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谢谢sob老师！请问一下sob老师我用qmmm计算重整能，分成两层，输入文件如下
%chk=13E1.chk
%nprocshared=8
# p oniom(b3lyp/6-311G**:amber=hardfirst) nosymm geom=connectivity iop(2/15=3) test opt=quadmac
remark line goes here
0 1 0 1 0 1
C-ca--0.136271 -1 4.74000000 8.42000000 0.17500000 L
C-ca--0.217096 -1 5.21000000 7.46000000 1.07500000 L
H-ha-0.191740 -1 4.46000000 6.94000000 1.67500000 L
H-ha-0.143885 -1 3.67000000 8.56000000 0.05500000 L
C-ca--0.095101 -1 5.57000000 9.28000000 -0.49500000 L
H-ha-0.137482 -1 5.19000000 10.11000000 -1.07500000 L
C-ca--0.235974 -1 6.96000000 9.13000000 -0.31500000 L
H-ha-0.163227 -1 7.57000000 9.82000000 -0.88500000 L
C-ca-0.068796 -1 7.48000000 8.17000000 0.53500000 L
C-ca-0.048744 -1 8.89000000 7.92000000 0.66500000 L
C-ca-0.081330 -1 9.36000000 6.92000000 1.53500000 L
.
.
.
HrmBnd1 ca cc cf 0.0000 0.0000
HrmBnd1 ca cc cc 67.7000 111.0400
HrmBnd1 cc cc cf 65.6000 123.9200
HrmBnd1 cc cf ha 49.3000 119.1600
HrmBnd1 cc cf cd 0.0000 0.0000
HrmBnd1 cf cd ss 0.0000 0.0000
HrmBnd1 cf cd cc 0.0000 0.0000
HrmBnd1 ha cf cd 0.0000 0.0000
HrmBnd1 cd ss cd 67.0000 89.9100
HrmBnd1 cd cc ha 48.4000 122.8900
HrmBnd1 cd cc cc 68.2000 114.1900
HrmBnd1 ss cd cc 63.6000 118.1700
HrmBnd1 ss cd cd 63.8000 115.0200
HrmBnd1 cc cc ss 63.8000 115.0200
HrmBnd1 ha cc cc 46.6000 123.7400
HrmBnd1 cc ss cc 67.0000 89.9100
HrmBnd1 cc cd cd 68.2000 114.1900
HrmBnd1 ss cc cd 63.6000 118.1700
HrmBnd1 ss cc ca 64.9000 111.0900
HrmBnd1 cd cd c3 64.7000 115.9700
HrmBnd1 cd c3 ca 65.2000 108.0800
HrmBnd1 cd cc ca 68.2000 113.5100
HrmBnd1 c3 cd cc 64.8000 119.4500
HrmBnd1 c3 ca ca 63.8000 120.6300
HrmBnd1 ca c3 ca 63.6000 112.4700
HrmBnd1 ca c3 hc 47.0000 110.1500
HrmBnd1 hc c3 hc 39.4000 108.3500
最后Log文件的报错信息如下：
1025 6 1000 1.160000 -3.250000 -19.235000
1026 7 90000002 0.650000 -2.280000 -19.045000
1027 6 1000 2.680000 -4.410000 -20.545000
1028 7 90000002 3.260000 -4.390000 -21.515000
1029 6 20001003 2.010000 -7.000000 -19.085000
1030 6 20001003 1.130000 -7.950000 -18.645000
1031 6 20001003 1.210000 -9.310000 -19.135000
1032 6 20001003 0.180000 -10.260000 -18.925000
1033 6 20001003 0.200000 -11.510000 -19.545000
1034 6 20001003 1.220000 -11.890000 -20.385000
1035 6 20001003 2.260000 -10.960000 -20.555000
1036 1 20001033 3.050000 -11.280000 -21.235000
1037 1 20001033 1.170000 -12.830000 -20.935000
1038 1 20001033 -0.720000 -12.090000 -19.465000
1039 1 20001033 -0.740000 -10.040000 -18.395000
1040 6 20001003 2.330000 -9.690000 -19.915000
1041 6 20001003 3.430000 -8.780000 -20.025000
1042 6 20001003 3.280000 -7.420000 -19.465000
1043 6 20001003 4.400000 -6.580000 -19.425000
1044 1 20001033 4.310000 -5.550000 -19.075000
1045 6 20001003 5.650000 -7.080000 -19.845000
1046 1 20001033 6.490000 -6.390000 -19.905000
1047 6 20001003 5.800000 -8.270000 -20.415000
1048 1 20001033 6.750000 -8.600000 -20.805000
1049 6 20001003 4.710000 -9.160000 -20.515000
1050 1 20001033 4.850000 -10.140000 -20.945000
---------------------------------------------------------------------
Rotational constants (GHZ): 0.0006196 0.0003837 0.0003317
ONIOM: saving gridpoint 17
ONIOM: generating point 3 -- low level on real system.
Standard basis: Dummy (5D, 7F)
1050 basis functions, 1050 primitive gaussians, 1050 cartesian basis functions
833 alpha electrons 833 beta electrons
nuclear repulsion energy 53619.7660872314 Hartrees.
NAtoms= 1050 NActive= 1050 NUniq= 1050 SFac= 1.00D+00 NAtFMM= 60 NAOKFM=T Big=T
Integral buffers will be 131072 words long.
Raffenetti 1 integral format.
Two-electron integral symmetry is turned off.
AMBER calculation of energy and first derivatives.
Atom I= 27 has type= 1000.
MMAtmC is confused about an atom type.
Error termination via Lnk1e in /opt/soft/g09/G09E/g09e/l402.exe at Wed Jan 2 23:00:29 2019.
根据这个报错信息，找到是氰基里面的碳原子（gaff力场定义为ch）有问题,请问一下我可以怎么改
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发表于 Post on 2019-1-21 14:53:45
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发表于 Post on 2019-1-21 15:34:02
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