Here is the simplest explanation of the double exchange I've ever seen.

http://en.wikipedia.org/wiki/Double-exchange_mechanism

Double exchange mechanism was first introduced in Zener's paper in 1951(C. Zener, Phys. Rev. 81, 440). According to which in Mn4+-Mn3+ mixed valence complexes, two electrons are transferred simultaneously,one from a bridging O2- anion to the MnIV centre and one from a MnIII centre to the O2- anion and during these transfers electrons retain their spins.These electron transfers are only possible if the unpaired electrons of two metal centers are in high spin state.Due to this, high spin state gets extra stabilization than other spin states and complexes show ferromagnetic character. For more detail information about double exchange in Mn(III-IV) system please look into Chem. Eur. J. 2002, 8,No . 21.

in the modern interpretation, in is a generic property of high-spin magnetic compounds (say S=2 or so in the case of transition-metal oxides): the electron transferring to certain central site from the neighboring sites should have the same direction of spin as the central site, otherwise you lose in the exchange interaction (Hund's) energy. Therefore, the magnetic coupling should be ferromagnetic. It is not necessary to have simultaneously Mn3+ and Mn4+ or the intermediate oxygen states. Mathematically, it implies that exchange splitting between the majority- and minority-spin states is so large that one can eliminate the latter, that is equivalent to the redefinition of the transfer integrals (multiplication by certain factor, depending on the relative direction of spins).

Look for more details book "Quantum theory of magnetism" by Wolfgang Nolting.

http://books.google.rs/books?id=vrcHC9XoHbsC&printsec=frontcover&dq=Quantum+theory+of+magnetism&hl=sr&sa=X&ei=7cDZUYPvHcqS4AThsYCQBg&redir_esc=y#v=onepage&q=Quantum%20theory%20of%20magnetism&f=false

Double exchange interaction is due to basically an exchange of electrons between Mn3+ and Mn4+ (For your case) though another paramagnetic ion (O2- in your case). It is very similar to super exchange interaction where the exchange occurs with simillar charge species but in this case the exchange occurs only when an extra electron is present in one case. To learn more about it, you can follow the electron hoping mechanism in Fe3O4 nanoparticles where exchange interactions occurs between Fe2+ and Fe3+

Thank you

The physics of multiferroics RMn2O5 cannot be governed by double exchange mechanism because the interaction between Mn spins (that form a ring of five Mn^3+ and Mn^4+ ) dominantly antiferromagnetic. The double exchange mechanism of zener leads to ferromagnetic interaction.

@Chatterji Sir......I beg your pardon, but in the literature for RMn2O5, it has been insisted that the ferromagnetism in RMn2O5 is due to the double exchange mechanism between Mn spins. Please tell us, if not double exchange mechanism, then which phenomenon governs the multiferroic behavior of RMn2O5. Hope you have something new for us.

Dinesh, to speed up the discussion, would you provide the references that make your point? thanks

Even when two metal centers are antiferromagnetically coupled they can give ferromagnetic ground state if the double exchange parameter(B) >> magnetic coupling constant (J) and double exchange parameter is strongly dependent on some factors like bridging ligands, metal-metal distance. So, systems must have favorable conditions for double exchange to operate.

So far as I know the RMn2O5 with all rare earth orders antiferromagnetically at first to an incommensurate phase IC1 below T_N. At lower temperature below T1 below T1 there is a transition to a commensurate phase with the propagation vector k = (1/2,0,1/4). At further low temperature below T2 there is again a transition to an incommensurate phase. This is the general tendency and the typical examples are YMn2O5 and TbMn2O5. There is no ferromagnetism. Now please tell me where did you read that RMn2O5 is ferromagnetic?

Reference for the magnetic structures of RMn2O5:

L.C. Chapon et al., Phys. Rev. Lett. 93, 177402 (2004).

C. Vecchini et al., Phys. Rev. B 77, 134434 (2008).

P.G. Radaelli et al., Phys. Rev. Lett. 101, 067205 (2008).

C. Wilkinson, P.J. Brown and T. Chatterji, Phys. Rev. B 84, 224422 (2011).

P.J. Brown and T. Chatterji, J. Phys.: Cndens. Matter. 25, 236004 (2013).

You may also consult the references listed in these papers.

@Tapan Sir.....I apologize for that. Yeah RMn2O5 except YMn2O5 show antiferromagnetism. I did not have doubt on that. Actually I misunderstood your comment. You are right Sir. Thank you so much for your valuable comments.

@Kai Fauth Sir.....actually I misunderstood the comment of Tapan Chaterji Sir. I thought that according to his comment the magnetism is not because of the double exchange mechanism. I missed that word ferro and antiferro. Sorry for that.

Thank you so much your attention and valuable suggestions.

@Soumen Ghosh......thanks for your participation in the healthy discussion. You provided very nice information but I don't know about bridging ligands.

@Dinesh in chemistry literature I have seen oxo bridged mixed valence complexes have ferromagnetic ground states only when they have very short metal metal distances(around 3 angstrom) but some pi bridged complexes are found to show ferromagnetic ground states even when metals are 6-7 Angstrom apart. So, I think bridging ligands and metal metal distance are very important here.

@Soumen Ghosh......Here in RMn2O5 we have many types of bonds like Mn3+ - O - Mn4+, Mn3+ - O - Mn3+, Mn4+ - O - Mn4+. So no metal - metal bonding is present here. The structure of RMn2O5 is very complex.

Is the concept of bridging legends applicable here?

I don't know much about chemistry. Pardon me if this is a silly question.

metal-metal bonding is definitely not there as it would make the complex diamagnetic but when metals are very close to each other metal to metal direct electron transfer happens. That kind of interaction can also help double exchange to operate without involving bridging ligand. I think oxo bridging ligand is not good for electron delocalization, so metal-metal interaction is the only way. for oxo complexes. Now, as you have said there can be different kinds of interactions in the complex. Like Mn3+ - O - Mn3+, Mn4+ - O - Mn4+ systems can be antiferromagnetically coupled but Mn3+ - O - Mn4+ system can be ferromagnetic due to double exchange. Overall ground state of the cluster is determined by no of Mn3+ and Mn4+ ions and how they are connected.

The known RMn2O5 crystal structure gives you the position of Mn3+ and Mn4+ positions. They occupy different crystallographic sites and are fully known. The structure consists of equal number of Mn3+ and Mn4+ ions in definite positions in the unit cell. It is not like doped manganites La1-xCaxMnO3 where Mn3+ and Mn4+ are distributed with definite probabilities only. Instead of speculating and trying to find parallel in metalloorganic chemistry or cluster chemistry just look at the references I have already given. The physics here is very much different.

I like to see the Reference where anyone has discussed double exchange mechanism in these materials. Instead of talking vague things just provide references.

(Mn3+ - O - Mn4+) is double exchange - ferromagnetic for angle above ~120, (Mn3+ - O - Mn3+) is super exchange - antiferromagnetic for angle above ~120 , but (Mn3+ - O - Mn3+) can be super exchange - feromagnetic if angle les than ~120.

@Michael......I did not understand your comment properly. Could you please elaborate a little? Which angle you are talking about?

Thank you.

The angle between Mn - O - Mn. As in (a) the angle is 180 degree with Mn3+ - O - Mn3+, it's call super-exchange, give rise to antiferromagnetic arrangement. (b) the angle is 180 but the the right side Mn less 1 electron (Mn4+), so Mn3+ - O - Mn4+, it's call double-exchange, give rise to ferromagnetic spin arrangement. For (c) the configuration is Mn3+ - O - Mn3+, it's super-exchange and gotta be antiferomagnetic but because the angle is 90 degree than it become ferromagnetic.

This base on the electron density of Mn 3d orbital and Oksigen 2p orbital that act as a bridge between the two Mn ions.

Double exchange

Angle betwen Mn3+ - O -Mn4+ is 132.6

Super exchange

Angle between Mn3+ - O Mn3+ is 94.3

Here is the complete 1 unit cell spin arrangement of RMn2O5

Mn3+ occupied octahedral site (rhombus)

Mn4+ occupied tetrahedral site / pyramid (triangle)

@Tapan Sir......would you like to comment something on the above comments by Michael Tem?

Thank you.

As we known, the double-exchange (DE) is a type of a magnetic interaction between two transition metal ions via oxygen ions. For example, in the electronic structure of the ideal lattice of AMnO3 perovskite oxides, the Mn (3d) orbital splits into two states, a triple t2g3 and double eg2 sub-orbitals due to the crystal field (CF) created by the cubic symmetry surrounding the Mn (3d) sites. The doublet eg sub-orbitals lie a few eV above the triplet t2g sub-orbitals in energy. Therefore, if O2– (2p6) gives up its spin-up electron to Mn4+ (3d3) ion, it is vacant 2p orbital can then be filled by an electron hopping from Mn3+ (3d4) ion. At the end of this long-rang magnetic process, an electron has hopped between the neighboring transition metal ions, retaining its spin. The double-exchange theory predicts that the electron movement from one species to another will be facilitated more easily if the hopping electrons do not have to change their spin direction in order to obey the Hund's rule when on the accepting species. The ability to hop, or to localized, reduces the kinetic energy, hence the overall energy saving can lead to ferromagnetic (FM) alignment of the neighboring transition metal ions.

Who can plot an asymmetric super exchange and double exchange paths in Fe3O4 with indication of orbital hybritization?

Fe2+ (Oh)-O- Fe2+(Oh) DE

Fe2+(Oh)-O2- Fe3+(Td) SE

Fe3+(Oh)-O2- Fe3+(Td) SE

@ I. V. Solovyev, I'm quite confused about that in your description, the double exchange can be achieved even without the different valence states as well as the itinerant electron???

@Shan wan Fei: Yes. This is the definition of double exchange by P.-G. de Gennes, Phys. Rev. 118, 141 (1960).

@ I. V. Solovyev: It's very nice of you to reply so quickly, I'll check the reference you offered here. Thank you!

Is this double exchange mechanism only possible with Oxygen atoms or also possible with other p-block atoms ?

Dibya Prakash Rai Of course it could happen with other 2p orbitals(e.g. S, Se).

thank you Wanfei Shan

Double exchange interaction was proposed by Zener (1951) to account for the interaction between adjacent ions of parallel spins through the neighboring oxygen ion.This model is more restrictive than the super exchange interaction and requires the presence of ions of the same element, but in different valence states; for example, in magnetite Fe+2 and Fe+3. The excitation of a d-electron from the cation with the largest number of electrons, in the magnetite of the Fe+2 ion, into an overlapping anionic orbital (oxygen ion) with the simultaneous transfer of a p-electron with the same spin from an anion to a neighboring cation (Fe+3 ion)