# Oxidation States of Actinides [closed]

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

## closed as off-topic by Jannis Andreska, NotEvans., Jan, paracetamol, Todd MinehardtSep 16 '17 at 14:44

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• 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. – andselisk Sep 16 '17 at 15:09
• @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 – NotEvans. Sep 16 '17 at 16:32
• @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. – andselisk Sep 16 '17 at 16:44

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 $\ce{XeF2}$.