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Date: Sat, 18 May 2002 22:22:36 -0400 (EDT)
From: "Raghavan B. Sunoj" <rbsunoj@chemistry.ohio-state.edu>
To: chemistry@ccl.net
cc: rbsunoj@hotmail.com
Subject: Orbital switching 
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> --------------------------------------------------------------
>
> How can one achieve the best control over the electronic state of a
> molecule (within Gaussian program)?.
>
> When the orbital switching fails to give the correct state [instead it
> collapse to the lowest energy state during geometry optimization], what
one can
do
> about it?.

Also, which are the recommended method for excited state calculations
(I am interested in the excited state geometry).

>
> Thank you
>
> Sunoj
> ---------------------------------------------------------------
>


From chemistry-request@server.ccl.net Sat May 18 01:40:12 2002
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From: Roy Jensen <royj@uvic.ca>
To: chemistry@ccl.net
Subject: SUMMARY: Maximum in PES scans...
Date: Fri, 17 May 2002 22:41:27 -0700
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All

I received many replies to this question. Thanks to everyone who
responded. There are several possible answers to this observation,
which are summarized below. The original responses are at the end of
this message.

For this specific case, I highly suspect 6 is the explanation, since
N2 is one product.

Roy Jensen


ORIGINAL EMAIL
*************************************************************************

> PES scans often show a maximum along the dissociation coordinate.
> What is the physical interpretation of this maximum?
>
> In the _dissociation_ of a diatomic molecule, there is sometimes a
> maximum along the dissociation coordinate. I am observing this in a
> relaxed PES scan of a tetra-atomic dissociation. The dissociation goes
> to the proper, neutral fragments. The maximum, and all points beyond,
> have only one negative frequency, but I cannot rationalize the drop
> after the maximum. 
> 
>   X
>   X
>   X
>    X
>    X             XXXX
>    X            X    XXX
>     X          X        XXXX
>     X         X             XXXXXXXXXXXXXXXXXXXXXXXXX	(A+B)
>      X       X
>      X       X
>       X     X
>        X   X
>         XXX
>         (AB)
> 
> Your help is appreciated!


SUMMARY OF POSSIBILITIES
*************************************************************************

1. improper wave function behavior (poor choice: RHF, DFT)

2. activation energy; transition state

3. another path with a lower maximum

4. [improper] geometrical constraint in your scan

5. avoided crossing 

6. bonding change in fragments during dissociation (valid only for
polyatomics)

7. Etot = V + Erot. As A-B separation increases, Erot decreases; V
increases


COMPLETE RESPONSES
*************************************************************************

Alan Shusterman <Alan.Shusterman@directory.reed.edu>
> Can you provide an example? It sounds like a case of improper wave function behavior.


Sergiusz Kwasniewski <sergiusz.kwasniewski@luc.ac.be>

> that should represent the activation barrier or energy ..


Christoph van W黮len <Christoph.vanWullen@TU-Berlin.De>

> if your reaction coordinat is well chosen, this maximum is a (decent) 
> approximation to a transition state (at least a good guess).
> However you do not know beforehand if the choice of the reaction coordinate
> was good -- that is, if there is another path with a lower maximum.


Pascal <p.boulet@qmul.ac.uk>

> If it's a first order maximum (only one imaginary frequency) it's a transition
> state from the reactants to the products. 
> 
> If it's a higher order maximum some people agrees to say that there's no
> particular meaning (the scan is meaningless).


E. Lewars <elewars@trentu.ca>

> If you mean a max *along the reaction coordinate* for dissociation, I think this
> should correspond to a first-order saddle point, i.e. a transition state. You
> are probably cocerned because a dissociation reaction (=  a recombination
> reaction; I presume the two fragments are radicals) is not supposed to have a
> barrier. I think in some cases a small barrier can arise due to geometry
> readjustment.


Frank Jensen <frj@dou.dk>

> If these are DFT calculations, the maximum is probably an artifact. 


Li Zhenhua <li_zhenhua@hotmail.com>

> I don't know what molecule you are studying. For the dissociation of a simple diatomic molecule into two neutral atoms, the use of RHF wavefunction should be incorrect somewhere when the bond is broking. It is multiconfiguration in nature. The simplest way to get the correct wavefunction is to allow it to break down symmetry. In Gaussian, use guess=mix and scf=qc to let the symmetry break and keep it during the PES scan. Maybe in this case, the wavefunction around the maximum is not stable regarding to symmetry break. You can use stable=opt to check the wavefunction of your maximum geometry.
> But if your molecule is not so simple, the maximum maybe is a transition sate.
> Maybe after the maximum point, the symmetry of the wavefunction changed. You can check it.


Xavier ASSFELD <Xavier.Assfeld@lctn.uhp-nancy.fr>

> This reaction path is certainly not the minimum energy path (MEP).
> Perhaps you have a geometrical constrain in your scan that 
> creates the bump?
> Well, I don't know if I am of great help...
> Could you please keep me inform of any progress you will made?


Jim Kress <kresslists@kressworks.com>

> As I remember my Kinetics class, what you are observing is the activation
> barrier between reactants and products.  This is a common feature of many
> chemical reactions.  You can find it discussed in any Kinetics text.


Juan Pablo Senosiain <senos@Stanford.EDU>

> I can rationalize a maximum in a  dissociation curve in 2 ways:
> 
> a) a maximum in the potential energy (V)may occur if the bonding structure
> of A is such, that by breaking the A-B bond a new bond will form within A
> e.g.  O=C=O  --> CO + O
> 
> b) a maximum in the total energy (i.e. Etot=V+Erot) occurs due to
> rotational effects, that is, as the A-B distance is increased, the moment
> of inertia  increases with the square of this quantity. Angular
> momentum conservation mandates a change in the rotational velocity
> such that Erot decreases. Thus the net effect is that as the A-B
> separation increases Etot increases, reaching a (rotational) maximum and
> then decreases slightly.
> 
> Assuming a potential given by the (attractive part) of a Lennard-Jones
> potential, it is straight forward to demonstrate that this maximum will
> occur at
> 
> Rmax=sigma*(12 De/kB*T)^1/6
> 
> where V(r)=-4De(sigma/r)^6


Alan Shusterman <Alan.Shusterman@directory.reed.edu> 

> I was hoping that your problem was a simple one, i.e., dissociation of a ground-state HF wave function. Your difficulty may be related to this, but I rather doubt it.
> 
> Just in case you are not familiar with the dissociation problem, I'll describe the basics (skip this if this is too basic). HF (and for that matter DFT) wave functions cannot describe homolytic bond cleavages properly. The bonding pair of electrons are locked into the same MO even as the bond gets longer, which means (according to the wave function) there is a significant probability of both electrons being near one of the two fragments simultaneously. The only solution is to use a wave function that forces the electrons to correlate their behavior. GVB and fancier types of MCSCF will do this. I would think CISD would do this too, hence my confusion.
> 
> One thought - perhaps the "maximum" that you find at intermediate distances represents an avoided crossing of two electronic states? You might detect this by comparing electron distributions for structures on each side of the maximum. Are they substantially different?


JENS SPANGET-LARSEN <spanget@ruc.dk>

> you probably have a case of 'avoided crossing' of two electronic 
> states. Perhaps corresponding to the diagram below, where Y is a 
> dissociative state that becomes the ground state for large 
> internuclear distances.
> 
> Yours, Jens >--<
> 
>               Y
>                Y
>    X           Y              XXXXXXX
>    X            Y         XXX
>    X             YY    XX
>    X               YYX   
>     X
>     X             XXYY
>     X            X    YYY
>      X          X        YYY
>      X         X             YYYYYYYYYYYYYYYYYYYY (A+B)
>       X       X
>       X       X
>        X     X
>         X   X
>          XXX
>          (AB)


John Bushnell <bushnell@chem.ucsb.edu>

> I would look at the change in bond distances between
> the four atoms for a clue.  You would have to consider
> the mixing of higher (diabatic) states if this were
> in fact the dissociation of a diatomic molecule (which
> I guess it isn't actually).


Theis <theis.ivan.soelling@risoe.dk>

> How exactly did you determine that all the points beyond the saddle point
> have imaginary frequencies?
> 
> When you have a saddlepoint it will be proceeded by a minimum - a species
> that is often referred to as "an exit-channel complex". The reason why you
> do not see it in the specfic case is probably due to the fact that both of
> the products are neutral species.  Your exit-channel complex is held
> together by dipole-dipole or even weaker interactions so the dip will be
> very shallow indeed.


Shobe, Dave <dshobe@sud-chemieinc.com>

> Maybe there is an avoided crossing?
> 
>   X    Y
>   X     Y                  YYYYYYYYYYYYYYYYYYYYYYYYYYY
>   X      YYY          YYYYY
>   X          YYYYYYYYY
>    X
>    X             XXXX
>    X            X    XXX
>     X          X        XXXX
>     X         X             XXXXXXXXXXXXXXXXXXXXXXXXX	(A+B)
>      X       X
>      X       X
>       X     X
>        X   X
>         XXX
>         (AB)
> 
> Or perhaps a significant change in bonding (besides the obvious one) between
> AB and A+B, so that it's not a simple bond cleavage?



From chemistry-request@server.ccl.net Sat May 18 07:23:35 2002
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 Dir sir

   I am doing my research on the MD study of
interface.
And I need a very important paper about the interface
eagerly,which name is "Chirality induction from chiral
molecules to adsorbed monolayer   J. Chem. Phys 2001,
114(21):9532-9538 " I can't find this paper in our
library .But I need it eagerly.Can you help me?Maybe
you can send  a .PDF file to me?

  thank you 

  Y Pei 

_________________________________________________________
Do You Yahoo!? 
摊开你的掌心 让我看看你
http://sweepstakes.yahoo.com/2002cnuser


From chemistry-request@server.ccl.net Sat May 18 01:38:36 2002
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From: Roy Jensen <royj@uvic.ca>
To: chemistry@ccl.net
Subject: SUMMARY: MRCI in Gaussian?
Date: Fri, 17 May 2002 22:39:52 -0700
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Thanks to those who responded.

Original Question

> Is it possible to do a multireference CI calculation using Gaussian 98? If
> so, could someone provide a typical input file.


Summarized Answers

> No. Use MOLPRO (fastest?), MOLCAS, or GAMMES (free).





From chemistry-request@server.ccl.net Sat May 18 15:01:34 2002
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Date: Sat, 18 May 2002 21:03:21 +0200
From: "Tamas E. Gunda" <tamasgunda@tigris.klte.hu>
Subject: Re: CCL:Ligand preparation
To: amor san juan <amor_sanjuan@yahoo.com>, chemistry@ccl.net
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References: <1021360120.3ce0b7f8a9c1b@mail.up.edu.ph>

The file you produced this way is a 2D drawing. Therefore addition of
hydrogens and any further manipulations are nonsense. You should use a real
modelling application.

Tamas

----- Original Message -----
From: "amor san juan" <amor_sanjuan@yahoo.com>
To: <chemistry@ccl.net>
Sent: Tuesday, May 14, 2002 9:08 AM
Subject: CCL:Ligand preparation


> Dear AutoDocker's community:
>
> Can someone recommend any other softwares for ligand preparation (aside
> from
> chemskestch5, babel &/or ADT). I need your help.
>
> Using Chemsketch5 I draw the benzamidine(ben.mol) structure then, used
Babel to
> add  explicit hydrogens & ben.mol2 conversion. The result is flawed
ben.pdbq
> file as viwed below:
>
> REMARK  2 active torsions (identified by new id numbers):
> REMARK  status: ('A' for Active; 'I' for Inactive)
> REMARK       I      3-  7 between atoms:  C3   and  C7
> REMARK    1  A      7-  8 between atoms:  C7   and  N1
> REMARK    2  A      7- 10 between atoms:  C7   and  N2
> ROOT
> ATOM      1  A1  <1>     1      -4.129   2.188   0.000  0.00  0.00
0.000   2
> ATOM      2  A2  <1>     1      -4.129   1.362   0.000  0.00  0.00
0.000   2
> ATOM      3  A3  <1>     1      -3.417   0.954   0.000  0.00  0.00
0.000   3
> ATOM      4  A4  <1>     1      -2.705   1.362   0.000  0.00  0.00
0.000   2
> ATOM      5  A5  <1>     1      -2.705   2.188   0.000  0.00  0.00
0.000   2
> ATOM      6  A6  <1>     1      -3.417   2.604   0.000  0.00  0.00
0.000   2
> ATOM      7  C7  <1>     1      -3.417   0.129   0.000  0.00  0.00
0.000   3
> ENDROOT
> BRANCH 7 8
> ATOM      8  N1  <1>     1      -4.132  -0.283   0.000  0.00  0.00
0.000   2
> ATOM      9  H6  <1>     1      -5.107  -0.581   0.000  0.00  0.00
0.000   1
> ENDBRANCH 7 8
> BRANCH 7 10                                                      ATOM
10
> N2  <1>     1      -2.703  -0.283   0.000  0.00  0.00     0.000   3
> ATOM     11  H7  <1>     1      -2.703  -1.303   0.000  0.00  0.00
0.000   1
> ATOM     12  H8  <1>     1      -1.819   0.227   0.000  0.00  0.00
0.000   1
> ENDBRANCH 7 10
> TDOF 3
>
>  ------------------
> Diagramatically, the steps I did to prepare ligand is:
>
>   ben.mol -----> ben.mol2 ---------> ben.pdbq
>            babel            autotors3
>
> Alternatively, I used ADT yet, the same zeros in last columns appeared (no
> charge present).
>
> Amor
>
>
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