From owner-chemistry@ccl.net Sun Dec 2 04:32:01 2007 From: "akef afaneh akef_afnh : yahoo.com" To: CCL Subject: CCL: basis set erroneous concept Message-Id: <-35747-071202043016-4735-Qn44Ow1BkozJxo+aF3a7ng%a%server.ccl.net> X-Original-From: akef afaneh Content-Transfer-Encoding: 8bit Content-Type: multipart/alternative; boundary="0-1568555098-1196587802=:11805" Date: Sun, 2 Dec 2007 01:30:02 -0800 (PST) MIME-Version: 1.0 Sent to CCL by: akef afaneh [akef_afnh:_:yahoo.com] --0-1568555098-1196587802=:11805 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit This is a good question. In the ab initio calculations, the choice of basis sets is, as summarized by Hehre et al. (Hehre, W. J.; Radom, L.; Schleyer, P. v. R. and pople, J. A., Ab Initio Molecular Orbital Theory, Wiley, New York, 1986), "a compromise between accuracy and efficiency". Generally speaking, there are three types of AO basis sets can be used in the ab initio quantum chemistry: (1) Minimal Basis Sets (2) Multiple-Zeta Basis Sets, and (3) Polarized Basis Sets. Now, let us go directly to your question. 6-31G basis set is a multiple-zeta basis set, which we call it double zeta or split valence basis set. We can write out the functional form of the 1s, 2s and 2p orbitals of the C-atom in the 6-31G basis set: 6 ƒ³1s(r) = ‡”d1si ƒ³GF1s(r,ƒ¿1si) i=1 3 ƒ³2s(r) = ‡”d2si ƒ³GF2s(r,ƒ¿2si) + d'2s ƒ³GF2s(r,ƒ¿'2s) i=1 3 ƒ³2p(r) = ‡”d2pi ƒ³GF2p(r,ƒ¿2pi) + d'2p ƒ³GF2p(r,ƒ¿'2p) i=1 where the values of d1si, ƒ¿1si, d2si, ƒ¿2si, ƒ¿'2s, d2pi, ƒ¿2pi and ƒ¿'2p are fixed in the program. The value of the coefficients d'2s and d'2p are optimized as a part of the Roothaan-Hartree-Fock procedure and their values depend on the system at hand. I hope this answer will correct your conceptJ. Good Luck Akef "mohamed aish mhmdaish!^!yahoo.com" wrote: Hi All; I am a new in computational chemistry field. Now, I am reading about the basis sets. There are something confused me and would be appreciative if anyone helps me to correct my erroneous concepts. Suppose that we have a compound containing carbons with sp2 and sp3 hybridization, for example CH3CH=CH2, and the basis set is 6-31G. Now my question is: 6-31G basis set for Csp3 is similar to 6-31G basis set for Csp2. Is this right??!! Please correct my concept. Thanks a Lot --------------------------------- Be a better sports nut! Let your teams follow you with Yahoo Mobile. Try it now. --------------------------------- Never miss a thing. Make Yahoo your homepage. --0-1568555098-1196587802=:11805 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: 8bit
This is a good question. In the ab initio calculations, the choice of basis sets is, as summarized by Hehre et al. (Hehre, W. J.; Radom, L.; Schleyer, P. v. R. and pople, J. A., Ab Initio Molecular Orbital Theory, Wiley, New York, 1986), "a compromise between accuracy and efficiency". Generally speaking, there are three types of AO basis sets can be used in the ab initio quantum chemistry:
(1)  Minimal Basis Sets
(2)  Multiple-Zeta Basis Sets, and
(3)  Polarized Basis Sets.
Now, let us go directly to your question. 6-31G basis set is a multiple-zeta basis set, which we call it double zeta or split valence basis set. We can write out the functional form of the 1s, 2s and 2p orbitals of the C-atom in the 6-31G basis set:
               6
ƒ³1s(r) = ‡”d1si ƒ³GF1s(r,ƒ¿1si)
              i=1
               3
ƒ³2s(r) = ‡”d2si ƒ³GF2s(r,ƒ¿2si) + d'2s ƒ³GF2s(r,ƒ¿'2s)
              i=1
               3
ƒ³2p(r) = ‡”d2pi ƒ³GF2p(r,ƒ¿2pi) + d'2p ƒ³GF2p(r,ƒ¿'2p)
              i=1
where the values of d1si, ƒ¿1si, d2si, ƒ¿2si, ƒ¿'2s, d2pi, ƒ¿2pi and ƒ¿'2p are fixed in the program. The value of the coefficients d'2s and d'2p are optimized as a part of the Roothaan-Hartree-Fock procedure and their values depend on the system at hand. I hope this answer will correct your conceptJ.
 
Good Luck
Akef
 
 
 

"mohamed aish mhmdaish!^!yahoo.com" <owner-chemistry(0)ccl.net> wrote:
Hi All;
I am a new in computational chemistry field. Now, I am reading about the basis sets. There are something confused me and would be appreciative if anyone helps me to correct my erroneous concepts.
Suppose that we have a compound containing carbons with sp2 and sp3 hybridization, for example CH3CH=CH2, and the basis set is 6-31G. Now my question is: 6-31G basis set for Csp3 is similar to 6-31G basis set for Csp2. Is this right??!! Please correct my concept.
 
Thanks a Lot
 
Be a better sports nut! Let your teams follow you with Yahoo Mobile. Try it now.

Never miss a thing. Make Yahoo your homepage. --0-1568555098-1196587802=:11805-- From owner-chemistry@ccl.net Sun Dec 2 05:51:00 2007 From: "akef afaneh akef_afnh]|[yahoo.com" To: CCL Subject: CCL: basis set erroneous concept(Corrected symbols) Message-Id: <-35748-071202054603-28179-Pta1e2IITl73tnqPI1HRBg!=!server.ccl.net> X-Original-From: akef afaneh Content-Transfer-Encoding: 8bit Content-Type: multipart/alternative; boundary="0-1270294729-1196592352=:40650" Date: Sun, 2 Dec 2007 02:45:52 -0800 (PST) MIME-Version: 1.0 Sent to CCL by: akef afaneh [akef_afnh||yahoo.com] --0-1270294729-1196592352=:40650 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit This is a good question. In the ab initio calculations, the choice of basis sets is, as summarized by Hehre et al. (Hehre, W. J.; Radom, L.; Schleyer, P. v. R. and pople, J. A., Ab Initio Molecular Orbital Theory, Wiley, New York, 1986), "a compromise between accuracy and efficiency". Generally speaking, there are three types of AO basis sets can be used in the ab initio quantum chemistry: (1) Minimal Basis Sets (2) Multiple-Zeta Basis Sets, and (3) Polarized Basis Sets. Now, let's go directly to your question. 6-31G basis set is a multiple-zeta basis set, which we call it double zeta or split valence basis set. We can write out the functional form of the 1s, 2s and 2p orbitals of the C-atom in the 6-31G basis set: 6 A1s(r) = SUM d1si AGF1s(r,a1si) i=1 3 A2s(r) = SUM d2si AGF2s(r, a2si) + d'2s AGF2s(r, a'2s) i=1 3 A2p(r) = SUM d2pi AGF2p(r, a2pi) + d'2p AGF2p(r, a'2p) i=1 where the values of d1si, a1si, d2si, a2si, a'2s, d2pi, a2pi and a'2p are fixed in the program. The value of the coefficients d'2s and d'2p are optimized as a part of the Roothaan-Hartree-Fock procedure and their values depend on the system at hand. I hope this answer will correct your conceptJ. Good Luck Akef "mohamed aish mhmdaish!^!yahoo.com" wrote: Hi All; I am a new in computational chemistry field. Now, I am reading about the basis sets. There are something confused me and would be appreciative if anyone helps me to correct my erroneous concepts. Suppose that we have a compound containing carbons with sp2 and sp3 hybridization, for example CH3CH=CH2, and the basis set is 6-31G. Now my question is: 6-31G basis set for Csp3 is similar to 6-31G basis set for Csp2. Is this right??!! Please correct my concept. Thanks a Lot --------------------------------- Be a better sports nut! Let your teams follow you with Yahoo Mobile. Try it now. --------------------------------- Never miss a thing. Make Yahoo your homepage. --0-1270294729-1196592352=:40650 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: 8bit
This is a good question. In the ab initio calculations, the choice of basis sets is, as summarized by Hehre et al. (Hehre, W. J.; Radom, L.; Schleyer, P. v. R. and pople, J. A., Ab Initio Molecular Orbital Theory, Wiley, New York, 1986), "a compromise between accuracy and efficiency". Generally speaking, there are three types of AO basis sets can be used in the ab initio quantum chemistry:
(1)   Minimal Basis Sets
(2)   Multiple-Zeta Basis Sets, and
(3)   Polarized Basis Sets.
Now, let's go directly to your question. 6-31G basis set is a multiple-zeta basis set, which we call it double zeta or split valence basis set. We can write out the functional form of the 1s, 2s and 2p orbitals of the C-atom in the 6-31G basis set:
                 6
A1s(r) = SUM d1si AGF1s(r,a1si)
                i=1
                 3
A2s(r) = SUM d2si AGF2s(r, a2si) + d'2s AGF2s(r, a'2s)
                i=1
                 3
A2p(r) = SUM d2pi AGF2p(r, a2pi) + d'2p AGF2p(r, a'2p)
                i=1
where the values of d1si, a1si, d2si, a2si, a'2s, d2pi, a2pi and a'2p are fixed in the program. The value of the coefficients d'2s and d'2p are optimized as a part of the Roothaan-Hartree-Fock procedure and their values depend on the system at hand. I hope this answer will correct your conceptJ.
 
Good Luck
Akef


"mohamed aish mhmdaish!^!yahoo.com" <owner-chemistry!A!ccl.net> wrote:
Hi All;
I am a new in computational chemistry field. Now, I am reading about the basis sets. There are something confused me and would be appreciative if anyone helps me to correct my erroneous concepts.
Suppose that we have a compound containing carbons with sp2 and sp3 hybridization, for example CH3CH=CH2, and the basis set is 6-31G. Now my question is: 6-31G basis set for Csp3 is similar to 6-31G basis set for Csp2. Is this right??!! Please correct my concept.
 
Thanks a Lot
 
Be a better sports nut! Let your teams follow you with Yahoo Mobile. Try it now.

Never miss a thing. Make Yahoo your homepage. --0-1270294729-1196592352=:40650-- From owner-chemistry@ccl.net Sun Dec 2 20:43:01 2007 From: "Naser Eltaher Eltayeb nasertaha90-$-yahoo.co.uk" To: CCL Subject: CCL:G: Band gap Message-Id: <-35749-071202203652-21198-iZl89DEAgTRr054uam0rDA###server.ccl.net> X-Original-From: "Naser Eltaher Eltayeb" Date: Sun, 2 Dec 2007 20:36:49 -0500 Sent to CCL by: "Naser Eltaher Eltayeb" [nasertaha90 ~ yahoo.co.uk] Dear All 1-Could any one give me details about How can I calculate the band gap (HOMO-LUMO) energy,using Gaussian 03. 2- I am tring to calculate the band gap by generating the molecular orbital after optimization, then I looked to energy of HOMO and LUMO , I found the energy of HOMO= -0.261 and the energy of LUMO = -0.221. First what the unit of energy of HOMO and LUMO in Gasssian 03, secondly is the band gap 0.04 acceptable?