4
$\begingroup$

Cerium in $0$ oxidation state has electronic configuration

$$[\ce{Xe}]\mathrm{(4f)^1(5d)^1(6s)^2}$$

But when it gets oxidised to $+2$ state, it becomes

$$[\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$

This is a change as in other elements in the lanthanides series get their $\mathrm{6s}$ electrons abstracted and the other shells remain at liberty. Could you please explain this phenomenon to me? And also correct me if anything unsaid above was incorrect.

Improve this question
$\endgroup$
2
  • $\begingroup$ Related, but having no answers with positive scores: chemistry.stackexchange.com/questions/83170/…. $\endgroup$
    – Oscar Lanzi
    Mar 9, 2019 at 10:33
  • $\begingroup$ I think it's because that when we go across the lanthanides, at the first d orbital is penetrated more, but then the f orbital penetrates more and d becomes the outer orbital after s, and electrons like to be closer to the nucleus duo to columbic forces, and that's what's happened there. $\endgroup$
    – user130315
    Apr 7 at 14:53

1 Answer 1

Reset to default
7
$\begingroup$

The lanthanides are most stable in their 3+ oxidation state. For example, lanthanum is perfectly happy changing from

$$\mathrm{La}^0 = [\ce{Xe}]\mathrm{(5d)^1(6s)^2}$$ to

$$\mathrm{La}^{3+} = [\ce{Xe}]\mathrm{(5d)^0(6s)^0}$$

Then, as you go along the lanthanide series, you're adding more 4f electrons. For example:

$$\mathrm{Ce}^{3+} = [\ce{Xe}]\mathrm{(4f)^1(5d)^0(6s)^0}$$ $$\mathrm{Pr}^{3+} = [\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$ $$\mathrm{Nd}^{3+} = [\ce{Xe}]\mathrm{(4f)^3(5d)^0(6s)^0}$$

etc.

This electronic configuration is extremely stable for all lanthanides, and there are very few exceptions to it.

Ce(II) is extremely unstable, and if there's one way to stabilise it, that would be adoption of the Pr(III) electronic structure. This way Ce has no 5d or 6s electrons, which makes lanthanides happy (apologies for the non-scientific way of framing this).

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Calcium in the zero oxidation state is highly unstable, whereas titanium in the +2 state is more stable as titanium monoxide. So why does the calcium not adopt the $d^2$ valence electron structure seen in the latter? $\endgroup$
    – Oscar Lanzi
    Mar 9, 2019 at 13:07
  • $\begingroup$ @OscarLanzi because Ca prefers losing the two s electrons rather than having a partially filled d shell? Whereas lanthanides are perfectly happy with partially filled 4f shells, as long as their 6s and 5d are empty. $\endgroup$
    – Gimelist
    Mar 9, 2019 at 22:10
  • 1
    $\begingroup$ In the "Chemistry of the Elements", By N. N. Greenwood and A. Earnshaw, it is stated that the only lanthanides which exist under the oxidation state $+2$ are Samarium, Europium and Ytterbium. Only a couple of cerium(II) compounds are known. But sulfide for example is probably made of Ce(III) plus one electron present in the conduction band. The most stable compound is the iodide $\ce{CeI2}$. But this diiodide is very easily oxidized : It liberates hydrogen from water. $\endgroup$
    – Maurice
    Apr 7 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.