From owner-chemistry@ccl.net Fri Jun 24 07:31:00 2016 From: "Tandon Swetanshu tandons:-:tcd.ie" To: CCL Subject: CCL: Coupling constant (Jab) - Why more unpaired electrons means a smaller coupling? Message-Id: <-52248-160624064544-3354-TTkZo/TsQ5kevjFhmn2C9A_-_server.ccl.net> X-Original-From: Tandon Swetanshu Content-Type: multipart/alternative; boundary=001a114168d6f6c4be053603de7f Date: Fri, 24 Jun 2016 11:45:36 +0100 MIME-Version: 1.0 Sent to CCL by: Tandon Swetanshu [tandons[A]tcd.ie] --001a114168d6f6c4be053603de7f Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Hi All, I have a question somewhat related to this topic. When working out the J values in a system with more than 3 metal atoms, there are many different solutions. Each solutions is obtained by choosing a different set of equations. From the above discussion it seems to me that the large differeence in the solutions are due to the large number of unpaired electrons and the reduction in the spacing between levels at higher energy. Due to this, depending upon the states under consideration the J values obtained would differ (please correct me if I am wrong). But how does one decide as to which set of solution is appropriate. Thanks, Swetanshu. On 12 June 2016 at 00:27, James Buchwald buchwja/rpi.edu < owner-chemistry#%#ccl.net> wrote: > Hi Henrique, > The diminishing Jab that you're predicting assumes that (E[HS] - E[BS]) > does not grow as quickly as the spin term in the denominator. Depending = on > the system, this is not necessarily the case, and the energy spacing can > grow faster. > > The reason that the equations appear to cause this is that the > Heisenberg-Dirac-van Vleck Hamiltonian (which the three equations were > derived from) has a "spin ladder" of solutions ranging from the low-spin = to > the high-spin states. If your low-spin state is a singlet, you'll also > have triplets, pentets, and so on until you reach the high-spin state. > Similarly, if you start from a doublet, you'll have intermediate quartets= , > etc. > > As you introduce more and more unpaired electrons, the spin of the > high-spin state increases - but all of the intermediate states between th= e > high-spin and low-spin limits still exist. You can work out the splittin= g > between these individual states in terms of J, and what ends up happening > is that the states spread out. The denominator essentially corrects for > that spacing, rather than saying anything about the strength of the > magnetic coupling. > > Best, > James > > On 06/11/2016 05:54 PM, Henrique C. S. Junior henriquecsj-x-gmail.com > wrote: > > I hope this is not a "homework" question, but I'm having a bad time tryin= g > to figure this out. > Available literature proposes 3 equations to calculate the coupling > constant during a Broken-Symmetry approach: > > J(1) =3D -(E[HS]-E[BS])/Smax**2 > J(2) =3D -(E[HS]-E[BS])/(Smax*(Smax+1)) > J(3) =3D -(E[HS]-E[BS])/(HS-BS) > > I'm intrigued by the fact that, from the equations, the more the system > have unpaired electrons, the minor will be Jab. Why does this happen? > Doesn't more unpaired electrons increase magnetic momenta (and an increas= e > in magnetic coupling)? > > -- > *Henrique C. S. Junior* > Qu=C3=ADmico Industrial - UFRRJ > Mestrando em Qu=C3=ADmica Inorg=C3=A2nica - UFRRJ > Centro de Processamento de Dados - PMP > > > -- > James R. Buchwald > Doctoral Candidate, Theoretical Chemistry > Dinolfo Laboratory > Dept. of Chemistry and Chemical Biology > Rensselaer Polytechnic Institutehttp://www.rpi.edu/~buchwj > > --001a114168d6f6c4be053603de7f Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable
Hi All,

I have a question somewhat rela= ted to this topic. When working out the J values in a system with more than= 3 metal atoms, there are many different solutions.=C2=A0 Each solutions is= obtained by choosing a different set of equations. From the above discussi= on it seems to me that the large differeence in the solutions are due to th= e large number of unpaired electrons and the reduction in the spacing betwe= en levels at higher energy. Due to this, depending upon the states under co= nsideration the J values obtained would differ (please correct me if I am w= rong). But how does one decide as to which set of solution is appropriate. =

Thanks,
Swetanshu.

On 12 June 2016 at 00:27, J= ames Buchwald buchwja/rpi.edu<= /a> <owner-chemistry#%#ccl.net> wrote:
=20 =20 =20

Hi Henrique,

The diminishing Jab that you're predicting assumes that (E[HS] - E[BS]) does not grow as quickly as the spin term in the denominator.=C2=A0 Depending on the system, this is not necessarily the case, and the energy spacing can grow faster.

The reason that the equations appear to cause this is that the Heisenberg-Dirac-van Vleck Hamiltonian (which the three equations were derived from) has a "spin ladder" of solutions ranging f= rom the low-spin to the high-spin states.=C2=A0 If your low-spin state is a singlet, you'll also have triplets, pentets, and so on until you reach the high-spin state.=C2=A0 Similarly, if you start from a doublet= , you'll have intermediate quartets, etc.

As you introduce more and more unpaired electrons, the spin of the high-spin state increases - but all of the intermediate states between the high-spin and low-spin limits still exist.=C2=A0 You can wo= rk out the splitting between these individual states in terms of J, and what ends up happening is that the states spread out.=C2=A0 The denominator essentially corrects for that spacing, rather than saying anything about the strength of the magnetic coupling.

Best,
James

On 06/11/2016 05:54 PM, Henrique C. S. Junior h= enriquecsj-x-gmail.com wrote:
I hope this is not a "homework" question, but I'm= having a bad time trying to figure this out.
Available literature proposes 3 equations to calculate the coupling constant during a Broken-Symmetry approach:

J(1) =3D -(E[HS]-E[BS])/Smax**2
J(2) =3D -(E[HS]-E[BS])/(Smax*(Smax+1))
J(3) =3D -(E[HS]-E[BS])/(<S**2>HS-<S**2>BS)

I'm intrigued by the fact that, from the equations, the more the system have unpaired electrons, the minor will be Jab. Why does this happen? Doesn't more unpaired electrons increase magnetic momenta (and an increase in magnetic coupling)?

--
<= font face=3D"monospace, monospace">Henrique C. S= . Junior
Qu=C3=ADmico Industrial - UFRRJ
<= font face=3D"monospace, monospace">Mestrando em Qu=C3=ADmica Inorg=C3=A2nica - UFRRJ
Centro de Processamento de Dados - PMP

--=20
James R. Buchwald
Doctoral Candidate, Theoretical Chemistry
Dinolfo Laboratory
Dept. of Chemistry and Chemical Biology
Rensselaer Polytechnic Institute
http://www.rpi.edu=
/~buchwj

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