From owner-chemistry@ccl.net Fri Jun 9 09:53:00 2017 From: "Ulrike Salzner salzner*_*fen.bilkent.edu.tr" To: CCL Subject: CCL: symmetry breaking Message-Id: <-52832-170609085932-23714-hM0A5thQV7mCOX3UXEbo3Q*o*server.ccl.net> X-Original-From: Ulrike Salzner Content-Type: multipart/alternative; boundary="94eb2c0877fcffbed80551868906" Date: Fri, 9 Jun 2017 15:59:25 +0300 MIME-Version: 1.0 Sent to CCL by: Ulrike Salzner [salzner###fen.bilkent.edu.tr] --94eb2c0877fcffbed80551868906 Content-Type: text/plain; charset="UTF-8" Hello, I am comparing three large conjugated pi-systems with density functional theory (global and range-separated hybrids). The difference between them is that one contains phenyl rings, one thiophene, and the last one selenophene. All three of molecules have UDFT instability with B3P86 with 30% HF exchange. With open-shell singlet calculations and 50:50 HOMO-LUMO mixing, the phenyl and seleophene systems converge to open-shell biradical ground states that lie 0.02 and 0.12 eV below the closed-shell states. The thiophene system does not break the symmetry. This is odd because selenophene and thiophene system behave usually very similar. In addition, excited calculations with UTDDFT give negative triplet excitation energies for all of them when the closed-shell wave are used. As triplet excitation energies are probably underestimated with UTDDFT, I checked the ground state triplet energies. The triplets lie 1.24 eV (phen), 0.79 eV (thio), and 0.61 eV (seleno) above the closed-shell singlets. Again there is no reason why the thiophene system should differ > from the other two. Is there any way other way to find the broken the symmetry ground state? Or - is the lack of a broken symmetry solution after 50:50 HOMO-LUMO mixing sufficient to prove that the ground state is closed-shell? Thanks for suggestions, Ulrike -- Assoc. Prof. Ulrike Salzner Department of Chemistry Bilkent University 06800 Bilkent, Ankara --94eb2c0877fcffbed80551868906 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable
Hello,
I am comparing three large conju= gated pi-systems with density functional theory (global and range-separated= hybrids). The difference between them is that one contains phenyl rings, o= ne thiophene, and the last one selenophene.=C2=A0

All three of mole= cules have UDFT instability with B3P86 with 30% HF exchange.

With o= pen-shell singlet calculations and 50:50 HOMO-LUMO mixing, the phenyl and s= eleophene systems converge to open-shell biradical ground states that lie = 0.02 and 0.12 eV below the closed-shell states. The thiophene system does n= ot break the symmetry. This is odd because selenophene and thiophene system= behave usually very similar.

In addition, excited calcu= lations with UTDDFT give negative triplet excitation energies for all of th= em when the closed-shell wave are used.

As triplet excitation energi= es are probably underestimated with UTDDFT, I checked the ground state trip= let energies.=C2=A0 The triplets lie 1.24 eV (phen), 0.79 eV (thio), and 0.= 61 eV (seleno) above the closed-shell singlets. Again there is no reason wh= y the thiophene system should differ from the other two.

= Is there any way other way to find the broken the symmetry ground state? Or - is the lack o= f a broken symmetry solution after 50:50 HOMO-LUMO mixing sufficient to pro= ve that the ground state is closed-shell?

Thanks for sugg= estions,
Ulrike
--
Assoc. Prof. Ulrike Salzner
Department of Chemistry
Bilkent Un= iversity
06800 Bilkent, Ankara
--94eb2c0877fcffbed80551868906--