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?


Are you sure $\ce {Ce}$ (III) is really considered more stable than $\ce {Ce}$ (IV)? Since the most common cerium based compound is its dioxide so $\ce {Ce}$ (IV) is actually more stable.

In addition to that, $\ce{Ce}$ has 4 electrons in its outer orbital, thus leaving only one unpaired electron on the $\ce{6s}$ orbital doesn't sound really stable to me.

  • $\begingroup$ Yes, +3 is more stable than +4. My best guess is it is based on experimental results. $\endgroup$ – Shubham Dec 3 '14 at 5:23
  • $\begingroup$ Then explain me why Ce (IV) is very commun while Ce (III) is almost never met in standard state ? Or tell me on what you base this statement (because actually you need reductive condition and hight temperature to stabilize it) $\endgroup$ – Babounet Dec 3 '14 at 6:59
  • $\begingroup$ I think both the answers are acceptable to this question and it depends on the medium you're using. In gas +4 is stable but gaseous compounds of Ce are not that common and in aqueous solution +3 is more stable. Note : In aqueous medium many factors apart from Ionization Energy are taken into account (Hydration Enthalpy, Sublimation Energy etc). Also Cu(II) is more stable in aqueous medium than Cu(I). $\endgroup$ – Shubham Dec 4 '14 at 11:12
  • $\begingroup$ Then precise in wich condition you think Ce (III) to be more stable than Ce (IV) $\endgroup$ – Babounet Dec 4 '14 at 11:14
  • $\begingroup$ Im sorry Im on mobile and SE doesn't work good on my phone. Please reread my previous comment. $\endgroup$ – Shubham Dec 4 '14 at 11:16

Well, as far my syllabus is concerned about, inspite of having F° configuration in $Ce(IV)$ it reduces itself easily to $Ce(III)$ as +3 O.S is predominantly stable in inner transitional metal series.


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