(26 Nov 96)
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.
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
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
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
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 $ENDnormally a GVB-PP calculation will use GUESS=MOREAD
$GUESS GUESS=HUCKEL $END
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 $ENDlook at all state symmetries, by using C1 symmetry
$CIDRT GROUP=C1 IEXCIT=2 NFZC=1 NDOC=3 NVAL=3 $ENDground state is 1-A-1, 1st excited state is 1-B-1
$GUGDIA NSTATE=2 $ENDcompute properties of the 1-B-1 state
$GUGDM NFLGDM(1)=1,1 IROOT=2 $ENDcompute 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
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
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
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
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
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
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
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
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
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
$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
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
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
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
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
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
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 $ENDsinglet-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 $ENDSince 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
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
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
GAMESS obtains
$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
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
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 | 0.0000270 | 0.0000294 | 0.0000015 |
r(SiH) | 1.42119 | 1.45813 | 1.52104 |
a(HSiH) | 101.956 | 120.000 | 96.280 |
$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
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
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 | 0.0000278 | 50 |
dangling Z-mat | 0.0600637 | ... | OO | bond crashes on 1st step |
good Z-matrix | 0.0232915 | -48.7022508 | 0.0000299 | 20 |
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 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
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
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
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
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
The scan is done over a 3x3 grid centered on the SCF
transition state for the SN2 type reaction
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
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