# Determination of spinel structure by CFSE

$$\ce{CuFe2O4}$$ has inverse spinel structure, but $$\ce{ZnFe2O4}$$ has normal spinel structure. Explain.

I know that normal spinels have a structure like $$\ce{(A^{II})^{tet}(B^{III})^{oct}_2O_4}$$-like structure, while inverse spinels have a $$\ce{(B^{III})^{tet}(A^{II}B^{III})^{oct}O4}$$-like structure.

Also, I know that if $$\ce{B^{III}}$$ ion has greater CFSE gain in octahedral site than that of $$\ce{A^{II}}$$ ion, a normal spinel structure is expected. If $$\ce{A^{II}}$$ ion has greater CFSE gain in octahedral geometry than the trivalent $$\ce{B^{III}}$$ ion, then we get an inverse spinel structure.

For the given question, $$\ce{Fe}$$ has a $$\mathrm{3d^6}$$ configuration having a structure $$\mathrm{t_{2g}^4 e_{g}^2},$$ giving a CFSE

$$2(0.6 \Delta_\mathrm{o}) + 4(-0.4 \Delta_\mathrm{o}) = -0.4 \Delta_\mathrm{o},$$

while $$\ce{Zn}$$ has a $$\mathrm{3d^{10}}$$ structure having 0 CFSE. So, CFSE of $$\ce{Zn}$$ is greater than CFSE of $$\ce{Fe}.$$ So, according to the rule stated before, it gives me an inverse spinel structure.

Similarly for $$\ce{Cu},$$ it should have a $$\mathrm{3d^9}$$ $$(\mathrm{t_{2g}^6 e_g^3})$$ structure which gives a CFSE of

$$3(0.6 \Delta_\mathrm{o}) + 6(-0.4 \Delta_\mathrm{o}) = -0.6 \Delta_\mathrm{o} < -0.4 \Delta_\mathrm{o},$$

which gives a normal spinel structure.

What is wrong with this calculation so that I'm getting opposite structures to those predicted in the question?

• Your calculations are wrong because you should be looking at $Fe^{3+}$ which is $d^5$ and you are using $d^6$ – Gwyn Jul 17 at 19:16

In your situation the iron is Fe(III), so a $$d^5$$ system. For $$d^5$$ it does not matter if it's in an octahedron or a tetrahedron since the energies will just cancel out. So in both situations it will be determined by the other ion. For Zn(II) it also doesn't really matter. So I guess here the regular spinel dominates as Zn(II) just prefers more tetrahedral coordinations in general, especially in solid states while Fe(III) is happy with an octahedron.
For the Cu(II) however you've got a $$d^9$$ system. So the octahedron will give you something like $$(6*2/5)-(3*3/5) = 3/5$$ (ignoring things like spin pairing energ here). But in the tetrahedral case you also have to consider that the ligand field splitting for a tetrahedron is only $$4/9$$. So it will be something like $$4/9*[(4*3/5)-(5*2/5)] = 8/45$$ and therefore lower.