D19=-111.54855725 D20=-163.81739212 D21=-177.01784202 D22=116.24854025 D23=84.73759842 D24=117.02628819 D25=-54.44419854 D26=120.14740904 D27=-120.73901029 D28=-120.36008831 D29=64.32343577 D30=119.79274556 D31=-121.08069298 D32=-179.41915959 D33=0.20082332 D34=-179.92188311 D35=-0.19458212
优化正常结束，频率计算时出错. .out文件末尾显示： Error #1 in AlGdDF.
Error termination via Lnk1e in d:\\chemistry\\gaussian98w\\l1002.exe. Job cpu time: 0 days 2 hours 34 minutes 4.0 seconds.
File lengths (MBytes): RWF= 129 Int= 0 D2E= 0 Chk= 13 Scr= 1
1。上面我贴的那个问题经过我将内存设定从256M改为192M后频率计算正常结束了。为什么你会想到把内存变小？ 2。你试试不设定内存看看。我觉得你的分子较大，算频率需要内存应该很多，所以不如不设。
输入文件： %chk=t_d %mem=250mb
# rb3lyp/gen opt=(calcfc,ts,noeigentest) freq scf=maxcyc=500 pseudo=read
C60O C2v Epoxide PM3 Optimization 0,1
74 0 0.248370 0.014156 0.033227 6 0 0.403673 -0.936350 1.733392 6 0 2.200917 0.171752 -0.336405 6 0 -0.270860 2.061381 -0.203209 6 0 -0.180546 -1.298567 -1.594701 1 0 0.913675 -1.906338 1.762663 1 0 -0.056469 -0.642812 2.681098 1 0 2.947154 -0.605352 -0.128354 1 0 2.704061 1.060569 -0.742512
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1 0 -0.333335 2.571891 0.768423 1 0 -1.257830 2.123101 -0.686918 1 0 0.480105 2.561522 -0.830148 1 0 0.004133 -0.784037 -2.549068 1 0 -1.248033 -1.562489 -1.535515 1 0 0.435959 -2.206120 -1.587137 8 0 -2.716252 -0.092883 -0.267346 8 0 -1.769735 -0.112964 0.616623 C H O 0 6-31+G(d) **** W 0 sddall **** W 0 sddall
--Link1-- %NoSave %Chk=t_d %mem=250mb
#T b3lyp/gen irc=(rcfc,stepsize=20) geom=check guess=read pseudo=read
C60O Cs Open Isomer PM3 Optimization 0,1 C H O 0 6-31+G(d,p] **** W 0 sddall **** W 0 sddall
错误信息：过渡态已经找到，在计算IRC中，
Rotational constants (GHZ): 2.3655799 1.4105816 1.3561379
Isotopes: W-184,C-12,C-12,C-12,C-12,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H-1,O-16,O-16 QPERR --- A SYNTAX ERROR WAS DETECTED IN THE INPUT LINE. 6-31+G(d,p]
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Last state=\
TCursr= 1032 LCursr= 10
Error reading general basis specification. Error termination via Lnk1e in F:\\G98W\\l301.exe. Job cpu time: 0 days 0 hours 0 minutes 4.0 seconds.
File lengths (MBytes): RWF= 16 Int= 0 D2E= 0 Chk= 12 Scr= 1 请帮忙！！ 基组错误：6-31+G(d,p]，改成6-31+G(d,p)
个人感觉，做计算还是linux下比较好。windows下的计算我只做小的test，所以没有大体系的经验。不过看别人的帖子，好像主要是内存和硬盘文件分割的问题，你要是能试出解决办法就好了，这样也为大家总结了经验。 输入文件为：%chk=qwq %mem=200mb
# b3lyp/sto-3g opt scf(maxcyc=200)
calculate 0 2
Cu 0-0.6710.9510.000 O -1-0.6712.7810.000 O -1-0.671-1.0090.000 O -1-0.6711.1391.951 O -1-0.6710.763-1.951 C 01.2690.9510.000 C 02.039-0.3810.064 C 01.269-1.7130.128 H 01.2691.0021.069 H 01.8041.876-0.044 H 01.2690.899-1.069 H 02.671-0.3400.926 H 02.671-0.423-0.798 H 02.302-1.9900.141 H 00.270-2.0970.146 H 01.325-2.7810.179 O -1-2.6710.9510.000 输出文件为：
Standard orientation:
--------------------------------------------------------------------- Center Atomic Atomic Coordinates (Angstroms) Number Number Type X Y Z --------------------------------------------------------------------- 1 29 0 0.589550 -0.004850 0.014248 2 8 0 1.785028 1.345274 0.325533 3 8 0 -0.690853 -1.450884 -0.319149 4 8 0 0.702649 0.580333 -1.852973
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5 8 0 0.476451 -0.590033 1.881469 6 6 0 -0.879246 1.227196 0.311403 7 6 0 -2.332692 0.748137 0.140471 8 6 0 -2.620186 -0.708937 -0.266347 9 1 0 -0.851253 1.509461 -0.720539 10 1 0 -0.679810 2.239331 0.593525 11 1 0 -0.907893 0.944194 1.343175 12 1 0 -2.788695 1.377020 -0.594863 13 1 0 -2.834331 0.921252 1.069246 14 1 0 -3.583302 -0.254293 -0.167892 15 1 0 -2.114776 -1.622563 -0.502207 16 1 0 -3.360528 -1.449643 -0.489082 17 8 0 2.103773 -1.275000 -0.292097 --------------------------------------------------------------------- Rotational constants (GHZ): 1.8760398 1.0372847 0.9889064
Isotopes: Cu-63,O-16,O-16,O-16,O-16,C-12,C-12,C-12,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H -1,O-16
Standard basis: STO-3G (5D, 7F)
There are 66 symmetry adapted basis functions of A symmetry. Crude estimate of integral set expansion from redundant integrals=1.000. Integral buffers will be 262144 words long. Raffenetti 2 integral format.
Two-electron integral symmetry is turned on. 66 basis functions 201 primitive gaussians 48 alpha electrons 47 beta electrons
nuclear repulsion energy 822.6634918295 Hartrees. One-electron integrals computed using PRISM. NBasis= 66 RedAO= T NBF= 66 NBsUse= 66 1.00D-04 NBFU= 66 Projected Huckel Guess. of initial guess= 0.7500
Requested convergence on RMS density matrix=1.00D-08 within 200 cycles. Requested convergence on MAX density matrix=1.00D-06. Virtual orbitals will be shifted by 0.200 hartree.
Keep R1 and R2 integrals in memory in canonical form, NReq= 5677382. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation. Restarting incremental Fock formation.
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Restarting incremental Fock formation. >>>>>>>>>> Convergence criterion not met.
SCF Done: E(UB+HF-LYP) = -2003.48215555 A.U. after 201 cycles Convg = 0.1927D+00 -V/T = 1.9356 S**2 = 0.7500
Annihilation of the first spin contaminant:
S**2 before annihilation 0.7500, after 0.7500 Convergence failure -- run terminated.
Error termination via Lnk1e in F:\\G98W\\l502.exe. Job cpu time: 0 days 0 hours 25 minutes 31.0 seconds.
File lengths (MBytes): RWF= 6 Int= 0 D2E= 0 Chk= 1 Scr= 1 请问是拿出错了？是不是初始构行不好？
有时自旋多重度的设定也会影响收敛，你是不是换换多重度的数值试试
输入文件为：%chk=qwq %mem=200mb
# b3lyp/sto-3g opt scf(maxcyc=200) ...
先试试scf=qc看看
我认为你可以尝试把SCF的收敛标准降低，比如5，然后在此结果的基础上，恢复SCF的默认收敛标准，继续计算。
下面引用由zjx1103在 2004/04/27 09:48am 发表的内容： 输入文件为：%chk=qwq %mem=200mb
# b3lyp/sto-3g opt scf(maxcyc=200) ...
对Cu用sto－3g？ 可行吗？可信吗？
下面引用由helpme在 2004/05/08 09:01am 发表的内容： 先试试scf=qc看看
初始构型太差的缘故，建议调整构型。
我的输入 %chk=al
#hf/6-31++g(3df) opt=(qst3,bimolecular) Guess=Alter freq test 输出
Rotational constants (GHZ): 1.5082594 0.6204235 0.5203045 我的输入应该怎样改阿？还望各位高手们不吝赐教
GAUSSIAN输出文件中出错信息大汇集
G98的出错讯息都在用户手册中的‘Gaussian 98的链接’一节中就给出了，以前版面上也曾有人贴出来过，你仔细看看。 计算千差万别，出错的原因也各有不同，一时是总结不了的。但大部分都是： 1。输入有误，这没话说了。一般都是L1，L101出错
2。基组有误。特别是刚开始用自定义基组的，或者不知道基组适合范围的，这是L301出错
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