Why are superoxide ions stable only in presence of large cations such as $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$?

  • 1
    $\begingroup$ I think it has a lot to with the heat of formation/..; See this link for some research on lithium superoxide: google.be/… $\endgroup$
    – user2117
    Commented Feb 9, 2014 at 17:16

3 Answers 3


According to the Wikipedia page Alkali metal oxides, the superoxide of $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ form when these metals are burnt in air, and according to the Bodner Research Group page Active Metal Reactions, they

react so rapidly with oxygen they form superoxides, in which the alkali metal reacts with $\ce{O2}$ in a 1:1 mole ratio.

Given, according to the Chem-Guide blog article Alkali metals that

The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of these oxides.

Which, according to the blog, results in the $\ce{Li+}$ ion (which is small, possessing a larger positive field around it) only being able to naturally combine with a small anion $\ce{O^{2-}}$ forming $\ce{Li2O}$, whereas the substantially larger $\ce{K+}$, $\ce{Rb+}$ and $\ce{Cs+}$ have weaker positive fields and can stabilise the larger superoxide anion $\ce{O2^{-}}$ forming superoxides.

  • $\begingroup$ This is hardly an explanation; I don't why how this is the accepted answer. Stating a general rule from some blog is not the same as an intuitive reason, which is what the OP was looking for. $\endgroup$ Commented Mar 13, 2021 at 5:46

The word "stability" itself isn't very clear, so you have to figure out exactly what it is stable with respect to. If you look up the data, the reaction

$$\ce{NaO2 (s) -> Na(s) + O2(g)}$$

is not enthalpically favourable. So, $\ce{NaO2}$ is actually stable with respect to decomposition to the constituent elements.

Instead, it is actually likely to be unstable with respect to decomposition to the oxide/peroxide. Since sodium is known to burn in oxygen to form the peroxide, this hints that we should be looking at the energetics of the reaction

$$\ce{2NaO2 (s) -> Na2O2 (s) + O2 (g)} \qquad \Delta H_1$$ $$\ce{2CsO2 (s) -> Cs2O2 (s) + O2 (g)} \qquad \Delta H_2$$

If cesium superoxide is "stable" in this respect, then we would expect $\Delta H_2 > 0$. And if sodium superoxide is "unstable" in this respect, then we would expect $\Delta H_1 < 0$.

The controlling factor here is likely to be the lattice energies (which are directly related to the size of the cation). Using the Kapustinskii equation we have

$$\Delta H_\mathrm{L}(\ce{M2O2}) \propto \frac{3\cdot 1\cdot 2}{r_\ce{M} + r_\ce{O2}} = \frac{6}{r}$$


$$\Delta H_\mathrm{L}(\ce{MO2}) \propto \frac{2 \cdot 1 \cdot 1}{r_\ce{M} + r_\ce{O2}} = \frac{2}{r}$$

where $r = r_\ce{M} + r_\ce{O2}$ (I have assumed the radii of the superoxide and peroxide ions to be the same). In the enthalpies of the decomposition reactions above $\Delta H_i$ ($i = 1,2$), the lattice energy contribution is

$$\begin{align} \Delta H_i &= 2\Delta H_\mathrm{L}(\ce{MO2}) - \Delta H_\mathrm{L}(\ce{M2O2}) + \cdots \\ &= 2\left(\frac{2}{r}\right) - \frac{6}{r} + \cdots \\ &= -\frac{2}{r} + \cdots \end{align}$$

Sodium is smaller than cesium, and therefore we expect the value of $r$ to be smaller. This clearly translates into a more negative value of $\Delta H$, as expected.

Why should $\Delta H_2$ be positive, then? It is to do with the other $\cdots$ terms that were omitted. These terms basically involve the production of $\ce{O2^2- + O2}$ from $\ce{2O2-}$, which is enthalpically unfavourable and hence have a positive contribution to the enthalpy of reaction.

In the case of sodium, the $-2/r$ term is large enough to outweigh this positive term. In the case of cesium, it isn't.

Looking to other aspects of inorganic chemistry one might recall that carbonates of heavier alkaline earths are more stable than carbonates of light alkaline earths (e.g. $\ce{MgCO3}$ vs $\ce{BaCO3}$). The reaction $\ce{MCO3 -> MO + CO2}$ can be treated very similarly and the same conclusion will be reached.

The thermodynamics of the alkali metal oxides/peroxides/superoxides is treated more thoroughly in the following question: Why do the alkali metals form different products upon combustion in air?


Well according to my book 'General and inorganic chemistry volume 2, by Dr. Ivan Filipovic & Dr. Stjepan Lipanovic ', the oxidation state of superoxides is -1/2. $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ react with oxygen and form type $\ce{MO2}$ compounds called superoxides. They are ionic crystalised compounds containing cation $\ce{M+}$ and anion $\ce{O^2-}$, and since the oxygen anion has 13 valent electrons, superoxides are colorful, paramagnetic, and unstable compounds, and they also exist only in solid state.


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