From owner-chemistry@ccl.net Mon May 3 02:06:01 2010 From: "Joaquin Barroso Flores joaco_barroso---yahoo.com" To: CCL Subject: CCL:G: Sigma and Pi orbitals via NBO Message-Id: <-41783-100503020419-10220-YCwBaDNTD828zT9H2EJ4Kw|*|server.ccl.net> X-Original-From: Joaquin Barroso Flores Content-Type: multipart/alternative; boundary="0-388302785-1272866640=:84981" Date: Sun, 2 May 2010 23:04:00 -0700 (PDT) MIME-Version: 1.0 Sent to CCL by: Joaquin Barroso Flores [joaco_barroso,yahoo.com] --0-388302785-1272866640=:84981 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable Within that same section (Second order perturbation theory analysis...) you= can see to which Cu-Cu bond the "back bonding" is occurring. Read the numb= er assigned to it, you will find teh acceptor bonds arent listed numericall= y and then go to the section labeled "Bond orbital/ Coefficients/ Hybrids" = and find the bond, it will describe the composition of the bond.=0A=0Ahope = it helps=0A=0A =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=0AJoaquin Barroso-Flores, Ph. D.=0AFacultatea de Chim= ie, Universitatea Babes-Bolyai=0ACluj-Napoca, Romania=0A=0A=0A-> http://joa= quinbarroso.com=0A-> http://joaquinbarroso.wordpress.com=0A=0A=0Ajoaquinbar= roso''a''chem.ubbcluj.ro=0Ajoaquin.barroso''a''gmail.com =0A=0A=0A"Blogastr= onom=EDa": http://joaquinbarroso.blogspot.com =0A=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=0A=0A=0A=0A=0A___= _____________________________=0ADe: Comp Chem compchemgroup1#gmail.com =0APara: "Flores, Joaquin Barroso " =0AEnviado: domingo, 2 de mayo, 2010 16:39:21=0AAsunto: CCL:= G: Sigma and Pi orbitals via NBO=0A=0A=0ASent to CCL by: "Comp Chem" [comp= chemgroup1#gmail.com]=0ADear All,=0AHow I can identify the sigma and pi orb= ital from NBO analysis using Gaussian 09? =0A=0AAnother question, I have di= nuclear complex includes -Cu1-O(H)-Cu2- (the oxygen of hydroxyl group is co= ordinated with both Cu-atom). I did NBO analysis and I found the second per= terbution energy of LP(3)O-->BDCu1-Cu2 =3D 72 kJ/mol. The question is how I= know which orbital in both Cu1 and Cu2 interact with the lone pair of oxyg= en (2Pz??=0AThanks =0ACCG1=0A=0A=0A=0A-=3D This is automatically added to e= ach message by the mailing script =3D-=0ATo recover the email address of th= e author of the message, please change=0Athe strange characters on the top = line to the ++ sign. You can also=0Alook up the X-Original-From: line in the= mail header.=0A=0A=0A ==0A=0AE-mail to administrat= ors: CHEMISTRY-REQUEST++ccl.net or use=0A http://www.ccl.net/cgi-bin/cc= l/send_ccl_message=0A=0A=0A http://www.ccl.net/= chemistry/sub_unsub.shtml=0A=0ABefore posting, check wait time at: http://w= ww.ccl.net=0A=0A=0AConferences: http://server.= ccl.net/chemistry/announcements/conferences/=0A=0ASearch Messages: http://w= ww.ccl.net/chemistry/searchccl/index.shtml=0A=0AIf your mail bounces from C= CL with 5.7.1 error, check:=0A=0A=0ART= FI: http://www.ccl.net/chemistry/aboutccl/instructions/=0A=0A=0A --0-388302785-1272866640=:84981 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable
Within that same section (Second order perturbation theory ana= lysis...) you can see to which Cu-Cu bond the "back bonding" is occurring. = Read the number assigned to it, you will find teh acceptor bonds arent list= ed numerically and then go to the section labeled "Bond orbital/ Coefficien= ts/ Hybrids" and find the bond, it will describe the composition of the bon= d.

hope it helps
 
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Joaquin Barroso-Flor= es, Ph. D.
Facultatea de Chimie, Universitatea Babes-Bolyai
Cluj-Napo= ca, Romania



joaquinbarroso''a''chem.ubbcluj.ro
j= oaquin.barroso''a''gmail.com


"Blogastronom=EDa": <= a target=3D"_blank" href=3D"http://joaquinbarroso.blogspot.com">http://joaq= uinbarroso.blogspot.com
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D


De: Comp Chem compchemgroup1#gmail.com <owner-chemistry= ++ccl.net>
Para: "Flo= res, Joaquin Barroso " <joaco_barroso++yahoo.com>
Enviado: domingo, 2 de mayo, 2010 1= 6:39:21
Asunto: CCL:G: Sigma and Pi orbitals via NBO
<= br>
Sent to CCL by: "Comp  Chem" [compchemgroup1#gmail.com]
Dear= All,
How I can identify the sigma and pi orbital from NBO analysis usin= g Gaussian 09?

Another question, I have dinuclear complex includes = -Cu1-O(H)-Cu2- (the oxygen of hydroxyl group is coordinated with both Cu-at= om). I did NBO analysis and I found the second perterbution energy of LP(3)= O-->BDCu1-Cu2 =3D 72 kJ/mol. The question is how I know which orbital in= both Cu1 and Cu2 interact with the lone pair of oxygen (2Pz??
Thanks CCG1



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=0A=0A=0A=0A   --0-388302785-1272866640=:84981-- From owner-chemistry@ccl.net Mon May 3 06:58:00 2010 From: "CompChem Group compchemgroup1[*]gmail.com" To: CCL Subject: CCL:G: Sigma and Pi orbitals via NBO Message-Id: <-41784-100503065402-4110-RsUCIPoDAgb6kyAJv6+kWw_-_server.ccl.net> X-Original-From: CompChem Group Content-Type: multipart/alternative; boundary=0016367fae4995d6bb0485ad7cda Date: Mon, 3 May 2010 10:47:52 +0100 MIME-Version: 1.0 Sent to CCL by: CompChem Group [compchemgroup1|-|gmail.com] --0016367fae4995d6bb0485ad7cda Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Dear Joaquin and CCLers, Thanks for your reply. However, there something misunderstanding. So I will show you here what I meant for more help from you and CCL subscribers as well. * N.B. This calculation is done using B3LYP/6-311G(d,p). * *The Second order perturbation theory analysis section is*: 178. LP ( 3) O3 /***. BD (1)Cu1 -Cu 2 72.50 0.10 0.088 Please observe here that the donor is the third lone pair of Oxygen, while the accepptor part is Cu-Cu bond. By the way there is no backdonation at al= l here. Also, please see the acceptor is BD not BD* * Bond orbital/ Coefficients/ Hybrids* *section is:* 36. (1.99380) BD*( 1)Cu 1 -Cu 2 ( 49.90%) 0.7064*Cu 1 s( 0.09%)p 0.00( 0.00%)d99.99( 99.91%) f 0.00( 0.00%) 0.0000 0.0000 0.0001 -0.0294 -0.0001 0.0010 0.0001 0.0000 0.0000 0.0000 0.0001 0.0014 -0.0002 0.0000 0.0000 0.0000 0.0001 0.0002 0.0000 0.0000 0.0000 -0.0005 0.0000 0.0000 -0.0265 0.0006 -0.0001 -0.1173 -0.0005 -0.0001 0.0878 0.0006 0.0001 0.8712 0.0018 0.0003 -0.4669 -0.0008 -0.0002 -0.0001 -0.0002 -0.0001 -0.0004 -0.0001 0.0003 0.0003 ( 50.10%) -0.7078*Cu 2 s( 0.08%)p 0.00( 0.00%)d99.99( 99.92%) f 0.00( 0.00%) 0.0000 0.0000 -0.0001 0.0285 0.0010 -0.0007 -0.0001 0.0000 0.0000 0.0000 0.0001 0.0011 -0.0003 0.0000 0.0000 0.0000 -0.0012 -0.0002 0.0000 0.0000 0.0000 -0.0010 0.0001 0.0000 0.1990 0.0012 0.0000 0.0823 -0.0001 0.0001 0.0717 0.0005 0.0000 -0.8666 -0.0017 -0.0004 0.4435 0.0008 0.0002 -0.0002 -0.0003 -0.0001 0.0001 0.0000 0.0001 -0.0005 *I did not find any analysis for BDCu1-Cu2 (without star *) in the **bond orbital/ Coefficients/ Hybrids* *section. The quesitions are: 1- Why I did not find any analysis for **BDCu1-Cu2? 2-What are the d-orbital in both Cu atoms which is primaryly interact with the 2Pz of Oxygen? 3- Is reseanable if we say that there is bond between the two Cu atoms even though the distance between them 4.31 Angestrom. Thanks in advance. CCG1 ** ** * * * 2010/5/3 Joaquin Barroso Flores joaco_barroso---yahoo.com < owner-chemistry[-]ccl.net> > Within that same section (Second order perturbation theory analysis...) y= ou > can see to which Cu-Cu bond the "back bonding" is occurring. Read the num= ber > assigned to it, you will find teh acceptor bonds arent listed numerically > and then go to the section labeled "Bond orbital/ Coefficients/ Hybrids" = and > find the bond, it will describe the composition of the bond. > > hope it helps > > =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D > Joaquin Barroso-Flores, Ph. D. > Facultatea de Chimie, Universitatea Babes-Bolyai > Cluj-Napoca, Romania > > -> http://joaquinbarroso.com > -> http://joaquinbarroso.wordpress.com > > > joaquinbarroso''a''chem.ubbcluj.ro > joaquin.barroso''a''gmail.com > > > "Blogastronom=EDa": http://joaquinbarroso.blogspot.com > =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D > > > ------------------------------ > *De:* Comp Chem compchemgroup1#gmail.com > *Para:* "Flores, Joaquin Barroso " > *Enviado:* domingo, 2 de mayo, 2010 16:39:21 > *Asunto:* CCL:G: Sigma and Pi orbitals via NBO > > > Sent to CCL by: "Comp Chem" [compchemgroup1#gmail.com] > Dear All, > How I can identify the sigma and pi orbital from NBO analysis using > Gaussian 09? > > Another question, I have dinuclear complex includes -Cu1-O(H)-Cu2- (the > oxygen of hydroxyl group is coordinated with both Cu-atom). I did NBO > analysis and I found the second perterbution energy of LP(3)O-->BDCu1-Cu2= =3D > 72 kJ/mol. The question is how I know which orbital in both Cu1 and Cu2 > interact with the lone pair of oxygen (2Pz?? > Thanks > CCG1 > > > > -=3D This is automatically added to each message by the mailing script = =3D-> the strange characters on the top line to the (!) sign. You can also > E-mail to subscribers: CHEMISTRY(!)ccl.net or > use:> > E-mail to administrators: CHEMISTRY-REQUEST(!)ccl.netor use> > > > > --0016367fae4995d6bb0485ad7cda Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable
Dear Joaquin and CCLers,
Thanks for your reply. However= , there something misunderstanding. So I will show you here what I meant fo= r more help from you and CCL subscribers as well.

N.B. This calcu= lation is done using B3LYP/6-311G(d,p).

=A0The Second order perturbation theory analysis section is:=
178. LP (=A0=A0 3) O3 =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 /***. = BD (1)Cu1 -Cu= =A0 2 =A0=A0=A0=A0=A0=A0=A0=A0=A0 72.50=A0=A0=A0 0.10=A0=A0=A0 0.088

Please observe here that the donor is the third lone pair of Oxygen, wh= ile the accepptor part is Cu-Cu bond. By the way there is no backdonation a= t all here.
=A0
Also, please see the acceptor is BD not BD*

Bond orbital/ Coefficients/ Hybrids section is:
36. (1.99380) BD*( 1)Cu=A0 1 -Cu=A0 2=A0 =
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0 ( 49.90%)=A0=A0 0.7064*Cu=A0 1 s(=A0 0.09%)p 0.00(= =A0 0.00%)d99.99( 99.91%)
=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 f 0.00(=A0 0.0= 0%)
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0000=A0 0.0000=A0 0.0001 -0.0294 -0.000= 1
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0 0.0010=A0 0.0001=A0 0.0000=A0 0.0000=A0 0.0000 =A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0001=A0 0.0014 -0.0002=A0 0.0000=A0 0.0= 000
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0 0.0000=A0 0.0001=A0 0.0002=A0 0.0000=A0 0.0000
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0000 -0.0005=A0 0.0000=A0 0.0000 -0.026= 5
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0 0.0006 -0.0001 -0.1173 -0.0005 -0.0001
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0878=A0 0.0006=A0 0.0001=A0 0.8712=A0 0= .0018
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0 0.0003 -0.4669 -0.0008 -0.0002 -0.0001
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 -0.0002 -0.0001 -0.0004 -0.0001=A0 0.0003
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0 0.0003
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0 ( 50.10%)=A0 -0.7078*Cu=A0 2 s(=A0 0.08%)p 0.00(= =A0 0.00%)d99.99( 99.92%)
=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 f 0.00(=A0 0.0= 0%)
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0000=A0 0.0000 -0.0001=A0 0.0285=A0 0.0= 010
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0 -0.0007 -0.0001=A0 0.0000=A0 0.0000=A0 0.0000<= br style=3D"background-color: rgb(255, 255, 255);"> =A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0001=A0 0.0011 -0.0003=A0 0.0000=A0 0.0= 000
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0 0.0000 -0.0012 -0.0002=A0 0.0000=A0 0.0000<= br style=3D"background-color: rgb(255, 255, 255);"> =A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0000 -0.0010=A0 0.0001=A0 0.0000=A0 0.1= 990
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0 0.0012=A0 0.0000=A0 0.0823 -0.0001=A0 0.0001
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 0.0717=A0 0.0005=A0 0.0000 -0.8666 -0.001= 7
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0 -0.0004=A0 0.4435=A0 0.0008=A0 0.0002 -0.0002
=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0 -0.0003 -0.0001=A0 0.0001=A0 0.0000=A0 0.000= 1
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0= =A0=A0=A0=A0=A0 -0.0005
I did not find any analysis for BDCu1-Cu2 (without star *) in the bond orbital/ Coefficients/ Hybrids section.

T= he quesitions are:
1- Why I did not find any analysis for BDCu1-Cu2?
2-What are the d-orbital in both Cu atoms which is primaryly interact with = the 2Pz of Oxygen?
3- Is reseanable if we say that there is bond between= the two Cu atoms even though the distance between them 4.31 Angestrom.
Thanks in advance.
CCG1
=A0


=

=A0








2010/5/3 Joaquin Barroso Flores joaco_barroso---yahoo.com <owner-chemistry[-]ccl.ne= t>
Wi= thin that same section (Second order perturbation theory analysis...) you c= an see to which Cu-Cu bond the "back bonding" is occurring. Read = the number assigned to it, you will find teh acceptor bonds arent listed nu= merically and then go to the section labeled "Bond orbital/ Coefficien= ts/ Hybrids" and find the bond, it will describe the composition of th= e bond.

hope it helps
=A0
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Joaquin Barroso-Flores, Ph. D= .
Facultatea de Chimie, Universitatea Babes-Bolyai
Cluj-Napoca, Roman= ia



joaqui= nbarroso''a''chem.ubbcluj.ro
joaquin.barroso''a''gmail.com


"Blogastronom=EDa"= : http://j= oaquinbarroso.blogspot.com
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D


De: Comp Chem compchemgroup1#gmail.com <owner-chemistry(!)ccl.net>
Para: "Flores, Joaqui= n Barroso " <joaco_barroso(!)yahoo.com>
Enviado= : domingo, 2 de mayo, 2010 16:39:21
Asunto: CCL:G: Sigma and P= i orbitals via NBO


Sent to CCL by: &quo= t;Comp=A0 Chem" [compchemgroup1#gmail.com]
Dear All,
How I can identify the sigma and pi orbital from NBO analysis = using Gaussian 09?

Another question, I have dinuclear complex inclu= des -Cu1-O(H)-Cu2- (the oxygen of hydroxyl group is coordinated with both C= u-atom). I did NBO analysis and I found the second perterbution energy of L= P(3)O-->BDCu1-Cu2 =3D 72 kJ/mol. The question is how I know which orbita= l in both Cu1 and Cu2 interact with the lone pair of oxygen (2Pz??
Thanks
CCG1



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=A0
--0016367fae4995d6bb0485ad7cda-- From owner-chemistry@ccl.net Mon May 3 08:29:01 2010 From: "=?ISO-8859-1?B?Vu1jdG9y?= victor.gil.sepulveda[*]gmail.com" To: CCL Subject: CCL: how I can model for a protein homology? Message-Id: <-41785-100502154610-16491-whEbRTasUyQKdlwREdCEJA ~~ server.ccl.net> X-Original-From: =?ISO-8859-1?B?Vu1jdG9y?= Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; charset=ISO-8859-1 Date: Sun, 2 May 2010 21:45:40 +0200 MIME-Version: 1.0 Sent to CCL by: =?ISO-8859-1?B?Vu1jdG9y?= [victor.gil.sepulveda#,#gmail.com] Maybe you can try with MODELLER (http://www.salilab.org/modeller/ ). You will need a base structure and a suitable alignment to get your final model. Cheers, V=EDctor 2010/4/29 Naudis Antonio De Voz ndev{:}unicartagena.edu.co : > > Sent to CCL by: "Naudis Antonio De Voz" [ndev^unicartagena.edu.co] > is that I've been reading about by homology modeling proteins , but I cou= ld not find a clear methodology ... and I really want to know how to build = a theoretical model, since I need one for a study... thanks very much > > > > -=3D This is automatically added to each message by the mailing script = =3D-> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_message> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_message> =A0 =A0 =A0http://www.ccl.net/chemistry/sub_unsub.shtml> =A0 =A0 =A0http://www.ccl.net/spammers.txt> > > --=20 V=EDctor Gil Sep=FAlveda E. Informatica - FIB - UPC From owner-chemistry@ccl.net Mon May 3 10:31:00 2010 From: "Sergio Emanuel Galembeck segalemb%%usp.br" To: CCL Subject: CCL:G: Sigma and Pi orbitals via NBO Message-Id: <-41786-100503072355-18246-v6lGafrV6RatJ+nivftnTA%%server.ccl.net> X-Original-From: Sergio Emanuel Galembeck Content-Disposition: inline Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes"; format="flowed" Date: Mon, 03 May 2010 07:23:37 -0300 MIME-Version: 1.0 Sent to CCL by: Sergio Emanuel Galembeck [segalemb(_)usp.br] Hello, You can identify sigma and pi orbitals by looking in the composition of NBOs. If the NBO has no contribution of s orbital, or this contribution is very small, the orbital in question is pi. You can also visualize the NBOs, by using some graphical software, as gOpenMol, for example. Hope this help you, Sergio Citando "Comp Chem compchemgroup1#gmail.com" : > > Sent to CCL by: "Comp Chem" [compchemgroup1#gmail.com] > Dear All, > How I can identify the sigma and pi orbital from NBO analysis using > Gaussian 09? > > Another question, I have dinuclear complex includes -Cu1-O(H)-Cu2- > (the oxygen of hydroxyl group is coordinated with both Cu-atom). I > did NBO analysis and I found the second perterbution energy of > LP(3)O-->BDCu1-Cu2 = 72 kJ/mol. The question is how I know which > orbital in both Cu1 and Cu2 interact with the lone pair of oxygen > (2Pz?? > Thanks > CCG1> > > From owner-chemistry@ccl.net Mon May 3 11:32:00 2010 From: "CARBONNIERE Philippe philippe.carbonniere+*+univ-pau.fr" To: CCL Subject: CCL: conference of theoretical chelists of Latin Expression (chitel 2010) Message-Id: <-41787-100503052517-13878-EsrbTOuQoObhzitdoXSUTg()server.ccl.net> X-Original-From: CARBONNIERE Philippe Content-Transfer-Encoding: 8bit Content-Type: text/plain; charset=windows-1252; format=flowed Date: Mon, 03 May 2010 10:23:11 +0200 MIME-Version: 1.0 Sent to CCL by: CARBONNIERE Philippe [philippe.carbonniere-,-univ-pau.fr] The XXXVI Conference of Theoretical Chemists of Latin Expression (CHITEL 2010) will be held in Anglet (near Biarritz) in the South-West of France, from September 18 to September 24, 2010. As in the previous editions, the conference: - will provide the state of the art in theoretical chemistry through seminars given by leading theoreticians; - is directed towards researchers in the fields of theoretical chemistry, in the beginning of their careers (senior PhD students, Postdoctoral fellows, and Junior researchers); - aims to enable wide discussion and contacts between the participants. Contributions in all areas of methods development and applications in theoretical chemistry are welcome. As usual, the Proceedings of the CHITEL 2010 (with full papers) will be published. Location : The Conference will be held at the Auditorium “Espace de l’Océan” in Anglet (Pyrénées Atlantiques), a city in the “Basque” coast at 110 km > from Pau and 5 km from Biarritz. Accomodation: Hotel Belembra, close to the Auditorium, will receive the participants for accomodation and meals. The Special meeting prices including full board and accommodation are available for 5 or 6 days from September 18^th evening or 19^th morning to September 24^th afternoon. The rooms will be attributed on a ‘first registered first served’ basis. Non-resident fees do not include room, breakfast, lunch and dinner. Registration: The fees include board for five or six days, free use of facilities of the CHITEL 2010 premises, coffee breaks, the booklet of abstracts, and a social programme involving a guided tour, a banquet dinner with the award ceremony. Registration and Communication demands start on March 1^st . All-inclusive attendance rates about the CHITEL are given on the web site: http://quitel.univ-pau.fr . Overseas participants could combine their coming to CHITEL 2010 with a MOLEC meeting near Coimbra, Portugal, between September 5 and 10, or with the QSCP-XV workshop in Cambridge between August 31 and September 5.