Is there are any other factors that determines what type of spinel will form ?
I think that experience shows that the answer of Alas Zaben is a popular misconception since the CN of the tetrahedral site is only 4, whereas that of the octahedral site is 6 so for a classic spinel structure where there is no LFSE/CFSE argument to use, the divalent, and usually larger, cation takes the tetrahedral sites and the the trivalent one takes the octahedral sites. The classic example is MgAl2O4 where Mg is the M(II) cation and Al the M(III) cation. The relative ionic radii are: Mg(II) = 0.74 Angstrom and Al(III) = 0.54 Angstrom. It is more or less fair to say that the ionic radii of Mg(Ii), Mn(II), Fe(II) and Co(II) are all very close to circa 0.70 Angstrom and yet the spinel versus inverse spinel structures of Mn3O4, Fe3O4 and Co3O4 can be very nicely explained by simple ligand field considerations.
So, the classic spinel structure of formula M(II)M'(III)2O4 has M(II) in the tetrahedral sites and M(III) in the octahedral sites. In the inverse structure, half of the M(III) ions are found in the tetrahedral sites because it is energetically more favourable to have the M(II) ions in octahedral sites. Usually LFSE arguments can be used to justify this, but probably these represent limiting cases and for systems such as Mn3O4 it is difficult to assign defined oxidation states.
Note that this is bearing in mind that ionic radii are tricky to determine for cations since the whole thing is determined by the close packing of the anions - but spinels are a nice example to use since the close pack g of the oxides stays mostly constant until we get to defect structures such as are seen in manganites.
Back to the transition metal ion cases, which are more easily explained using LFSE arguments but can be contrary to what is expected.
Here the classic examples are to compare the mixed valent Fe3O4, (inverse spinel) and Co3O4 (normal) and Mn3O4 (complicated by defects/oxidation state assignments of the metal ions) structures. In all cases the transition metal ions are high spin, so the main factor is the LFSE of the tetrahedral site versus the octahedral site for the M(II) versus M(III) case to determine whether the structure moves from spinel to inverse spinel.
Here it is important to look at the octahedral site first since this has the larger LFSE.
For the inverse spinel structure of Fe3O4 it is easy since the Fe(III) ion has no preference by virtue of it being d5 high spin - so no LFSE in any configuration - we can check this out for the two possibilities: Octahedral, LFSE = 3 x 4Dq for the stabilising t2g orbitals and 2 x 6 Dq for the destabilising eg. orbitals = (12-12) Dq = 0 and for tetrahedral LFSE = 2 x 6 Dq - 3 x 4 Dq = (12 - 12)Dq = 0
As an aside, this is actually obvious once you realise that the splitting of the orbitals has to be around a barycentre so that energy is neither lost or gained in relation to the situation in a spherical (isotropic) field - so one electron per orbital has to come to no change in the energy compared with this spherical field. That means for a splitting energy of 10 Dq the energy gain for the stabilised orbitals has to equal the energy loss for the destabilised orbitals. If the field is octahedral, 3 t2g orbitals are stabilised by 4Dq and 2 eg orbitals are destabilised by 6 Dq: (3 x 4 - 2 x 6)Dq = (12 - 12)Dq = 0 Dq. If it is tetrahedral there are two stabilised e orbitals and 3 destabilised t2 orbitals and we have (2 x 6)Dq - (3 x 4)Dq = (12 - 12)Dq = 0 Dq
For the inverse spinel Fe3O4 (magnetite) the situation is directed by the balance of what the Fe(II) ion prefers and here the stabilisation of the d6 high spin Fe(II) in an octahedral field is higher than that in a tetrahedral field as long as we remember that the splitting of the d orbitals is much smaller for a tetrahedral field at 4/9 of that for an octahedral field.
For an octahedral site you gain one extra electron in the t2g orbitals (so another 4 Dq compared with Fe(III)) and for the tetrahedral site you gain and extra electron in the energetically lower e orbital, which sounds good at a further 6 Dq compared with the Fe(III) high spin d5 configuration, but we have to consider the smaller energy gain compared with the octahedral field: Octahedral LFSE = (4 x 4 Dq - 2 x 6Dq) = 4 Dq and tetrahedral LFSE = (3 x 6 Dq - 3 x 4 Dq) = 6 Dq. Note that for Fe(II) d6 we did not have to pair any further electrons on moving between octahedral and tetrahedral geometry. However, the apparently better stabilsation of 6 Dq for the tetrahedral case is actually only 4/9 x 6 Dq compared with the octahedral case, so lower than the 4 Dq we calculated and about 2.7 Dq. So, energetically speaking the octahedral sites will be preferred by the Fe(II) ions since these have an energy advantage. Note that here the ionic radius is a smaller influence - theoretically, the larger M(II) ion should prefer the tetrahedral site. The end result is that half of the Fe(III) ions take the available tetrahedral sites and the remaining Fe(III) ions plus the Fe(II) ions take the available octahedral sites.
For the Co3O4 case, we have high spin Co(II) d7 versus high spin Co(III) d6. High spin octahedral Co(III) has 4 electrons in the stabilised orbitals and two in the destabilising t2g orbitals. That gives a LFSE of (4 x 4Dq - 2 x 6 Dq) = 4 Dq for the octahedral case. For the tetrahedral case there are 3 electrons in the e levels and 3 in the t2 levels, so (3 x 6Dq - 3 x 4Dq) = 6 Dq - but only 4/9 of what we gain for the octahedral case = 4/9 x 6Dq which is about 2.7 Dq compared with the octahedral case and we have not had to pair any extra electrons.
For the d7 Co(II) ion there are 5 electrons in the t2g and 2 in the e.g. levels for octahedral geometry - so we have a LFSE of (5 x 4Dq - 2 x 6Dq) = 8Dq which looks a better situation than for the octahedral Co(III) since we gain a stabilisation of 4 Dq, but we have to spend pairing energy (PE) in order to have one more electron pair compared with the Co(III) case. So, we should write this as LFSE = 8 Dq - 1PE. If the pairing energy is more than 4 Dq, then it is better for this larger ion to go to the tetrahedral site.
For the d7 ion in a tetrahedral situation we have 4 electrons in the e levels and 3 in the t2 levels and this gives us (4 x 6Dq - 2 x 4Dq) = 4/9x16 Dq - 1 PE which is better than for the octahedral case for this ion and also much better than for the Co(III) case as long as the pairing energy is not an issue. So we have 7 Dq - 1 PE versus 2.7 Dq and the structure is a normal spinel one with the M(II) in the tetrahedral site and all the M(III) in the octahedral sites..
If you do the same type of argument for Mn3O4 you now have Mn(III) versus Mn(II) - Mn(II) is the same bet as Fe(III) was being d5 high spin, so no advantage for any ligand field. So, the Mn(III) has to decide and the d4 situation is clearly better for an octahedral case at (3 x 4Dq - 1 x 6 Dq) = 6 Dq versus (2 x 6 Dq - 2 x 4 Dq) = 4 Dq and then with the 4/9 factor giving only about 1.8 Dq stabilisation for the tetrahedral case compared with the 6 Dq for the octahedral one. So here, the M(III) ions take all of the octahedral sites.
After all of that, maybe we should return to the possible reason for asking the question on the first place, and maybe it was in view of when these structures are no longer valence localised.
On some timescale it is likely that these systems cannot be adequately described in terms of valence localised structures which can be partly rationalised by the fact that there is little energy difference (using our simple ligand field - more accurately, crystal field) arguments. Indeed, the Mn3O4 structure is a defect one with interesting electronic properties and Fe3O4 is famous as lodestone, or magnetite, the first magnetic material used by mankind and its microcrystalline black appearance and shiny macrocrystalline states suggest that the electrons are fairly mobile within the structure.
Structures based on the parent Mn, Fe and Co cases with small amounts of other metal ions doped into them provide a fascinating field of research with potential uses in many fields of materials chemistry.
They also provide textbook examples for teaching some basic principles of crystal field theory - even though their real electronic structures are probably much more complicated!
There are several factors that may affect the spinel structure ; normal or inverse structure .
In addition to LFSE factor , the relative size of divalent and trivalent cations will play an important role , the smaller the cation , the lower coordination site will be occupied i.e.,tetrahedral site .
I think that experience shows that the answer of Alas Zaben is a popular misconception since the CN of the tetrahedral site is only 4, whereas that of the octahedral site is 6 so for a classic spinel structure where there is no LFSE/CFSE argument to use, the divalent, and usually larger, cation takes the tetrahedral sites and the the trivalent one takes the octahedral sites. The classic example is MgAl2O4 where Mg is the M(II) cation and Al the M(III) cation. The relative ionic radii are: Mg(II) = 0.74 Angstrom and Al(III) = 0.54 Angstrom. It is more or less fair to say that the ionic radii of Mg(Ii), Mn(II), Fe(II) and Co(II) are all very close to circa 0.70 Angstrom and yet the spinel versus inverse spinel structures of Mn3O4, Fe3O4 and Co3O4 can be very nicely explained by simple ligand field considerations.
So, the classic spinel structure of formula M(II)M'(III)2O4 has M(II) in the tetrahedral sites and M(III) in the octahedral sites. In the inverse structure, half of the M(III) ions are found in the tetrahedral sites because it is energetically more favourable to have the M(II) ions in octahedral sites. Usually LFSE arguments can be used to justify this, but probably these represent limiting cases and for systems such as Mn3O4 it is difficult to assign defined oxidation states.
Note that this is bearing in mind that ionic radii are tricky to determine for cations since the whole thing is determined by the close packing of the anions - but spinels are a nice example to use since the close pack g of the oxides stays mostly constant until we get to defect structures such as are seen in manganites.
Back to the transition metal ion cases, which are more easily explained using LFSE arguments but can be contrary to what is expected.
Here the classic examples are to compare the mixed valent Fe3O4, (inverse spinel) and Co3O4 (normal) and Mn3O4 (complicated by defects/oxidation state assignments of the metal ions) structures. In all cases the transition metal ions are high spin, so the main factor is the LFSE of the tetrahedral site versus the octahedral site for the M(II) versus M(III) case to determine whether the structure moves from spinel to inverse spinel.
Here it is important to look at the octahedral site first since this has the larger LFSE.
For the inverse spinel structure of Fe3O4 it is easy since the Fe(III) ion has no preference by virtue of it being d5 high spin - so no LFSE in any configuration - we can check this out for the two possibilities: Octahedral, LFSE = 3 x 4Dq for the stabilising t2g orbitals and 2 x 6 Dq for the destabilising eg. orbitals = (12-12) Dq = 0 and for tetrahedral LFSE = 2 x 6 Dq - 3 x 4 Dq = (12 - 12)Dq = 0
As an aside, this is actually obvious once you realise that the splitting of the orbitals has to be around a barycentre so that energy is neither lost or gained in relation to the situation in a spherical (isotropic) field - so one electron per orbital has to come to no change in the energy compared with this spherical field. That means for a splitting energy of 10 Dq the energy gain for the stabilised orbitals has to equal the energy loss for the destabilised orbitals. If the field is octahedral, 3 t2g orbitals are stabilised by 4Dq and 2 eg orbitals are destabilised by 6 Dq: (3 x 4 - 2 x 6)Dq = (12 - 12)Dq = 0 Dq. If it is tetrahedral there are two stabilised e orbitals and 3 destabilised t2 orbitals and we have (2 x 6)Dq - (3 x 4)Dq = (12 - 12)Dq = 0 Dq
For the inverse spinel Fe3O4 (magnetite) the situation is directed by the balance of what the Fe(II) ion prefers and here the stabilisation of the d6 high spin Fe(II) in an octahedral field is higher than that in a tetrahedral field as long as we remember that the splitting of the d orbitals is much smaller for a tetrahedral field at 4/9 of that for an octahedral field.
For an octahedral site you gain one extra electron in the t2g orbitals (so another 4 Dq compared with Fe(III)) and for the tetrahedral site you gain and extra electron in the energetically lower e orbital, which sounds good at a further 6 Dq compared with the Fe(III) high spin d5 configuration, but we have to consider the smaller energy gain compared with the octahedral field: Octahedral LFSE = (4 x 4 Dq - 2 x 6Dq) = 4 Dq and tetrahedral LFSE = (3 x 6 Dq - 3 x 4 Dq) = 6 Dq. Note that for Fe(II) d6 we did not have to pair any further electrons on moving between octahedral and tetrahedral geometry. However, the apparently better stabilsation of 6 Dq for the tetrahedral case is actually only 4/9 x 6 Dq compared with the octahedral case, so lower than the 4 Dq we calculated and about 2.7 Dq. So, energetically speaking the octahedral sites will be preferred by the Fe(II) ions since these have an energy advantage. Note that here the ionic radius is a smaller influence - theoretically, the larger M(II) ion should prefer the tetrahedral site. The end result is that half of the Fe(III) ions take the available tetrahedral sites and the remaining Fe(III) ions plus the Fe(II) ions take the available octahedral sites.
For the Co3O4 case, we have high spin Co(II) d7 versus high spin Co(III) d6. High spin octahedral Co(III) has 4 electrons in the stabilised orbitals and two in the destabilising t2g orbitals. That gives a LFSE of (4 x 4Dq - 2 x 6 Dq) = 4 Dq for the octahedral case. For the tetrahedral case there are 3 electrons in the e levels and 3 in the t2 levels, so (3 x 6Dq - 3 x 4Dq) = 6 Dq - but only 4/9 of what we gain for the octahedral case = 4/9 x 6Dq which is about 2.7 Dq compared with the octahedral case and we have not had to pair any extra electrons.
For the d7 Co(II) ion there are 5 electrons in the t2g and 2 in the e.g. levels for octahedral geometry - so we have a LFSE of (5 x 4Dq - 2 x 6Dq) = 8Dq which looks a better situation than for the octahedral Co(III) since we gain a stabilisation of 4 Dq, but we have to spend pairing energy (PE) in order to have one more electron pair compared with the Co(III) case. So, we should write this as LFSE = 8 Dq - 1PE. If the pairing energy is more than 4 Dq, then it is better for this larger ion to go to the tetrahedral site.
For the d7 ion in a tetrahedral situation we have 4 electrons in the e levels and 3 in the t2 levels and this gives us (4 x 6Dq - 2 x 4Dq) = 4/9x16 Dq - 1 PE which is better than for the octahedral case for this ion and also much better than for the Co(III) case as long as the pairing energy is not an issue. So we have 7 Dq - 1 PE versus 2.7 Dq and the structure is a normal spinel one with the M(II) in the tetrahedral site and all the M(III) in the octahedral sites..
If you do the same type of argument for Mn3O4 you now have Mn(III) versus Mn(II) - Mn(II) is the same bet as Fe(III) was being d5 high spin, so no advantage for any ligand field. So, the Mn(III) has to decide and the d4 situation is clearly better for an octahedral case at (3 x 4Dq - 1 x 6 Dq) = 6 Dq versus (2 x 6 Dq - 2 x 4 Dq) = 4 Dq and then with the 4/9 factor giving only about 1.8 Dq stabilisation for the tetrahedral case compared with the 6 Dq for the octahedral one. So here, the M(III) ions take all of the octahedral sites.
After all of that, maybe we should return to the possible reason for asking the question on the first place, and maybe it was in view of when these structures are no longer valence localised.
On some timescale it is likely that these systems cannot be adequately described in terms of valence localised structures which can be partly rationalised by the fact that there is little energy difference (using our simple ligand field - more accurately, crystal field) arguments. Indeed, the Mn3O4 structure is a defect one with interesting electronic properties and Fe3O4 is famous as lodestone, or magnetite, the first magnetic material used by mankind and its microcrystalline black appearance and shiny macrocrystalline states suggest that the electrons are fairly mobile within the structure.
Structures based on the parent Mn, Fe and Co cases with small amounts of other metal ions doped into them provide a fascinating field of research with potential uses in many fields of materials chemistry.
They also provide textbook examples for teaching some basic principles of crystal field theory - even though their real electronic structures are probably much more complicated!
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