******************************
*
*
* 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.
Example Description
------- -----------
1 CH2 RHF geometry
optimization
2 CH2 UHF + gradient
3 CH2 ROHF + gradient
4 CH2 GVB + gradient
5 CH2 RHF + CI gradient
6 CH2 MCSCF geometry
optimization
7 HPO RHF + gradient
8 H2O RHF + MP2 gradient
9 H2O MCSCF + MCQDPT
energy correction
10 H2O RHF + hessian
11 HCN RHF IRC
12 HCCH closed shell DFT
geometry opt.
13 H2O RHF properties
14 H2O CI transition
moment
15 C2- GVB/ROHF on
2-pi-u state
16 Si GVB/ROHF
on 3-P state
17 CH2 GVB/ROHF +
hessian
18 P2 RHF +
hessian, effective core pot.
19 NH spin-orbit
coupling
20 I- exponent
TRUDGE optimization
21 CH3 OS-TCSCF hessian
22 H3CN UHF + UMP2 energy
23 SiH3- PM3 geometry optimization
24 H2O SCRF test
case
25 ?
internal coordinate example
26 H3PO localized orbital
test
27 NH3 DRC example
28 H2O-NH3 Morokuma decomposition
29 FNH2OH surface scan
30 HCONH2(H2O)3 effective fragment solvation
31 H2O PCM test case
1
! 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
1
! 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 <S**2> = 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
1
! EXAM03.
!
3-B-1 CH2 ROHF calculation on methylene ground state.
!
The wavefunction is a pure triplet state (<S**2> = 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
1
! 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
1
! 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 = -38.4270674136, c(1) = 0.970224
!
State 2 EIGENvalue = -38.3130036824, 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
1
! 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=300000 $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.
! must find
at least two roots since ground state is 3-B-1
!
$DET
NCORE=3 NACT=2 NELS=2 NSTATE=2 $END
1
! 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
1
! 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 10 iterations.
!
E(MP2) is -75.7060361996, RMS grad=0.017449524
!
dipole moments are SCF=2.435689, MP2=2.329368
!
$CONTRL
SCFTYP=RHF MPLEVL=2 RUNTYP=GRADIENT $END
$SYSTEM
TIMLIM=2 MEMORY=100000 memddi=1 parall=.true. $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
1
! 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 225 determinants,
!
which includes 105 singlet configurations.
!
The second order perturbation theory correction
!
to the MCSCF energy is then obtained, using the
!
slower code that produces information about
!
what weight percent the MCSCF comprises in the
!
first order wavefunction.
!
!
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
!
the 2nd order correction has 2 terms, semi-internal
!
is -0.03446957 and external -0.03787916 Hartree,
!
and the MCSCF reference weight is 98.4%.
!
$CONTRL
SCFTYP=MCSCF MPLEVL=2 $END
$SYSTEM
TIMLIM=8 MEMORY=300000 $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
$DET
NCORE=1 NACT=6 NELS=8 $END
$MCQDPT
NSTATE=1 ISTSYM=1 INORB=0 REFWGT=.TRUE. $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
1
! 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
1
! 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) Energy
distance
! T.S.
1.22136 1.43764 52.993 -91.5648510 0.0
!
1 1.22533 1.33296 58.476 -91.5673097
0.29994
!
2 1.22802 1.23827 64.747 -91.5735346
0.59986
!
3 1.22974 1.16350 72.039 -91.5814775
0.89968
!
$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
1
! EXAM 12.
!
This job illustrates linear bends, for acetylene, and
!
tests the closed shell LDA density functional program.
!
!
At the input geometry,
!
the FINAL E= -76.5352218424 after 12 iterations,
!
and the RMS gradient is 0.0944646.
!
!
At the final geometry, 5 steps later,
!
the FINAL E= -76.5841366158, RMS gradient=0.0000007,
!
R(CC)=1.2119150 and R(CH)=1.0779838.
!
$CONTRL
SCFTYP=RHF RUNTYP=OPTIMIZE NZVAR=5 $END
$SYSTEM
TIMLIM=20 MEMORY=500000 $END
$DFT
DFTTYP=SVWN $END
$BASIS
GBASIS=N31 NGAUSS=6 NDFUNC=1 $END
$GUESS
GUESS=HUCKEL $END
$STATPT
OPTTOL=0.00001 $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
! EXAM 13.
! This
run duplicates the POLYATOM calculation of
! D.Neumann
+ J.W.Moskowitz, J.Chem.Phys. 49,2056(1968)
! SCF
convergence is a bit better today, so some of the
! results
are not precisely the same.
!
!
V(NE) = -199.1343264099
!
V(EE) = 37.8955167210 T
= 75.9557584991
!
V(NN) = 9.2390200836 E(TOT)= -76.0440311061
! Mulliken
charge(O)=-0.647397 Bond Order=0.905
! Density:
O=286.491824 H=0.404989
! Moments:
DZ= 2.093290
!
QXX=-2.388658 QYY= 2.495388 QZZ=-0.106730
!
OXXZ=-0.890362 OYYZ= 2.186853 OZZZ=-1.296490
! Electric
field/gradient: H(YZ)=+/-0.365168
!
O(Z)=-0.060033 H(Y)=+/-0.006572 H(Z)=0.001233
!
O(XX)=1.904867 O(YY)=-1.735891 O(ZZ)=-0.168977
!
H(XX)=0.301208 H(YY)=-0.258153 H(ZZ)=-0.043055
! Potential:
V(O)=-22.330374 V(H)=-1.006648
!
EXAM13 is continued on the next page...
1
$CONTRL
SCFTYP=RHF RUNTYP=ENERGY UNITS=BOHR ISPHER=1 $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
$END
1
! 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 <Q>=0.392614
! Dipole
VELOCITY is <d/dQ>=0.368205
!
$CONTRL
SCFTYP=NONE CITYP=GUGA RUNTYP=TRANSITN UNITS=BOHR $END
$SYSTEM
TIMLIM=1 MEMORY=100000 $END
$BASIS
GBASIS=STO NGAUSS=3 $END
!
standard SD-CI calculation
$DRT1
GROUP=C2V IEXCIT=2 NFZC=1 NDOC=4 NVAL=2 $END
$TRANST
NFZC=1 IROOTS(1)=2 $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
--- 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
1
! 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
1
! 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
1
! 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
1
! EXAM 18.
! effective
core potential...diatomic P2...RHF/CEP-31G*
! See
Stevens,Basch,Krauss, J.Chem.Phys. 81,6026-33(1984).
! GAMESS
FINAL E= -12.6956518702, FREQ=913.17
! A separate
run gives E(P)= -6.32635, so De= 26.95 kcal/mol
!
$CONTRL
SCFTYP=RHF RUNTYP=HESSIAN ECP=SBKJC NZVAR=1 $END
$SYSTEM
TIMLIM=15 MEMORY=900000 $END
$GUESS
GUESS=HUCKEL $END
$ZMAT
IZMAT(1)=1,1,2 $END
$DATA
diatomic phosphorous
Dnh 4
PHOSPHORUS 15.0
0.0000000000 0.0000000000 0.9393077548
SBKJC
D 1
1 0.45 1.0
$END
1
! EXAM 19.
! Spin-orbit
coupling example.
! This run duplicates
the results shown in Table 3 of
! T.R.Furlani,
H.F.King, J.Chem.Phys. 82, 5577-83(1985),
! GAMESS 1e-=
114.3851, 2e-= -49.4168, lit=114.38,-49.42
!
! Energies for
the singlet CI are
!
State= 1 Energy = -54.868531216 (1-delta)
!
State= 2 Energy = -54.868531216 (1-delta)
!
State= 3 Energy = -54.798836731 (1-sigma-plus)
! Energies for
the triplet CI are
!
State= 1 Energy = -54.938225701 (3-sigma-minus)
! Final energies
of all 6 levels in the pi**2 configuration,
! after diagonalization
of the spin-orbit Hamiltonian, are
!
BREIT RELATIVE E= -15296.570, -15296.432, -15296.432,
!
BREIT RELATIVE E= 0.0, 0.0, and +15296.570 wavenumbers.
! If run as
OPERAT=HSO1, with ZEFF taken as true atomic Z,
! then inclusion
of only the 1e- operator is 114.3851, and
!
ZEFF RELATIVE E= -15296.859, -15296.432, -15296.432,
!
ZEFF RELATIVE E= 0.0, 0.0, and +15296.859 wavenumbers.
!
! Why are there
six levels? The singlet-delta is two roots,
! the singlet-sigma-plus
is a third. During the CI, the
! spatial 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, namely adding the
! ab+ba and
beta/beta triplets. So, 2+1+3=6 levels. You
! can work out
for yourself these have the quantum number
! omega=0,0,1,2.
Only the omega=0 states can interact,
! raising the
triplet's degeneracy and slightly affecting
! the singlet-sigma-plus
state's position.
!
! Note that
the lower multiplicity DRT1 is done in C1
! symmetry to
generate both components of the delta state.
!
$CONTRL
SCFTYP=NONE MULT=3 CITYP=GUGA RUNTYP=TRANSITN
UNITS=BOHR $END
$SYSTEM
TIMLIM=2 MEMORY=900000 $END
$BASIS
GBASIS=N31 NGAUSS=6 $END
$TRANST
OPERAT=HSO2 NFZC=3 NOCC=5 NUMVEC=1 NUMCI=2
IROOTS(1)=3,1 $END
$DRT1
GROUP=C1 IEXCIT=2 NFZC=3 NDOC=1 NVAL=1 $END
$DRT2
GROUP=C4V IEXCIT=2 NFZC=3 NALP=2 $END
$DATA
Imidogen radical
Cnv 4
Nitrogen
7.0
Hydrogen
1.0 0.0 0.0 1.9748
$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
...orbitals
omitted to save space...
$END
1
! EXAM 20.
! Optimize
an orbital exponent.
! The
SBKJC 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=SBKJC $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
SBKJC
L 1
1 0.02 1.0
$END
1
! 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
1
... EXAM21 continued from previous page
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
! 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
1
! EXAM23.
!
semiempirical calculation, using the MOPAC/GAMESS combo
!
AM1 gets the geometry disasterously wrong!
!
!
initial geometry,
MNDO AM1
PM3
!
FINAL HEAT OF FORMATION 105.14088 93.45997
46.89387
!
RMS gradient
0.0818157 0.1008587 0.0366232
!
final geometry (# steps), 8
11 10
!
FINAL HEAT OF FORMATION 46.45649 -1.81716
-2.79647
!
RMS gradient
0.0000246 0.0000294 0.0000015
!
r(SiH)
1.42117 1.45813 1.52104
!
a(HSiH)
101.962 120.000 96.280
!
!
At the final PM3 geometry, the charge on Silicon is -.4681,
!
and the dipole moment is 2.345322 Debye.
!
$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
1
! 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
1
! 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 RMS final E final RMS #steps
!
Cartesians 0.0200113 -48.7022520
0.0000304 50
!
dangling Z-mat 0.0600637 ... OO bond crashes on 1st step
!
good Z-matrix 0.0232915 -48.7022510 0.0000285
21
!
deloc. coords. 0.0176452 -48.7022537 0.0000267
22
!
nat. internals 0.0209442 -48.7022570 0.0000183
15
!
$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
1
...EXAM25 continues
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 a dangling O-O bond:
--- $contrl
nzvar=33 $end
To use well
chosen internals, with all 5 ring bonds defined:
--- $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 delocalized
coordinates:
--- $contrl
nzvar=33 $end
--- $zmat
dlc=.true. auto=.true. $end
1
...EXAM25 continues
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
1
! 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
1
! 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
1
! EXAM28.
Morokuma energy decomposition.
! This run duplicates
a result from Table 16 of
! H.Umeyama,
K.Morokuma, J.Am.Chem.Soc. 99,1316(1977)
!
!
GAMESS literature
!
ES= -14.02 -14.0
!
EX= 8.98 9.0
!
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
1
! 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= -229.0368324615, E(MP2)= -229.3873302375
! FINAL
E= -229.0356378402, E(MP2)= -229.3866642673
! FINAL
E= -229.0309266321, E(MP2)= -229.3822094777
! FINAL
E= -229.0372146702, E(MP2)= -229.3923234074
! FINAL
E= -229.0385440296, E(MP2)= -229.3936486644
! FINAL
E= -229.0367369562, E(MP2)= -229.3913683073
! FINAL
E= -229.0328601144, E(MP2)= -229.3918932009
! FINAL
E= -229.0364643934, E(MP2)= -229.3948325500
! FINAL
E= -229.0372478250, E(MP2)= -229.3943498144
!
! 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 memddi=2 $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
1
!
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
1
! EXAM31.
! Water
in PCM water...RHF geometry optimization
!
! FINAL
E= -74.9677081860, 12 iters, RMS GRAD = 0.0321490
! FINAL
E= -74.9494679452, 8 iters, RMS GRAD = 0.0647800
! FINAL
E= -74.9690975801, 7 iters, RMS GRAD = 0.0238593
! FINAL
E= -74.9712684509, 7 iters, RMS GRAD = 0.0081236
! FINAL
E= -74.9715453381, 8 iters, RMS GRAD = 0.0018667
! FINAL
E= -74.9714737628, 7 iters, RMS GRAD = 0.0049760
! FINAL
E= -74.9715558829, 7 iters, RMS GRAD = 0.0000526
! FINAL
E= -74.9715559513, 4 iters, RMS GRAD = 0.0000260
!
! The
final geometry is not symmetric,
!
O
!
H 1 .9873716
!
H 1 .9872902 2
100.0327788
!
!
----------------------------------------------
!
------- RESULTS OF PCM CALCULATION -------
!
----------------------------------------------
!
!
FREE ENERGY IN SOLVENT
= -74.9715559513 A.U.
!
INTERNAL ENERGY IN SOLVENT =
-74.9655730812 A.U.
!
DELTA INTERNAL ENERGY
= .0000000000 A.U.
!
ELECTROSTATIC INTERACTION =
-.0059828701 A.U.
!
PIEROTTI CAVITATION ENERGY =
.0000000000 A.U.
!
DISPERSION FREE ENERGY
= .0000000000 A.U.
!
REPULSION FREE ENERGY
= .0000000000 A.U.
!
TOTAL INTERACTION
= -.0059828701 A.U.
!
TOTAL FREE ENERGY IN SOLVENT =
-74.9715559513 A.U.
!
$contrl
scftyp=rhf runtyp=optimize coord=zmt $end
$system
memory=300000 $end
$basis
gbasis=sto ngauss=3 $end
$guess
guess=huckel $end
$pcm
solvnt=water $end
$data
a water molecule
solvated by PCM water
Cnv 2
O
H 1 rOH
H 1 rOH 2 aHOH
rOH=0.95
aHOH=104.5
$end