[molpro-user] Hello, I want to calculate the CAS(5,5) for CN

[xu_enhua]

Hello, I want to calculate the CAS(5,5) for CN without symmetry.
when C-N bong length is 1.25 Angstrom, the CAS(5,5) state is right, in which the Pz orbital is the open-shell orbital.
However, when the C-N bond length stretches to 1.3 Angstrom, the Px orbital becomes the open-shell orbital.


If I use the symmetry, the CAS(5,5) may be right.
However, I want to read all the MO integrals from FCIDUMP file.
So what shall I do?
Thank you!


This is my input file:


***, CN_13


memory,1000,m
symmetry,nosym
geometry={angstrom;
 C         0.00000000      0.00000000      0.000
 N         0.00000000      0.00000000      1.250
}


basis=cc-pvdz


hf


{multi;
closed,4;
occ,9;
orbprint;
}


memory,1000,m
symmetry,nosym
geometry={angstrom;
 C         0.00000000      0.00000000      0.000
 N         0.00000000      0.00000000      1.300
}


basis=cc-pvdz


{multi;
closed,4;
occ,9;
orbprint;
}


{fci,
orbit,2140.2;
core,0;
wf,2,1,0;
dump
}


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