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,