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I would like to ask a question about europium's stability in the $+2$ and $+3$ oxidation state.

The electronic configuration of europium in its neutral state is $\ce{[Xe] (4f)^7 (6s)^2}$.

Now, when in the $+2$ oxidation state, the electronic configuration is $\ce{[Xe] (4f)^7}$ and in the $+3$ oxidation state, it is $\ce{[Xe] (4f)^6}$.

Now, I thought the $+2$ oxidation state is more stable because it's a half-filled $\ce{f}$ sub-shell so there is less mutual repulsion between electrons in the same sub-shell.

However, an article from nature.com reads the following:

Europium metal is now known to be highly reactive; the element's most stable oxidation state is +3, but the +2 state also occurs in solid-state compounds and water.

Now, I would like to ask, how is it possible for europium to be most stable in its $+3$ oxidation state, knowing what I have mentioned above?

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  • $\begingroup$ We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. $\endgroup$ – andselisk Dec 10 '17 at 12:13
  • $\begingroup$ Somewhat related: Lanthanide property exceptions. It's worth noticing that the answers given there also use the same argumentation as you do, which I would also agree with. $\endgroup$ – andselisk Dec 10 '17 at 12:18
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This is going to be a rehash of Oscar Lanzi's answer, but this point needs to be driven home, so I make no apologies.

"Special" electron configurations - fully-filled or half-filled subshells - are only a relatively minor factor in determining the stability of oxidation states. I've written about this before in a slightly different context, but IMO it would be well worth reading this, as the exact same principles operate in this case: Cr(II) and Mn(III) - their oxidizing and reducing properties?

What determines the "most stable oxidation state"? We can consider what happens when we increase the oxidation state of a metal: firstly, we need to pay ionisation energies to remove electrons from the metal. On the other hand, though, metal ions in higher oxidation states can form stronger bonds (e.g. lattice energy in ionic compounds, solvation energy in solution, or more covalent bonds in molecular compounds). The balance between these two factors therefore leads to a "best" oxidation state.

For example, sodium could hypothetically go up to Na(II) and form $\ce{NaCl2}$. Theoretically, this compound would have a significantly larger lattice energy than $\ce{NaCl}$. However, sodium's second ionisation energy is prohibitively large, which means that it stays put in the +1 oxidation state. On the other hand, magnesium's second ionisation energy is not prohibitively large, and therefore Mg preferentially forms $\ce{MgCl2}$ over $\ce{MgCl}$.

For all lanthanides, the first three ionisation energies are all fairly comparable. There is a graph here which shows the variation in the ionisation energies of the lanthanides, which I reproduce below.

enter image description here

(source: Inorganic Chemistry 6ed, Weller et al., p 630)

For all lanthanides, these three ionisation energies are easily compensated for by the extra lattice energy / solvation energy derived from a more highly charged ion. The fact that Eu(II) has a $\mathrm{f^7}$ configuration only serves to make its third IE marginally larger than that of Gd. This difference is sufficient to make Eu(II) an accessible oxidation state (cf. different electronic behaviour of EuO and GdO), but not sufficient to make it the most stable oxidation state.

Note that "most stable oxidation state" depends on the conditions, too, but I assume we are talking about aqueous solution.

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Electron configuration is not the be-all and end-all of stability. It's just one term in the energy balance. Other factors like the electron affinity of the anion former (think of oxygen or fluorine versus sulfur or iodine), lattice or solvation energies, etc may override the electron configuration term; if so, then you find your europium in the +3 oxidation state after all.

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    $\begingroup$ That first sentence needs to be bolded in all caps. If there ever was a time to use obnoxious formatting for emphasis, this is it... $\endgroup$ – orthocresol Dec 10 '17 at 12:48
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One reason is that in the f orbital, the +2 oxidation state makes it that one pair of electron in the orbital is not filled, and will at least try to fill it. And the +3 orbitals filled those pairs, even though it didn't fill the whole orbital. Of course filling the whole orbital (like noble gases) will be great but it is easier for the metals to fill up it's pair.

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