11
$\begingroup$

I am trying to simulate the excitation state of acetone. I ran TDDFT for it both in gas phase and solvated state in water (both implicit and explicit water).

The experimental data say that acetone undergoes n->π* transition, which means it has a higher wavelength in gas phase ($\mathrm{\lambda_{max}}=$~$276~\mathrm{nm}$) and a shorter wavelength in the solvated state ($\mathrm{\lambda_{max}}=265~\mathrm{nm}$). I started with gas phase and expected something a wavelength of about $276~\mathrm{nm}$ ($4.49~\mathrm{eV}$ ), but surprisingly I got $\mathrm{\lambda_{max}}=136.172~\mathrm{nm}$ ($9.326~\mathrm{eV}$)! I really cannot understand why there is such a big discrepancy!

Here is my GAMESS input file. What is wrong with this molecule ?

!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NGAUSS=6 $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END
 $CONTRL ICHARG=0 MULT=1 $END
 $TDDFT NSTATE=9 $END
 $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END
 $SYSTEM MWORDS=1000 PARALL=.TRUE. $END
 $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END

 $DATA 
Title
C1
O     8.0     0.00000    -1.27900     0.00300
C     6.0    -0.00000    -0.05800     0.00100
C     6.0     1.29700     0.69100    -0.00000
C     6.0    -1.29800     0.69000    -0.00000
H     1.0     1.35900     1.32900    -0.90600
H     1.0     1.35900     1.33200     0.90300
H     1.0     2.15700    -0.01300     0.00100
H     1.0    -2.15700    -0.01400     0.00100
H     1.0    -1.35900     1.32900    -0.90600
H     1.0    -1.35900     1.33200     0.90300

 $END
$\endgroup$
  • $\begingroup$ 1) Is geometry optimized (at the same level of theory)? 2) Where is the output? 3) B3LYP is not the best choice for TD-DFT, basis set could also be better. $\endgroup$ – Wildcat May 12 '15 at 6:06
  • $\begingroup$ In the mean time, I suggest first to try the PBE0/6-311++G(d,p) level of theory instead of B3LYP/6-311G. $\endgroup$ – Wildcat May 12 '15 at 6:20
  • $\begingroup$ I did a few calculations: pbe0/def2tzvpp 143 nm; df-bp86/def2tzvpp 158 nm; pbe0/6-311++g(df) 147 nm. @Wildcat While I agree, that B3LYP is not the best choice, it seems like this is not a problem of only this method. $\endgroup$ – Martin - マーチン May 12 '15 at 10:07
  • $\begingroup$ @Martin-マーチン, ok. But now we are certain that it is not the failure of B3LYP. :) $\endgroup$ – Wildcat May 12 '15 at 10:16
  • $\begingroup$ @Wildcat Usually I like pointing out the problems of B3LYP, but in this case it seems to behave normally. Even stranger CIS//BP86/def2TZVPP gives $\pi\rightarrow\pi^*$ with 130 nm. $\endgroup$ – Martin - マーチン May 12 '15 at 11:07
8
$\begingroup$

It seems that the results of the calculations are more or less fine and the OP just misinterpreted the NIST data. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper[1] referenced on the NIST page it is said (emphasis mine):

There are two diffuse ultraviolet bands; the first at 2800 Å. is very weak with its oscillator strength $f \sim 0.0004$ and the second at about 1900 Å is moderately intense with a maximum extinction coefficient $\epsilon_{\mathrm{m}} \sim 1000$. McMurry has identified the 2800 Å. band as a forbidden $\pi^* \leftarrow n$ transition involving excitation of a non-bonding $\ce{O}$ electron to an anti-bonding $\pi$ orbital between the $\ce{C}$ and $\ce{O}$ of the carbonyl group.

A high-resolution photoabsorption spectrum in the energy range 3.7-10.8 eV can be found, for instance, in this recent study[2] and is more or less consistent with the result of the OP's calculations.

  1. Noel S. Bayliss, Eion G. McRae, J. Phys. Chem., 1954, 58 (11), 1006–1011.
  2. M. Nobre, A. Fernandes, F. Ferreira da Silva, R. Antunes, D. Almeida, V. Kokhan, S. V. Hoffmann, N. J. Mason, S. Eden, P. Limão-Vieira, Phys. Chem. Chem. Phys., 2008, 10, 550-560. (available at researchgate.net)
$\endgroup$
  • 1
    $\begingroup$ Interesting indeed. || I have included the complete references to the papers you cited, I find it better to cite like journals would do. $\endgroup$ – Martin - マーチン May 13 '15 at 5:32
2
$\begingroup$

this is your output, right?

TRANSITION DIPOLE, A.U.  OSCILLATOR
                    HARTREE          EV         X       Y       Z     STRENGTH
   0  A         -193.0290234748    0.000
   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000
   2  A         -192.7831335626    6.691     0.4051 -0.0004 -0.0000    0.027
   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002
   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042
   5  A         -192.7210047911    8.382    -0.0121 -0.0000 -0.0006    0.000
   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001
   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000
   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022
   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290

I am confused, why do you take the 9ths excitation and assume it is the experimentally measured one? The first excitation agrees reasonably quite well with the experimental data.

If you look at the exp spectrum eg here http://www.sciencedirect.com/science/article/pii/S0022407308000307 you can see that there is a 2nd absorption band below 200nm (6eV). This seems to be the 2nd excitation.

Did you look at the orbitals involved in the first excitation? I would assume it shows a n-pi* transition and then your are all good.

edit:

This paper contains some references (i cannot access the cited references therein)

They report the following

The n → π* electronic transition of acetone corresponds to the excitation of an electron from the nonbonding 2py orbital on the oxygen to the antibonding π* molecular orbital on the carbonyl. For an acetone molecule in its C2 or C2v geometry, the transition is symmetry-forbidden; as a consequence of vibronic coupling, in the gas phase, a weak band at ΔE = 4.4−4.5 eV is observed, with an oscillator strength of f ≈ 0.0004.

Do we see that in the TD-DFT output? Mostly, yes! The oscillator strength is smaller than the printout (bad program), but if you use e.g. ORCA you get a non-zero oscillator strength. It is still much smaller than the reported value but that is somewhat expected from TD-DFT in this case (partially CT-state!) The small oscillator strength is troublesome, but the excitation energy is ok. It could indicate a ghost state (=purely artificial state), but these are not often seen in hybrid-functional TD-DFT calculations. Likely all states are physical.

The first transition is a HOMO-LUMO transition and if you bother to look at the orbital it is a nice n-pi* transition as expected. HOMOLUMO

$\endgroup$
  • 4
    $\begingroup$ The first excitation has a strength of 0.000 and can therefore never be $\lambda_\mathrm{max}$. It also corresponds to the HOMO LUMO excitation, which is $\sigma\rightarrow\pi^*$, i.e. orthogonal orbitals. I am sorry, but I need to downvote this answer. $\endgroup$ – Martin - マーチン May 12 '15 at 13:09
  • $\begingroup$ @Martin-マーチン , so which one do you select as λmax ? is it just based on the largest oscillator strength or something else shoul be mentioned ? $\endgroup$ – Aug May 12 '15 at 14:14
  • 2
    $\begingroup$ @Aug, photoabsorption intensities are indeed directly related to TD-DFT oscillator strengths, so yes, the transition with the largest oscillator strength corresponds to the most intense line in the absorption spectrum. $\endgroup$ – Wildcat May 12 '15 at 14:47
  • $\begingroup$ @Martin-マーチン, There is something really interesting in your comment that can help me a lot: you said its HOMO-LUMO transition is σ -> π* which is definitely true, but how did you realize that? was it just based on experimental knowledge or there is something in the output file that you noticed ? This can be answer to my other question here ( I had to put a bounty on it becuase there is no satisfactory answer yet !): chemistry.stackexchange.com/questions/29464/… $\endgroup$ – Aug May 12 '15 at 18:52
  • $\begingroup$ @Aug The output gives you the orbitals that correspond to the transition. You just have to look at them, and you can identify them. It will get much harder for open shell problems, so natural transition orbitals give you a qualitative picture. $\endgroup$ – Martin - マーチン May 13 '15 at 5:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.