Ionization potential of Pd using ECP28MDF

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Dear all,
                 I've tried to calculate the ionization potential of Pd using the above-mentioned Stuttgart ECP (J. Chem. Phys. 126, 124101). So far I've calculated the % error as ~7.7%. The implementation of the ECP  is shown below. Can anyone verify that it's correct? Thank you for your help.

ecp
Pd nelec 28

Pd s
2 12.798825 240.262789
2 5.800528 34.729961


Pd p
2 11.874697 56.746929
2 11.474335 113.444417
2 5.515999 9.345639
2 5.248043 18.345447


Pd d
2 8.502212 28.595554
2 7.983324 43.453921
2 3.107628 1.852286
2 2.476734 1.406765


Pd f
2 9.679571 -10.987255
2 9.691349 -14.626190

end

SO

Pd p
2 11.874697 -113.493859
2 11.474335 113.444417
2 5.515999 -18.691279
2 5.248043 18.345447

Pd d
2 8.502212 -28.595554
2 7.983324 28.969281
2 3.107628 -1.852286
2 2.476734 0.937844


Pd f
2 9.679571 7.324837
2 9.691349 -7.313095

END

Regards,
Bryne

  • Bert Forum:Admin, Forum:Mod, NWChemDeveloper, bureaucrat, sysop
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The input is correct.

An error of 7-8% is not that odd. A 5% error with DFT is considered really good.

Bert



Quote:Brynetan Jul 20th 8:58 pm
Dear all,
                 I've tried to calculate the ionization potential of Pd using the above-mentioned Stuttgart ECP (J. Chem. Phys. 126, 124101). So far I've calculated the % error as ~7.7%. The implementation of the ECP  is shown below. Can anyone verify that it's correct? Thank you for your help.

ecp
Pd nelec 28

Pd s
2 12.798825 240.262789
2 5.800528 34.729961


Pd p
2 11.874697 56.746929
2 11.474335 113.444417
2 5.515999 9.345639
2 5.248043 18.345447


Pd d
2 8.502212 28.595554
2 7.983324 43.453921
2 3.107628 1.852286
2 2.476734 1.406765


Pd f
2 9.679571 -10.987255
2 9.691349 -14.626190

end

SO

Pd p
2 11.874697 -113.493859
2 11.474335 113.444417
2 5.515999 -18.691279
2 5.248043 18.345447

Pd d
2 8.502212 -28.595554
2 7.983324 28.969281
2 3.107628 -1.852286
2 2.476734 0.937844


Pd f
2 9.679571 7.324837
2 9.691349 -7.313095

END

Regards,
Bryne

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Thank you for the verification, Bert.

  • Niri Forum:Admin, Forum:Mod, NWChemDeveloper, bureaucrat, sysop
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Hi Bryne,

Which functional did you use for this calculation ?

Best,
-Niri

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ECP28MDF
I used the m06 functional. I have another question : for the Palladium dimer, by using ECP28MDF, NWChem counts the total number of electrons as 36 = 2*(Number of valence electrons=18) whereas for the AuPd system, where I used ECP60MDF for Au and ECP28MDF for Pd, the total number of electrons = 65 = 19(number of valence electrons of Au) + 46(atomic number of Pd). Does NWChem treat the dimer differently from AuPd or did I make a mistake in the code? Thanks.


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