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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.

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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).

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  • $\begingroup$ 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 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 at 22:10

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