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For the 4f inner transition metals, +3 is the most common oxidation state (OS). Why is the +3 OS of cerium considered more stable than +4, at which it attains noble gas configuration?

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The question is a problem, it contains a fact which can not be trusted. It has the assumption that one oxidation state is more stable than another oxidation state.

The problem is that the relative stability of oxidation states depends on the chemical environment of an element. If we consider cobalt as an example then it is clear that cobalt(II) in a simple aqueous phase (dilute perchloric acid) will be more stable than cobalt(III) as the cobalt(III) is able to oxidize water and then convert itself into cobalt(II). There the cobalt will be in the form of [Co(H2O)6]^2+

Howver cobalt(II) in an aqueous cyanide solution is in the form of [Co(CN)6]4- which is a very strong reducing agent, this will reduce water to hydrogen. Thus in cyanide media cobalt(III) is the more stable oxidation state.

If we consider solid oxides then the cobalt(III) oxidation state is much more stable, in general the oxide ligand is a harder (in terms of Pearson's HSAB theroy) ligand than a water. The hard ligands tend to increase the stability of metals in the higher oxidation states. I know that in a lithium ion battery that there is an equilibrium between LiCoO2 and CoO2 at one of the electrodes. Here the oxide ligands are making the higher oxidation states of the cobalt more stable than they would be in aqueous media.

In the same way lead(IV) is far more stable in the form of solid metal oxides than it is in aqueous solution. Consider for a moment the mixed oxidation state oxide known as "red lead" which is Pb3O4 and the lead dioxide found on one of the plates of a lead acid battery (used normally as a "car battery").

Now back to the element which the question was about, the problem with the question is that it did not state what environment the cerium was in. if the cerium is in the solid state as an oxide then cerium dioxide (ceria) CeO2 will be very nice and stable. The other lanthanides do not form dioxides with the same ease. The reason why I think that cerium forms the dioxide with such ease that it forms a +4 ion with a noble gas configeration.

In aqueous media cerium(IV) is possible to form but I am sure that it is less stable, it is a strong oxidant.

Equally sometimes in the aqueous media there are times where the higher oxidation state is more stable. If we consider plutonium. Just the other day I exposed a little plutonium to ozone, this converts it from Pu(IV) to Pu(VI). The +6 oxidation state is sufficently stable in aqueous media (nitric acid) for me to do wet chemistry with it. But if we consider the metal oxides then plutonium trioxide is not something which can be formed with ease. The normal oxide for plutonium is the dioxide. There have been atttempts to make plutonium trioxide but when I last checked the literature it was not clear if anyone has ever managed to make a sample of it.

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Ce has electronic configuration [Xe]$\ce {4f^1}\ce{5d^1}\ce{6s^2}$. Electronic configuration of the Ce$^{3+}$ is [Xe]$\ce{4f^1}$ while that of Ce$^{4+}$ is [Xe]$\ce{4f^0}\ce{5d^0}\ce{6s^0}$ or simply [Xe]. It is evident from this that Ce$^{4+}$ has a noble gas configuration which means it should be more stable than Ce$^{3+}$ in which there is one left over f electron.

But a 4f electron is the one that is significantly deeper inside the atom $i.e.$ it is considerably closer to the nucleus compared to a 5d or a 6s electron. Even if you attain $+4$ for Cerium, it would require a large amount of energy to extract it from this much inside. However, you would already have achieved the $+3$ state before that in which you have already removed 5d and 6s electrons.

In other words, we cannot compensate the tremendous surge in energy of Ce$^{4+}$ with the stability achieved by attaining noble gas configuration.

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    $\begingroup$ When applying MathJax, it is better to apply it to everything maths related and not use it as a markup to. For example $\ce{Ce^4+}$ for $\ce{Ce^4+}$. You'll find more information on Chemistry Meta. As a happy accident, you can use mhchem also for electron configurations: $\ce{[Xe] 4f^1 5d^1 6s^2}$ $\endgroup$ Commented May 6, 2023 at 8:49

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