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Alex Granovsky
gran@classic.chem.msu.su

with 6-31G, HOMO is orbital # 19

Kind regards,
Alex Granovsky


On Wed May 28 '14 8:24am, Siddheshwar Chopra wrote
--------------------------------------------------
>Dear Alex,
>A BIG Thanks for that explanation. The problem is solved..
>But could you try with 6-31G also. I got the same problem here too. Please have a look at that.

>Kind Regards,

>On Tue May 27 '14 3:28pm, Alex Granovsky wrote
>----------------------------------------------
>>The relevant lines of output are:

>> TOTAL NUMBER OF SHELLS              =   11
>> TOTAL NUMBER OF BASIS FUNCTIONS     =   50
>> NUMBER OF ELECTRONS                 =   38
>> CHARGE OF MOLECULE                  =    0
>> STATE MULTIPLICITY                  =    1
>>
>> NUMBER OF OCCUPIED ORBITALS (ALPHA) =   19
>> NUMBER OF OCCUPIED ORBITALS (BETA ) =   19
>>
>> TOTAL NUMBER OF ATOMS               =    3
>> THE NUCLEAR REPULSION ENERGY IS      124.9770929777

>>and

>>
>> THE ECP RUN REMOVES   14 CORE ELECTRONS, AND THE SAME NUMBER OF PROTONS.
>>
>> THE ADJUSTED NUCLEAR REPULSION ENERGY=       53.0405413922

>>So to get the HOMO # you should take 19 and then subtract 14/2  

>>Kind regards,
>>Alex Granovsky
>>
>>
>>
>>
>>On Tue May 27 '14 12:34pm, Alex Granovsky wrote
>>-----------------------------------------------
>>>Hi,

>>>You should take into account that ECP removes core electrons.
>>>As a result, HOMO becomes orbital number 12, while LUMO is orbital
>>>number 13. The HOMO-LUMO gap is then ca. 3 eV.

>>>Kind regards,
>>>Alex Granovsky
>>>
>>>
>>>
>>>
>>>On Tue May 27 '14 10:16am, Siddheshwar Chopra wrote
>>>---------------------------------------------------
>>>>Dear Users,
>>>>Please see the following code I am using to run for a single TiO2 molecule. I have used SBKJC and also 6-31G to get almost similar results for HOMO-LUMO gap. I am getting around 0.7 eV. However as far as literature goes, I should have got ~ 3 eV. As it a sngle TiO2 molecule, gap should have opened.

>>>> $SYSTEM MEMORY=34459319 aoints=dist TIMLIM=5295600 KDIAG=-1 $END
>>>> $CONTRL INTTYP=HONDO ICUT=11 ITOL=30 RUNTYP=OPTIMIZE SCFTYP=RHF $END
>>>> $CONTRL DFTTYP=B3LYP COORD=UNIQUE NZVAR=3 $END
>>>> $CONTRL MAXIT=10000 MOLPLT=.t. PLTORB=.t. ECP=SBKJC $END
>>>> $BASIS GBASIS=SBKJC $END
>>>> $STATPT opttol=10E-7 NSTEP=200 Method=GDIIS HSSEND=.TRUE. $END
>>>> $SCF dirscf=.t. fdiff= .f. diis=.t. $END  
>>>> $DFT HFX(1)=0.15 LMAX=41 NRAD=99 $end
>>>> $ZMAT DLC=.t. AUTO=.t. $END
>>>> $SCF NCONV=7 $END
>>>> $DATA
>>>>TiO2
>>>>C1
>>>> TI         22.0  -2.7439906036  -2.7281069598  -0.4118294433
>>>> O           8.0  -3.1097257096  -4.1115914499   0.4209344416
>>>> O           8.0  -3.1706230814  -1.3977023214   0.4765961088
>>>> $END

>>>>Please refer the following reference for the same:

>>>> J. Mater. Chem. A,2014, 2,637

>>>>Please help,

>>>>Kind Regards,