It is observed that actinides do not exhibit +2 oxidation state, contrary to that of some of the lanthanide elements. Why?

  • 3
    $\begingroup$ I vote to reopen the question as even though it contains traces of homework-alike style, it's quite fundamental and touches the chemistry of $\ce{f}$-elements which is not often discussed here. $\endgroup$
    – andselisk
    Commented Sep 16, 2017 at 15:09
  • $\begingroup$ @andselisk, I agree it could be a good question if edited, but as it stands there is no effort shown (it clearly falls within the scope of homework)- being interesting / niche doesn't really change this fact $\endgroup$
    – NotEvans.
    Commented Sep 16, 2017 at 16:32
  • $\begingroup$ @NotEvans. There are numerous similar questions falling into the same "Here is fact X. Why so?" pattern, though they and many others were accepted. I personally see some inconsistency in decision making. $\endgroup$
    – andselisk
    Commented Sep 16, 2017 at 16:44

1 Answer 1


See also: Why is WF6 stable whereas CrF6 is unknown?

Whenever one wants to compare oxidation states, there are a couple main factors to take into account. To reach a higher oxidation state, one obviously has to pay for it in the form of ionisation energy/energies. However, there is a compensatory effect in that elements in higher oxidation states generally get more out of bonding. For example, the lattice energy of $\ce{FeCl3}$ is larger than that of $\ce{FeCl2}$, or the four covalent bonds in $\ce{XeF4}$ are collectively stronger than the two covalent bonds in $\ce{XeF2}$.

The higher ionisation energies of the actinides tend to be smaller than those of the lanthanides, for the same reasons as explained in the linked question (5f orbitals have one radial node; 4f orbitals do not). This generally favours higher oxidation states and disfavours low oxidation states in actinides.


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