Hi=2C Swetanshu=2C

It is not an easy task to decide what configuration is = correct to describe the magnetic couplings in a polynuclear system. The bes= t approach is to compare the various solutions with an experimental magneti= c susceptibility curve using a statistical fit software (like origin).

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Hen= rique C. S. Junior

Qu=EDmico Industrial - UFRRJ
Mestrando em Qu=EDmica I= norg=E2nica - UFRRJ

Centro de Processamento de Dados - PMP

From: owner-chemistry() ccl.net
To: hen= riquecsj() gmail.com
Subject: CCL: Coupling constant (Jab) - Why more unpa= ired electrons means a smaller coupling?
Date: Fri=2C 24 Jun 2016 11:45:= 36 +0100

Hi All=2C

I have a ques= tion somewhat related to this topic. When working out the J values in a sys= tem with more than 3 metal atoms=2C there are many different solutions.&nbs= p=3B Each solutions is obtained by choosing a different set of equations. F= rom the above discussion it seems to me that the large differeence in the s= olutions are due to the large number of unpaired electrons and the reductio= n in the spacing between levels at higher energy. Due to this=2C depending = upon the states under consideration the J values obtained would differ (ple= ase correct me if I am wrong). But how does one decide as to which set of s= olution is appropriate.

Thanks=2C

S= wetanshu.

On 12 June 2016 at 00:27=2C James Buchwald buchwja/rpi.edu <=3Bowner-chemistry=2Cccl.net<= /a>>=3B wrote:

=0A= =0A= =0A= =0A=
=0A= Hi Henrique=2C
=0A=
=0A= The diminishing Jab that you're predicting assumes that (E[HS] -=0A= E[BS]) does not grow as quickly as the spin term in the=0A= denominator. =3B Depending on the system=2C this is not necessarily= the=0A= case=2C and the energy spacing can grow faster.
=0A=
=0A= The reason that the equations appear to cause this is that the=0A= Heisenberg-Dirac-van Vleck Hamiltonian (which the three equations=0A= were derived from) has a "spin ladder" of solutions ranging from the=0A= low-spin to the high-spin states. =3B If your low-spin state is a= =0A= singlet=2C you'll also have triplets=2C pentets=2C and so on until you= =0A= reach the high-spin state. =3B Similarly=2C if you start from a dou= blet=2C=0A= you'll have intermediate quartets=2C etc.
=0A=
=0A= As you introduce more and more unpaired electrons=2C the spin of the=0A= high-spin state increases - but all of the intermediate states=0A= between the high-spin and low-spin limits still exist. =3B You can = work=0A= out the splitting between these individual states in terms of J=2C and= =0A= what ends up happening is that the states spread out. =3B The=0A= denominator essentially corrects for that spacing=2C rather than=0A= saying anything about the strength of the magnetic coupling.
=0A=
=0A= Best=2C
=0A= James
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=0A=
On 06/11/2016 05:54 PM=2C Henrique C. S.=0A= Junior h= enriquecsj-x-gmail.com wrote:
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I=0A= hope this is not a "homework" question=2C but I'm having a=0A= bad time trying to figure this out.
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Available=0A= literature proposes 3 equations to calculate the coupling=0A= constant during a Broken-Symmetry approach:
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J(1)=0A= =3D -(E[HS]-E[BS])/Smax**2
=0A=
J(2)=0A= =3D -(E[HS]-E[BS])/(Smax*(Smax+1))
=0A=
J(3)=0A= =3D -(E[HS]-E[BS])/(<=3BS**2>=3BHS-<=3BS**2>=3BBS)
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I'm=0A= intrigued by the fact that=2C from the equations=2C the more= =0A= the system have unpaired electrons=2C the minor will be Jab.= =0A= Why does this happen? Doesn't more unpaired electrons=0A= increase magnetic momenta (and an increase in magnetic=0A= coupling)?
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Henr= ique C. S. Junior
=0A= Qu=EDmico Industrial - UFRRJ
=0A=
Mestrando em Qu=EDmica=0A= Inorg=E2nica - UFRRJ
=0A= Centro de Processamento de Dados - PMP
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-- =0A= James R. Buchwald=0A= Doctoral Candidate=2C Theoretical Chemistry=0A= Dinolfo Laboratory=0A= Dept. of Chemistry and Chemical Biology=0A= Rensselaer Polytechnic Institute=0A= http://www.rpi.edu= /~buchwj
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