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我投了一篇文章（Cell姊妹刊一区，Chem），利用B3LYP算了基态，利用TD-DFT/B3LYP算了激发态，但是发现算出来的激发态为S7到S0，审稿意见如下：简而言之，1. input构象没有搜索；2. S7到S0违反kasha规则，我想是不是泛函选择的原因，因为我的分子具有PET效应，选择哪种泛函呢，CAM-B3LYP嘛，那是否需要只重新计算激发态而不用计算基态了？
Theinclusion of calculations is great but raises some problems. I am not qualifiedto review this section and a reviewer who is should be identified (I willsuggest some names to the editor). Two key problems jump out though. Firstly, itis not clear if a thorough conformational search was carried out prior to DFToptimisation. Even though x-ray data is available for some compounds, this doesnot represent the solution state structure. DFT will only find a local minimaand so it is very sensitive to the input geometry. If this is arbitrary, theresults are also arbitrary. This problem is kind of acceptable - computationalchemistry in large complex systems requires approximations to be made. Myconcern is that the authors simply say "DFT optimised" withoutcommenting on how the input geometry was selected and whether this involves anycompromises/presents any issues. This can be fixed by simply being more openabout how the calculations were done.
Secondly,and more problematically, the authors suggest that catenane 7-H emits from S7to S0. As I understand it, this violates Kasha's rule? I think they haveproposed this because the oscillator strength from S1 to S0 is essentially 0 atthe excited state geometry. The problem is that even if the initial excitationis from S0S7, there will be no molecules in the S7 state soon after excitation- they will all rapidly relax to the lowest available excited state (the originof Kasha's rule). The problem is made worse by the fact the calculations suggestthat initial excitation is from S0S1 (highest oscillator strength in theground state geometry).
As anon-expert (either in photophysics or computational chemistry), myinterpretation of the computational data is that excitation is from S0 to S1(highest oscillator strength). It should be noted that this initial excitationwill be to a highly vibrationally excited S1, given how far the molecule isfrom its optimal S1 geometry (i.e the excitation is from S0 (v=0) to S1 (v = n)where n is large). The molecule then undergoes a relaxation from the groundstate geometry to the excited state geometry, which corresponds to vibrationaldeactivation. Since the oscillator strength from S1 to S0 is effectively 0(i.e. very slow) at the excited state geometry, the molecule must becomevibrationally excited in order to emit, i.e. it must adopt a geometry more similarto the ground state. I don't think any of this is problematic - this isessentially the Frank-Condon principle in action.
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发表于 Post on 2020-6-19 15:02:14
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发表于 Post on 2020-6-19 18:16:16
