3
$\begingroup$

I do TD-DFT studies in Orca 5.0.3 on porphyrin system and hence at the very beginning I need to pick functional that reproduce excitation energies.

The experimental reported values equal: 691, 656, 430 and 351 nm. I tried insofar two approaches: B3LYP/def2TZVP//B3LYP/def2SVP and B3LYP/def2TZVP, both without TDA.

Unfortunatelly, the "SPIN ORBIT CORRECTED ABSORPTION SPECTRUM VIA TRANSITION ELECTRIC DIPOLE MOMENTS" are entirely different.

For the former:

enter image description here

and for the latter

enter image description here

The input I use is:

! B3LYP DEF2-TZVP CPCM(CH2CL2) RI-SOMF(1X)
! MOREAD 
%TDDFT
    NROOTS 4
    DOSOC TRUE
    TDA FALSE
    DONTO TRUE
END
%MOINP "10-iso.old.gbw" 
%MAXCORE 6000 
%PAL NPROCS 24 END
xyzfile 0 1 10-iso.xyz

The system has no metal present.

$\endgroup$
5
  • $\begingroup$ +1 but please replace the screenshots with code blocks and text. $\endgroup$ Feb 4 at 13:12
  • $\begingroup$ Are you saying that you're using different geometries to calculate your energies? $\endgroup$ Feb 5 at 3:16
  • $\begingroup$ @isolatedmatrix No? $\endgroup$
    – farmaceut
    Feb 5 at 7:56
  • 1
    $\begingroup$ @farmaceut Sorry, I've only ever seen method/basis-set//method/basis-set used in the context of calculating energies at one level of theory on a geometry calculated at a different level of theory (energy-theory/energy-bs//geom-theory/geom-bs). If that's not the case here, what does "B3LYP/def2TZVP//B3LYP/def2SVP and B3LYP/def2TZVP" mean? $\endgroup$ Feb 6 at 8:10
  • 1
    $\begingroup$ Do you really need spin-orbit correction? Also, your number of roots is too small, i would do at least 10. Also, TD-DFT values are often shifted a lot compared to experiment. You would have to look at the fosc (oscillator strength) values and compare them to the experimental peak heights to do an approximate assignment. Additionally, try double hybrid functionals like B2PLYP if possible, they give very good results. $\endgroup$
    – S R Maiti
    Feb 7 at 15:43

1 Answer 1

2
$\begingroup$

There are lots of reasons that calculated excited state energies won't match experimental values. Have a look at a good textbook on TD-DFT for why, but in general calculated energies will be higher in energy than the experimental absorption peaks (which is what I think you are comparing to). For emission energies, the calculated vertical excited states can be well off (depending on the Stokes shift).

However, you have calculated some very low-energy excited states with energies well into the IR, which is unusual. Did you optimise the geometry before calculating the excited states? If so, I would check the structure, it is easy to draw a porphyrin ring with the wrong number of hydrogens.

That being said, you results don't look too bad. Remember that TD-DFT will calculate excited states with no (or very small) oscillator strengths that you are unlikely to observe in an experiment. If you consider which of the excited states have 'high' oscillator strengths you have:

735, 532, 499, and 432 nm

Which are in the correct ballpark for the experimentally observed peaks at:

691, 656, 430 and 351 nm

$\endgroup$
3
  • $\begingroup$ Thanks for the answer. The geometries are optimised and the structures are drawn correctly. I would appreciate explanations why you say theoretical are in correct ballpark for the experimental ones as I don't see it at all. Thanks in advance. $\endgroup$
    – farmaceut
    Mar 3 at 9:58
  • $\begingroup$ Ok, I would consider running a freq calc on the geometry to check you're at a real minimum, but otherwise the energies are probably fine. Regarding the comparison, recall that some of the peaks in the expt spectrum arise from vibrational (and other) coupling effects, and will not be visible in a vertical excited states calculation. If you compare the lowest energy peaks (735 Vs 691 nm) the difference is only 0.1 eV, which is not bad. As S R Maiti said, consider calculating more states (20-50 depending on available time) to better model the full absorption spectrum. $\endgroup$
    – leeman
    Mar 3 at 11:24
  • $\begingroup$ Sorry, 10-50 states is what I meant, but as many as you can afford is the main takeaway $\endgroup$
    – leeman
    Mar 3 at 11:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .