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Alex Granovsky

gran@classic.chem.msu.su

Hello,

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,

Wed May 28 '14 3:49pm

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