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I read that the heavier alkali metals, like K, Rb, and Cs all prefer to form superoxides. Since Fr is the heaviest alkali metal, I assumed it would follow the same trend as the previous alkali metals, forming $\ce{FrO2}$. However, I was told that the most stable oxide of Fr is $\ce{Fr2O}$. Why would this be the case?

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    $\begingroup$ You know that Fr is itself unstable? $\endgroup$ – Mithoron Apr 13 '16 at 23:28
  • $\begingroup$ @Mithoron Yes, I'm aware that Fr decays quickly, but I suppose this is somewhat of a hypothetical question. $\endgroup$ – carbenoid Apr 13 '16 at 23:30
  • $\begingroup$ en.wikipedia.org/wiki/Francium says this superoxide would have more covalent character, the cited source doesn't say much more. $\endgroup$ – Mithoron Apr 13 '16 at 23:40
  • $\begingroup$ All Alkali metal oxides below sodium are superoxides. This is because of the bigger alkali metal cations' extremely low charge density and the high polarizability of superoxide relative to peroxide and regular oxide. What this means in the real world, though, I do not know. I don't think Francium itself is stable, so what can we really say about the stability of its oxides? $\endgroup$ – gannex Apr 14 '16 at 2:57
  • $\begingroup$ Was that the correct answer to the local exam? I also chose the superoxide. $\endgroup$ – Yunfei Ma Apr 16 '16 at 19:13

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