ELMOM  IEMOM=3 $END
           $ELFLDG IEFLD=2 $END
           $ELPOT  IEPOT=1 $END
           $ELDENS IEDEN=1 $END
           $DATA
          Water...properties test...(10,5,2/4,1)/[5,3,2/2,1] basis
          Cnv      2
           
          Oxygen     8.0
            S    2
              1   31.3166          0.243991
              2   76.232           0.152763
            S    3
              1  290.785           0.904785
              2 1424.0643          0.121603
              3 4643.4485          0.029225
            S    2
              1    4.6037          0.264438
              2   12.8607          0.458240
            S    2
              1    0.9311          1.051534
              2    9.7044         -0.140314
            S    1
              1    0.2825          1.0
            P    3
              1    7.90403         0.124190
              2   35.1832          0.019580
              3    2.30512         0.394730
            P    1
              1    0.21373         1.0
            P    1
              1    0.71706         1.0
            D    1
              1    1.5             1.0
            D    1
              1    0.5             1.0
           
          Hydrogen   1.0      0.0  1.428036   1.0957706
            S    3
              1   0.65341          0.817238
              2   2.89915          0.231208
              3  19.2406           0.032828
            S    1
              1   0.17758          1.0
            P    1
              1   1.0              1.0
           
 

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EXAM 14.

CI transition moments. Water, using RHF/STO-3G MOs. All orbitals are occupied, transition is 1-1A1 to 2-1A1.

E(STATE 1)= -75.0101113548, E(STATE 2)= -74.3945819375
Dipole LENGTH is =0.392614
Dipole VELOCITY is =0.368205

           $CONTRL SCFTYP=NONE CITYP=GUGA RUNTYP=TRANSITN UNITS=BOHR $END
           $SYSTEM TIMLIM=1 MEMORY=100000 $END
           $BASIS  GBASIS=STO  NGAUSS=3 $END
           $DATA
          WATER MOLECULE...STO-3G...TRANSITION MOMENT
          CNV      2
          
          OXYGEN      8.0   0.0   0.0      0.0
          HYDROGEN    1.0   0.0   1.428   -1.096
           $END
          !            standard SD-CI calculation
           $CIDRT1 GROUP=C2V IEXCIT=2 NFZC=1 NDOC=4 NVAL=2 $END
           $TRANST $END
          
          --- RHF ORBITALS --- GENERATED AT 09:24:04    18-FEB-88
          WATER MOLECULE...STO-3G...TRANSITION MOMENT
          E(RHF)=  -74.9620539825, E(NUC)=    9.2384802989,    8 ITERS
           $VEC1
           1  1 9.94117078E-01 2.66680164E-02 0.00000000E+00 ...
          ... vectors deleted to save paper ...
           7  2-8.42653177E-01 8.42653177E-01
           $END

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EXAM 15.

C2- diatom, in the electronic state doublet-pi-u. This illustrates a open shell SCF calculation, using fed in coupling coefficients, and the GVB/ROHF code.

The FINAL energy is -75.5579181071 after 8 iterations.

           $CONTRL SCFTYP=GVB  MULT=2  ICHARG=-1  UNITS=BOHR  $END
           $SYSTEM TIMLIM=15 MEMORY=300000 $END
           $BASIS  GBASIS=DH NDFUNC=1 POLAR=DUNNING $END
           $DATA
          C2-...DOUBLET-PI-UNGERADE...OPEN SHELL SCF
          DNH      4
           
          CARBON      6.0     0.0  0.0  -1.233
           $END
           $GUESS  GUESS=MOREAD NORB=30
                   NORDER=1  IORDER(5)=7,5,6  $END
           $SCF    NCO=5  NSETO=1  NO=2  COUPLE=.TRUE.
                     F(1)=1.0, 0.75
                 ALPHA(1)=2.0, 1.5, 1.00
                  BETA(1)=-1., -.75, -0.5    $END
           
          --- RHF ORBITALS --- GENERATED AT 14:05:16THU MAR 24/88
           CC  R(C-C) = 2 * 1.233 BOHR   BAS=831+1D
          E(RHF)=  -75.3856001855, E(NUC)=   14.5985401460,   18 ITERS
           $VEC
           1  1-7.06500288E-01-1.39103044E-03-3.57452331E-04 ...
          ... vectors deleted to save paper ...
           $END

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EXAM 16.

ROHF/GVB on Si 3-P state, using Gordon's 6-31G basis.

The purpose of this example is two-fold, namely to show off the open shell capabilities of the GVB code, and to emphasize that the 6-31G basis for Si in GAMESS is Mark Gordon's version. The basis stored in GAMESS is completely optimized, whereas Pople's uses the core from from a 6-21G set, reoptimizing only the -31G part. The energy from Pople's basis would be only -288.828405.

Jacobi diagonalization is intrinsically slow, but in this case results in pure subspecies in degenerate p irreps. In fact, these may be labeled in the highest Abelian subgroup of the atomic point group Kh.

The FINAL energy is -288.8285729745 after 7 iterations.

           $CONTRL SCFTYP=GVB MULT=3 $END
           $SYSTEM TIMLIM=2 MEMORY=100000 KDIAG=3 $END
           $BASIS  GBASIS=N31 NGAUSS=6 $END
           $DATA
          Si...3-P term...ROHF in full Kh symmetry
          Dnh 2
          
          Silicon     14.
           $END
           $GUESS  GUESS=HUCKEL $END
           $SCF    NCO=6  NSETO=1  NO=3   COUPLE=.TRUE.
                   F(1)=1.0, 0.333333333333333
                   ALPHA(1)=2.0,  0.66666666666667,  0.16666666666667
                   BETA(1)=-1.0, -0.33333333333333, -0.16666666666667
           $END

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EXAM 17.

Analytic hessian for an open shell SCF function.
Methylene's 1-B-1 excited state.
FINAL energy= -38.3334724780 after 8 iterations.
The FREQuencies are 1224.19, 3563.44, 3896.23
The INTENSities are 0.13317, 0.21652, 0.14589
The mean POLARIZABILITY is 0.53018

           $CONTRL SCFTYP=GVB  MULT=1  RUNTYP=HESSIAN  UNITS=BOHR $END
           $SYSTEM TIMLIM=4 MEMORY=100000 $END
           $CPHF   POLAR=.TRUE. $END
           $BASIS  GBASIS=STO NGAUSS=3 $END
           $SCF    NCO=3  NSETO=2  NO(1)=1,1  NPAIR=0 $END
           $ZMAT   IZMAT(1)=1,1,2,   1,1,3,   2,2,1,3   $END
           $GUESS  GUESS=HUCKEL  $END
           $DATA
          METHYLENE...1-B-1 STATE...ROHF...STO-3G BASIS
          CNV      2
          
          CARBON      6.0    0.0   0.0            0.0041647278
          HYDROGEN    1.0    0.0   1.8913952563   0.7563907037
           $END

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EXAM 18.

effective core potential...diatomic P2...RHF/CEP-31G*
See Stevens,Basch,Krauss, J.Chem.Phys. 81,6026-33(1984).
GAMESS FINAL E= -12.6956517413, RMS gradient=0.000204749
A separate run gives E(P)= -6.32635, so De= 26.95 kcal/mol

           $CONTRL SCFTYP=RHF RUNTYP=GRADIENT ECP=SBK NZVAR=1 $END
           $SYSTEM TIMLIM=15 MEMORY=300000 $END
           $GUESS  GUESS=HUCKEL $END
           $ZMAT   IZMAT(1)=1,1,2 $END
           $DATA
          diatomic phosphorous
          Dnh      4
           
          PHOSPHORUS 15.0    0.0      0.0        0.9395
             SBK
             D 1
               1 0.45 1.0
           
           $END

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EXAM 19.

Spin-orbit coupling example.

This run duplicates the one electron result shown in Table 3 of T.R.Furlani,H.F.King, J.Chem.Phys. 82, 5577-83(1985), namely

E(3s-) = -54.9382257056
E(1s+) = -54.7988368491
SOC(z) = -114.3851

Final energies of all 6 levels in the pi**2 configuration:
RELATIVE E= -15296.847, -15296.419, -15296.419,
RELATIVE E= 0.0, 0.0, and +15296.847 wavenumbers.

Why are there six levels? The singlet-delta is two roots, the singlet-sigma-plus is one. During the CI, the spatial part of the triplet-sigma-minus is one CSF, with alpha-alpha spin, hence IROOTS=3,1. The final spin-orbit Hamiltonian includes all three triplet spin states. So, 2+1+3=6 levels, and you can work out or yourself these have quantum number omega=0,0,1,2. Note that the lower multiplicity CIDRT1 is done in C1 symmetry in order to generate both components of the delta state.

           $CONTRL SCFTYP=NONE MULT=3 CITYP=GUGA RUNTYP=SPINORBT UNITS=BOHR $END
           $SYSTEM TIMLIM=2 MEMORY=200000 $END
singlet-delta, singlet-sigma-plus, triplet-sigma-minus SOC
           $TRANST NFZC=3 NOCC=5 NUMVEC=1 NUMCI=2 IROOTS(1)=3,1 $END
           $CIDRT1 GROUP=C1  IEXCIT=2 NFZC=3 NDOC=1 NVAL=1 $END
           $CIDRT2 GROUP=C4V IEXCIT=2 NFZC=3 NALP=2 $END
Since the 1e- spin orbit integrals cannot use L shells, we must input the 6-31G set for nitrogen by hand.
           $DATA
          Imidogen radical
          Cnv 4
          
          NITROGEN    7.0
             S    6
              1  4173.511460      0.001834772
              2   627.457911      0.01399463
              3   142.902093      0.06858655
              4    40.234329      0.2322409
              5    12.820213      0.4690699
              6     4.390437      0.3604552
             S    3
              1    11.626362     -0.1149612
              2     2.716280     -0.1691175
              3     0.772218      1.145852
             P    3
              1    11.626362      0.06757974
              2     2.716280      0.3239073
              3     0.772218      0.7408951
             S    1
              1     0.212031      1.0
             P    1
              1     0.212031      1.0

          HYDROGEN    1.0     0.0   0.0   1.9748
             N31 6
          
           $END
          
          --- ROHF ORBITALS --- GENERATED AT 12:04:18 29 MAR 90 ( 88)
          IMIDOGEN RADICAL
          E(ROHF)= -54.9382257007, E(NUC)=    3.5446627507,    8 ITERS
           $VEC1
           1  1 9.97073281E-01 1.91214515E-02 0.00000000E+00 ...
          ... vectors deleted to save paper ...
          11  3-1.58015531E+00
           $END
 

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EXAM 20.

Optimize an orbital exponent.

The SBK basis for I consists of 5 gaussians, in a -41 type split. The exponent of a diffuse L shell for iodide ion is optimized (6th exponent overall). The optimal exponent turns out to be 0.036713, with a corresponding FINAL energy of -11.3010023066

           $CONTRL SCFTYP=RHF RUNTYP=TRUDGE ICHARG=-1 ECP=SBK $END
           $SYSTEM TIMLIM=30 MEMORY=300000 $END
           $TRUDGE OPTMIZ=BASIS NPAR=1 IEX(1)=6 $end
           $GUESS  GUESS=HUCKEL $END
           $DATA
          I- ion
          Dnh 2
          
          Iodine 53.0
             SBK
             L 1
               1 0.02 1.0
          
           $END

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EXAM 21.

Open shell two configuration SCF analytic hessian.

M.Duran, Y.Yamaguchi, H.F.Schaefer III J.Phys.Chem. 1988, 92, 3070-3075.

Least motion insertion of CH into H2, which leads to a 3rd order hypersaddle point on the 2-B-1 surface.

Literature values are

FINAL E=-39.25104, C1=0.801, C2=-0.598
FREQ= 4805i, 1793i, 1317i, 989, 2914, 3216
mean POLARIZABILITY=2.05

GAMESS obtains

FINAL E=-39.2510351249, C1=0.801141, C2=-0.598476
FREQ= 4805.53i, 1793.00i, 1317.43i,
FREQ= 988.81, 2913.52, 3216.42
INTENS= 4.54563, 0.09731, 0.00768
mean POLARIZABILITY=2.04655

           $CONTRL SCFTYP=GVB MULT=2 RUNTYP=HESSIAN $END
           $SYSTEM TIMLIM=25 MEMORY=100000 $END
           $CPHF   POLAR=.TRUE. $END
           $GUESS  GUESS=MOREAD NORB=16 NORDER=1 IORDER(4)=6,4,5 $END
           $SCF    NCO=3 NSETO=1 NO=1 NPAIR=1 CICOEF(1)=0.7,-0.7 $END
           $DATA
          Insertion of CH into H2...OS-TCSCF ansatz...DZ basis
          CNV 2
          
          CARBON   6.0   0.0000000000   0.0000000000  -0.0001357549
            S  6
              1 4232.61    0.002029
              2  634.882   0.015535
              3  146.097   0.075411
              4   42.4974  0.257121
              5   14.1892  0.596555
              6    1.9666  0.242517
            S  1
              1    5.1477  1.0
            S  1
              1    0.4962  1.0
            S  1
              1    0.1533  1.0
            P  4
              1   18.1557  0.018534
              2    3.9864  0.115442
              3    1.1429  0.386206
              4    0.3594  0.640089
            P  1
              1    0.1146  1.0
          
          HYDROGEN  1.0   0.0000000000   0.0000000000   1.0922959062
            DH  0  1.2 1.2

          HYDROGEN  1.0   0.0000000000   0.4152229538  -1.4824967459
            DH  0  1.2 1.2
          
           $END
          --- these are 2-A1 ROHF vectors ---
          --- ROHF ORBITALS --- GENERATED AT 08:23:42 27 JUN 90 (178)
          INSERTION OF CH INTO H2...OS-TCSCF ANSATZ...DZ BASIS
          E(ROHF)= -39.2316245004, E(NUC)= 8.0760320442, 12 ITERS
           $VEC
           1  1 6.01223299E-01 4.37813104E-01 ...
          ... vectors deleted to save paper ...
          16  4-2.12429766E-02
           $END

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EXAM22.

3-A-2 H3CN UMP2/6-31G*//UHF/6-31G*

The FINAL UHF energy= -94.0039683676 after 13 iters.
The E(MP2) energy= -94.2315757758

           $CONTRL SCFTYP=UHF MULT=3 RUNTYP=ENERGY MPLEVL=2
                   COORD=ZMT $END
           $SYSTEM TIMLIM=30 MEMORY=300000 $END
           $BASIS  GBASIS=N31 NGAUSS=6 NDFUNC=1 NPFUNC=0 $END
           $GUESS  GUESS=HUCKEL $END
           $DATA
          Methylnitrene...UHF/6-31G* structure
          Cnv 3
          
          N
          C  1  rCN
          H  2  rCH  1  aHCN
          H  2  rCH  1  aHCN  3  120.0
          H  2  rCH  1  aHCN  3 -120.0
          
          rCN=1.4329216
          rCH=1.0876477
          aHCN=110.21928
           $END

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EXAM23.

semiempirical calculation, using the MOPAC/GAMESS combo

AM1 gets the geometry disasterously wrong!

initial geometry,MNDOAM1PM3
FINAL HEAT OF FORMATION105.1408893.4599746.89387
RMS gradient0.08181570.10085870.0366232
final geometry (# steps),81110
FINAL HEAT OF FORMATION46.45649-1.81716-2.79647
RMS gradient0.00002700.00002940.0000015
r(SiH)1.421191.458131.52104
a(HSiH)101.956120.00096.280
           $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE COORD=ZMT ICHARG=-1 $END
           $SYSTEM TIMLIM=5 MEMORY=200000 $END
           $BASIS  GBASIS=PM3 $END
           $DATA
          Silyl anion...comparison of semiempirical models
          Cnv 3
          
          Si
          H  1  rSiH
          H  1  rSiH  2 aHSiH
          H  1  rSiH  2 aHSiH  3   aHSiH  -1
          
          rSiH=1.15
          aHSiH=110.0
           $END

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EXAM24.

Self-consistent reaction field test, of water in water. Cavity radius is calculated from the 1.00 g/cm**3 density.
FINAL energy is -74.9666740755 after 12 iterations
Induced dipole= -0.03663, RMS gradient= 0.033467686

           $contrl scftyp=rhf runtyp=gradient coord=zmt $end
           $system memory=300000 $end
           $basis  gbasis=sto ngauss=3 $end
           $guess  guess=huckel $end
           $scrf   radius=1.93 dielec=80.0 $end
           $data
          water in water, arbitrary geometry
          Cnv 2
          
          O
          H 1 rOH
          H 1 rOH 2 aHOH 
          
          rOH = 0.95
          aHOH = 104.5
           $end
 

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EXAM25.

Illustration of coordinate systems for geometry searches. Arbitrary molecule, chosen to illustrate ring, methyl on ring, methine H10, imino in ring, methylene in ring.

      H8 H9
        \|
      H7-C6  O1---O5   H13
          \ /       \ /
          C2       C4
          /  \     /  \
         H10    N3     H12
                |
                H11

The initial AM1 energy is -48.6594935

initial RMSfinal Efinal RMS#steps
Cartesians0.0200113-48.70225200.000027850
dangling Z-mat0.0600637...OObond crashes on 1st step
good Z-matrix0.0232915-48.70225080.000029920
nat. internals0.0209442-48.70225700.000018315
           $contrl scftyp=rhf runtyp=optimize coord=zmt $end
           $system memory=300000 $end
           $statpt hess=guess nstep=100 nprt=-1 npun=-2 $end
           $basis  gbasis=am1 $end
           $guess  guess=huckel $end
           $data
          Illustration of coordinate systems
          C1
          O
          C 1 rCOa
          N 2 rCNa 1 aNCO
          C 3 rCNb 2 aCNC  1 wCNCO
          O 4 rCOb 3 aOCN  2 wOCNC
          C 2 rCC  1 aCCO  5 wCCOO
          H 6 rCH1 2 aHCC1 1 wHCCO1
          H 6 rCH2 2 aHCC2 1 wHCCO2
          H 6 rCH3 2 aHCC3 1 wHCCO3
          H 2 rCHa 1 aHCOa 5 wHCOOa
          H 3 rNH  2 aHNC  1 wHNCO
          H 4 rCHb 5 aHCOb 1 wHCOOb
          H 4 rCHc 5 aHCOc 1 wHCOOc
          
          rCOa=1.43
          rCNa=1.47
          rCNb=1.47
          rCOb=1.43
          aNCO=106.0
          aCNC=104.0
          aOCN=106.0
          wCNCO=30.0
          wOCNC=-30.0
           rCC=1.54
          aCCO=110.0
          wCCOO=-150.0
          rCH1=1.09
          rCH2=1.09
          rCH3=1.09
          aHCC1=109.0
          aHCC2=109.0
          aHCC3=109.0
          wHCCO1=60.0
          wHCCO2=-60.0
          wHCCO3=180.0
          rCHa=1.09
          aHCOa=110.0
          wHCOOa=100.0
          rNH=1.01
          aHNC=110.0
          wHNCO=170.0
          rCHb=1.09
          rCHc=1.09
          aHCOb=110.0
          aHCOc=110.0
          wHCOOb=150.0
          wHCOOc=-100.0
           $end
          
          To use Cartesian coordinates:
          --- $contrl nzvar=0 $end
          
          To use conventional Z-matrix, with dangling O-O bond:
          --- $contrl nzvar=33 $end
          
          To use well chosen internals, with all ring bonds closed:
          --- $contrl nzvar=33 $end
          --- $zmat   izmat(1)=1,1,2,  1,2,3,  1,3,4,  1,4,5,  1,5,1,
               2,1,2,3,  2,5,4,3,  3,5,1,2,3,  3,1,5,4,3,
               1,6,2,  2,6,2,1,  3,6,2,1,5,  
               1,6,7,  1,6,8,  1,6,9,  2,7,6,2,  2,8,6,2,  2,9,6,2,  
               3,7,6,2,1,  3,8,6,2,1,  3,9,6,2,1,
               1,10,2,  2,10,2,1,  3,10,2,1,5,
               1,11,3,  2,11,3,2,  3,11,3,2,1,
               1,12,4,  2,12,4,5,  3,12,4,5,1,  
               1,13,4,  2,13,4,5,  3,13,4,5,1 $end
 
          To use natural internal coordinates:
           $contrl nzvar=44 $end
           $zmat   izmat(1)=1,1,2,  1,2,3,  1,3,4,  1,4,5,  1,5,1,   ! ring !
               2,5,1,2,   2,1,2,3,   2,2,3,4,   2,3,4,5,   2,4,5,1,
               3,5,1,2,3, 3,1,2,3,4, 3,2,3,4,5, 3,3,4,5,1, 3,4,5,1,2,
                  1,2,6,  2,6,2,1,  2,6,2,3,  4,6,2,1,3,        ! methyl C !
               1,6,7,  1,6,8,  1,6,9,                           ! methyl Hs !
               2,7,6,8,  2,8,6,9,  2,9,6,7,  2,9,6,2,  2,7,6,2,  2,8,6,2,
               3,7,6,2,1,
                     1,10,2,  2,10,2,1,  2,10,2,3,  2,10,2,6,   ! methine !
                     1,11,3,  2,11,3,2,  2,11,3,4,  4,11,3,2,4, ! imino !
               1,12,4,  1,13,4,                                 ! methylene !
               2,12,4,13,  2,12,4,3,  2,13,4,3,  2,12,4,5,  2,13,4,5
          
                  ijS(1)=1,1,  2,2,  3,3,  4,4,  5,5,           ! ring !
                    6,6, 7,6, 8,6, 9,6,10,6,
                         7,7, 8,7, 9,7,10,7,
                   11,8,12,8,13,8,14,8,15,8,
                   11,9,12,9,     14,9,15,9,
                         16,10,   17,11,18,11,   19,12,         ! methyl C !
                   20,13,  21,14,  22,15,                       ! methyl Hs !
                   23,16, 24,16, 25,16, 26,16, 27,16, 28,16,
                   23,17, 24,17, 25,17,
                          24,18, 25,18,
                                        26,19, 27,19, 28,19,
                                               27,20, 28,20,
                   29,21,
                      30,22,    31,23,32,23,33,23,   32,24,33,24,  ! methine !
                      34,25,  35,26,36,26,  37,27,              ! imino !
                   38,28,  39,29,                               ! methylene !
                   40,30, 41,30, 42,30, 43,30, 44,30,
                          41,31, 42,31, 43,31, 44,31,
                          41,32, 42,32, 43,32, 44,32,
                          41,33, 42,33, 43,33, 44,33
          
                  Sij(1)=1.0, 1.0, 1.0, 1.0, 1.0,               ! ring !
                      1.0, -0.8090, 0.3090,  0.3090, -0.8090,
                           -1.1180, 1.8090, -1.8090, 1.1180,
                      0.3090, -0.8090, 1.0, -0.8090, 0.3090,
                      -1.8090, 1.1180,      -1.1180, 1.8090,
                              1.0,   1.0,-1.0,   1.0,           ! methyl C !
                   1.0,   1.0,   1.0,                           ! methyl Hs !
                   1.0, 1.0, 1.0,-1.0,-1.0,-1.0,
                   2.0,-1.0,-1.0,
                        1.0,-1.0,
                                  2.0,-1.0,-1.0,
                                       1.0,-1.0,
                   1.0,
                      1.0,     2.0,-1.0,-1.0,   1.0,-1.0,       ! methine !
                      1.0,   1.0,-1.0,    1.0,                  ! imino !
                   1.0,  1.0,                                   ! methylene !
                   4.0, 1.0, 1.0, 1.0, 1.0,
                        1.0,-1.0, 1.0,-1.0,
                        1.0, 1.0,-1.0,-1.0,
                        1.0,-1.0,-1.0, 1.0    $end
 

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EXAM26

Localized orbital test...see J.Phys.Chem. 1984, 88, 382-389

FINAL Energy= -415.2660357363 in 11 iters

If you localize only the valence orbitals, by commenting
out the $LOCAL group below, the

Boys localization sum is 204.693589
Ruedenberg localization sum is 5.081667
population localization sum is 4.610528

The SCF localized charge decomposition forces all orbitals to be localized, so the final diagonal sum is 28.389125. The nuclear charge assigned to the oxygen "lone pairs" is redistributed so that the total nuclear P and O charges are correct. The energies computed for the PO bond, PH bonds, and O lone pairs are -37.273022, -27.364212, -26.363865. The corresponding dipoles are 2.041, 3.484, and 3.465.

To analyze the MP2 valence contributions, select MPLEVL=2, and turn EDCOMP and DIPDCM off. The results should be E(MP2)=-415.4952200908, and the contribution of the PO bond, PH bonds, and O lone pairs to the correlation energy are -0.0442096, -0.0237793, and -0.0378790, respectively.

           $contrl scftyp=rhf runtyp=energy local=ruednbrg mplevl=0 $end
           $system memory=750000 $end
           $mp2    lmomp2=.true. $end
           $local  edcomp=.true.  moidon=.true. dipdcm=.true.
                   ijmo(1)= 1,11, 2,11, 1,12, 2,12, 1,13, 2,13
                   zij(1)=1.666666667,0.333333333,1.6666666667,0.333333333,
                          1.666666667,0.333333333
                   moij(1)= 2,1,  2,1,  2,1
                   nnucmo(11)=2,2,2  $end
           $basis  gbasis=n21 ngauss=3 ndfunc=1 $end
           $data
          phosphine oxide...3-21G* basis...localized orbital test
          Cnv 3
          
          P 15.0
          O  8.0  0.0000000000  0.0   1.4701
          H  1.0  1.2335928631  0.0  -0.6421021244
           $end
 

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EXAM27.

NH3 semi-empirical DRC calculation

The dynamic reaction coordinate is initiated at the planar inversion transition state, with a velocity parallel to the mode with imaginary frequency. The reactive trajectory is given one kcal/mole energy in excess of the amount needed to traverse the barrier. The trajectory is analyzed in terms of the equilibrium geometry's coordinates and normal modes. Because this is a test run, the trajectory is stopped after a much too short time interval.

The last point on the trajectory has
T=0.00163, V=-9.12874, E=-9.12710,
q(L6)=-0.153112, p(L6)=-0.014313,
velocity(H,z)=0.028857623667

           $CONTRL SCFTYP=RHF RUNTYP=DRC $END
           $SYSTEM MEMORY=300000 $END
           $BASIS  GBASIS=AM1 $END
           $DATA
          ammonia...DRC starting from the planar transition state
          C1
          NITROGEN    7.0   0.0000000000   0.0000000000   0.0000000000
          HYDROGEN    1.0  -0.4882960784   0.8457536168   0.0000000000
          HYDROGEN    1.0  -0.4882960784  -0.8457536168   0.0000000000
          HYDROGEN    1.0   0.9765921567   0.0000000000   0.0000000000
           $END
           $DRC NPRTSM=1 NSTEP=10 DELTAT=0.1 NMANAL=.TRUE. EKIN=1.0 
                VEL(1)=0.0 0.0 -0.1128,  
                       0.0 0.0 0.5213, 
                       0.0 0.0 0.5213,
                       0.0 0.0 0.5213
                C0(1)=0.0000000000   0.0000000000   0.0291576578
                     -0.4692651161   0.8127910232  -0.3097192193
                     -0.4692651161  -0.8127910232  -0.3097192193
                      0.9385302321   0.0000000000  -0.3097192193 $END
           $HESS
          ENERGY IS       -9.1354556210 E(NUC) IS        6.8369847904
           1  1 6.16231432E-01 3.45452916E-11-1.03923982E-05 ...
          ... 2nd derivatives deleted to save paper ...
          12  3 1.38181166E-10 5.72335505E-02
           $END

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EXAM28.

Morokuma energy decomposition.

This run duplicates a result from Table 16 of H.Umeyama, K.Morokuma, J.Am.Chem.Soc. 99,1316(1977)

GAMESSliterature
ES=-14.02-14.0
EX=8.989.0
PL=-1.12-1.1
CT=-2.37-2.4
MIX=-0.43-0.4
total-8.96-9.0
           $contrl scftyp=rhf runtyp=morokuma coord=zmt $end
           $system memory=300000 timlim=5 $end
           $basis  gbasis=n31 ngauss=4 $end
           $guess  guess=huckel $end
           $morokm iatm(1)=3 $end
           $data
          water-ammonia dimer
          Cs
          
          H
          O 1 rOH
          H 2 rOH 1 aHOH
          N 2  R  1 aHOH    3    0.0
          H 4 rNH 3 aHNaxis 1  180.0
          H 4 rNH 3 aHNaxis 5 +120.0
          H 4 rNH 3 aHNaxis 5 -120.0
          
          rOH=0.956
          aHOH=105.2
          rNH=1.0124
          aHNaxis=112.1451  ! makes HNH=106.67
          R=2.93
           $end
 

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EXAM29.

surface scan

The scan is done over a 3x3 grid centered on the SCF transition state for the SN2 type reaction

F- + NH2OH -> F-NH2-OH anion -> FNH2 + OH-

Groups 1 and 2 are F and OH, and their distance from the N is varied antisymmetrically, which is more or less what the IRC should be like. The results seem to indicate that the MP2/3-21G saddle point would shift further into the product channel, since the higher MP2 energies occur at shorter r(NF) and longer r(NO):

FINAL E= -229.0368324609, E(MP2)= -229.3873302369
FINAL E= -229.0356378404, E(MP2)= -229.3866642674
FINAL E= -229.0309266309, E(MP2)= -229.3822094766
FINAL E= -229.0372146681, E(MP2)= -229.3923234053
FINAL E= -229.0385440291, E(MP2)= -229.3936486639
FINAL E= -229.0367369550, E(MP2)= -229.3913683060
FINAL E= -229.0328601143, E(MP2)= -229.3918932008
FINAL E= -229.0364643928, E(MP2)= -229.3948325495
FINAL E= -229.0372478241, E(MP2)= -229.3943498134

A more conclusive way to tell this would be to compute single point MP2 energies along the SCF IRC, since the true reaction path always curves, and thus does not lie along rectangular grid points.

           $contrl scftyp=rhf runtyp=surface
                   icharg=-1 coord=zmt mplevl=2 $end
           $system memory=500000 timlim=30 $end
           $surf   ivec1(1)=2,1 igrp1=1       
                   ivec2(1)=2,5 igrp2(1)=5,6  
                   disp1= 0.10 ndisp1=3 orig1=-0.10
                   disp2=-0.10 ndisp2=3 orig2= 0.10 $end
           $basis  gbasis=n21 ngauss=3 $end
           $guess  guess=huckel $end
           $data
          F-NH2-OH exchange (inspired by J.Phys.Chem. 1994,98,7942-4)
          Cs
          
          F
          N 1 rNF
          H 2 rNH   1  aFNH 
          H 2 rNH   1  aFNH   3 aHNH  +1
          O 2 rNO   3  aONH   4 aONH  -1
          H 5 rOH   2  aHON   1 180.0
          
          rNF=1.7125469
          rNH=0.9966981
          rNO=1.9359887
          rOH=0.9828978
          aFNH=90.18493
          aONH=79.34339
          aHON=100.78851
          aHNH=108.57000
           $end
 

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EXAM30

Test of water EFP ... formamide/three water complex

FINAL E= -169.0085352303 after 12 iterations
RMS gradient=0.008099469

The geometry below combines a computed gas phase structure for formamide, with three waters located in a cylic fashion whose positions approximate the minimum structure of W.Chen and M.S.Gordon. This approximate structure lies about 11 mHartee above the actual minimum.

           $contrl scftyp=rhf runtyp=gradient coord=zmt $end
           $system memory=300000 $end
           $basis  gbasis=dh npfunc=1 ndfunc=1 $end
           $data
          formamide with three effective fragment waters
          C1
          C
          O 1 rCO
          N 1 rCN  2 aNCO
          H 3 rNHa 1 aCNHa 2 0.0
          H 3 rNHb 1 aCNHb 2 180.0
          H 1 rCH  2 aHCO  4 180.0
          
          rCO=1.1962565
          rCN=1.3534065
          rNHa=0.9948420
          rNHb=0.9921367
          rCH=1.0918368
          aNCO=124.93384
          aCNHa=119.16000
          aCNHb=121.22477
          aHCO=122.30822
           $end
           $efrag
          coord=int
          fragname=H2Oef2
          O1  4 1.926      3 175.0     1 180.0
          H2  7 0.9438636  4 117.4     3 -175.0
          H3  7 0.9438636  8 106.70327 4 95.0
          fragname=H2Oef2
          O1  8 1.901      7 175.0     4 0.0
          H2 10 0.9438636  8 110.0     4 -5.0
          H3 10 0.9438636 11 106.70327 8 -95.0
          fragname=H2Oef2
          H2  2 1.951      1 150.0     3 0.0
          O1 13 0.9438636  2 177.0     3 0.0
          H3 14 0.9438636 13 106.70327 3 140.0
           $end

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