Here is modified input from first tutorial. The number of fragments was decreased from 3 to 2 for the purpose of avoidance of dividing of C-C bond in ethanol. That was made for simplicity.
So, here we've got alcohol and water as two fragments. Suddenly, an electron have passed away, therefore the cation-radical was formed (multiplicity=2, charge=+1). As I had some concerns about ability to calculate gradients with open-shell structures, I've decided to move from energy calculation to RUNTYP=OPTFMO.
$CONTRL SCFTYP=ROHF ICHARG=+1 MULT=2 RUNTYP=OPTFMO $END
$SYSTEM MWORDS=20 $END
$BASIS GBASIS=STO NGAUSS=3 $END
$FMOPRP MODORB=3 $END
$data
C2H5OH+H2O
c1
h 1
c 6
o 8
$end
$FMOXYZ
C 6 2.3410689175 -0.2869692888 -0.0074194092
H 1 3.0745859649 0.3772736987 0.4397744143
H 1 2.5665310430 -0.3924000324 -1.0640918137
H 1 2.4261794556 -1.2632979826 0.4595623356
C 6 0.9166901279 0.2761650904 0.1831975319
H 1 0.7235442032 0.4041423414 1.2567611875
H 1 0.8641656999 1.2758468598 -0.2685095421
O 8 -0.0215616632 -0.6201531625 -0.4156796115
H 1 -0.9026816335 -0.1944297425 -0.2534321184
O 8 -2.4493614824 0.5180105259 0.0102319306
H 1 -2.9309841137 0.6564728575 -0.8399969145
H 1 -3.0583517680 -0.1059613981 0.4726454459
$end
$fmo
modgrd=0
nfrag=2 indat(1)=1,1,1,1, 1,1,1,1,1, 2,2,2 $end
As for me, this input works. If we compare the energy of this system before and after electron withdrawal, we'll be able to have ionization energy. Unfortunately, as it depends on the basis set, it is useless as "thing-in-itself".
