From Wikipedia's article on sodium:

When burned in dry air, it forms primarily sodium peroxide with some sodium oxide.

We know that sodium has a strong reducing capacity, so why does it produce a compound in which the oxygen atom is not reduced to the fullest possible extent?

After posting the question, I read up some more and found this, on Chemguide's page describing the oxidation properties of Group 1 elements:

So why do any of the metals form the more complicated oxides? It is a matter of energetics.

In the presence of sufficient oxygen, they produce the compound whose formation gives out most energy. That gives the most stable compound.

The amount of heat evolved per mole of rubidium in forming its various oxides is:

$$\begin{array}{|c|c|} \hline & \text{enthalpy change (kJ / mol of Rb)} \\ \hline \ce{Rb2O} & -169.5 \\ \ce{Rb2O2} & -236 \\ \ce{RbO2} & -278.7 \\ \hline \end{array}$$

This is hardly an explanation, because it just says in a circular way that it's energetically profitable to form such compounds. Why is this energetically profitable?

Can part of the explanation be "because in Group 1, starting from $\ce{Na}$, ion charge densities are low enough to allow peroxides and superoxides to exist"?

I wonder if the following behaviour is related to this: lithium nitrate decomposes to lithium oxide (again the other participant has the oxidation -2), while the rest of the group's elements have their nitrates decompose to a nitrite (again the other participant's oxidation is -1):

From Chemguide:

$$\ce{4 LiNO3 (s) -> 2 Li2O (s) + 4 NO2 (g) + O2 (g)}$$

$$\ce{2 MNO3 (s) -> 2 MNO2 (s) + O2 (g)}$$

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    $\begingroup$ What oxygen compound is formed depends on temperature, pressure, and the amount of oxygen present $\endgroup$
    – aventurin
    Commented May 7, 2016 at 11:47
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    $\begingroup$ @aventurin - I fully believe that, but I would like to know why it forms. Why is it better to create a compound in which the oxygen atom has the oxidation number -1 and not its usual -2. $\endgroup$ Commented May 7, 2016 at 11:50
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    $\begingroup$ Oxygen is a strongly oxidising compound that is present in a large excess. Once there is no metallic sodium left to react with, it starts looking for some other prey. I don't see what is wrong with RbO2 being a stable compound. Some peroxides are more stable than others. Forming bonds is nearly always profitable, unless entropy is against it or you have to spend too much energy to break or weaken other bonds for it. $\endgroup$
    – Karl
    Commented May 7, 2016 at 12:49
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    $\begingroup$ @Karl - but, for instance, Mg does not form peroxides upon exposure to air in ordinary conditions. $\endgroup$ Commented May 7, 2016 at 12:52
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    $\begingroup$ Related: Why does potassium form peroxides but sodium does not?; When will peroxide form? $\endgroup$
    – Mithoron
    Commented May 7, 2016 at 13:41

2 Answers 2


Since I will deal with all of the alkali metals in this answer, I think the question should also be broadened. There is no point in covering one single metal (sodium) without touching the others since it is the trend going down the group that we are interested in.

All thermodynamic data is taken from Prof. M. Hayward's lecture notes at Oxford.

So, firstly, some data. There is an increase in ionic radius going down the group, which should not be surprising:

$$\begin{array}{cc} \hline \ce{M} & \text{Ionic radius of }\ce{M+}\text{ / pm} \\ \hline \ce{Li} & 76 \\ \ce{Na} & 102 \\ \ce{K} & 138 \\ \ce{Rb} & 152 \\ \ce{Cs} & 167 \\ \hline \end{array}$$

This leads to an decrease in the magnitude of the lattice enthalpies of the Group I superoxides, peroxides, and oxides, going down the group:

$$\begin{array}{cccc} \hline \ce{M} & \Delta H_\mathrm{L}(\ce{MO2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O})\mathrm{~/~kJ~mol^{-1}} \\ \hline \ce{Li} & -960 & -2748 & -3295 \\ \ce{Na} & -860 & -2475 & -2909 \\ \ce{K} & -752 & -2175 & -2503 \\ \ce{Rb} & -717 & -2077 & -2375 \\ \ce{Cs} & -683 & -1981 & -2250 \\ \hline \end{array}$$

As for why there is an increase in magnitude going across the table, we have to look at the factors controlling the magnitude of the lattice enthalpy:

$$\Delta H_\mathrm{L} \propto \frac{\nu z_+ z_-}{r_+ + r_-} \tag{1}$$

where $\nu$ is the number of ions in one formula unit, $z_+$ and $z_-$ are the charge numbers on the cation and anion, and $r_+$ and $r_-$ are the ionic radii:

$$\begin{array}{ccccc} \hline \text{Formula unit} & \nu & z_+ & z_- & \text{Numerator in eq. (1)} & r_-\text{ / pm} \\ \hline \ce{MO2} & 2 & 1 & 1 & 2 & 149 \\ \ce{M2O2} & 3 & 1 & 2 & 6 & 159 \\ \ce{M2O} & 3 & 1 & 2 & 6 & 120 \\ \hline \end{array}$$

The lattice energies of the peroxides and oxides are roughly 3 times those of the corresponding superoxides, because of the larger numerator. The lattice energies of the oxides have a slightly larger magnitude than those of the corresponding peroxides, because of the smaller anionic radius.

Just looking at the lattice enthalpies, we might think that all metal cations would form the oxides. However, this approach is flawed because it does not take into consideration the energy cost of forming the three different anions from molecular dioxygen. Recall that the lattice enthalpy is defined as $\Delta H$ for the reaction

$$\ce{M+ (g) + X- (g) -> MX (s)}$$

However when we burn a metal in oxygen we are starting from $\ce{M}$ and $\ce{O2}$. So, we have to figure out the energy needed to get from $\ce{M}$ to $\ce{M+}$, and from $\ce{O2}$ to the relevant anion ($\ce{O2-}$, $\ce{O2^2-}$, or $\ce{O^2-}$).

The analysis here is complicated by the fact that the charges on the anions are not the same. Therefore, the reaction

$$\ce{M(s) + O2 (g) -> MO2(s)}$$

cannot be directly compared with

$$\ce{2M(s) + O2(g) -> M2O2(s)}$$

In order to "standardise" the equations, we will consider the reactions per mole of metal. One can loosely think of the combustion reaction as "releasing" some kind of energy within the metal; the most favourable reaction will be that which "releases" the most energy per mole of metal. Therefore, the three reactions we are considering are:

$$\begin{align} \ce{M (s) + O2 (g) &-> MO2 (s)} & \Delta H_1 \\ \ce{M (s) + 1/2 O2 (g) &-> 1/2 M2O2 (s)} & \Delta H_2 \\ \ce{M (s) + 1/4 O2 (g) &-> 1/2 M2O (s)} & \Delta H_3 \\ \end{align}$$

Now, we construct Hess cycles for all three reactions.

Superoxides $$\require{AMScd}\begin{CD} \color{blue}{\ce{M(s) + O2(g)}} @>{\Large\color{blue}{\Delta H_1}}>> \color{blue}{\ce{MO2(s)}} \\ @V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\Delta H_\mathrm{L}(\ce{MO2})}A \\ \ce{M+(g) + e- + O2(g)} @>>{\Large\Delta H_\mathrm{f}(\ce{O2^-})}> \ce{M+(g) + O2^-(g)} \end{CD}$$

Peroxides $$\require{AMScd}\begin{CD} \color{blue}{\ce{M(s) + 1/2O2(g)}} @>{\Large\color{blue}{\Delta H_2}}>> \color{blue}{\ce{1/2M2O2(s)}} \\ @V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O2})}A \\ \ce{M+(g) + e- + 1/2O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O2^2-})}> \ce{M+(g) + 1/2O2^2-(g)} \end{CD}$$

Oxides $$\require{AMScd}\begin{CD} \color{blue}{\ce{M(s) + 1/4O2(g)}} @>{\Large\color{blue}{\Delta H_3}}>> \color{blue}{\ce{1/2M2O(s)}} \\ @V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O})}A \\ \ce{M+(g) + e- + 1/4O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O^2-})}> \ce{M+(g) + 1/2O^2-(g)} \end{CD}$$

Now we need more data. $\Delta H_\mathrm{f}(\ce{M+})$ is simply the sum of the atomisation energy and first ionisation energy:

$$\begin{array}{cc} \hline \ce{M} & \Delta H_\mathrm{f}(\ce{M+})\text{ / }\mathrm{kJ~mol^{-1}} \\ \hline \ce{Li} & 679 \\ \ce{Na} & 603 \\ \ce{K} & 508 \\ \ce{Rb} & 484 \\ \ce{Cs} & 472 \\ \hline \end{array}$$

and the enthalpies of formation of the anions are

$$\begin{array}{cc} \hline \ce{X} & \Delta H_\mathrm{f}(\ce{X})\text{ / }\mathrm{kJ~mol^{-1}} \\ \hline \ce{O2-} & -105 \\ \ce{O2^2-} & +520 \\ \ce{O^2-} & +1020 \\ \hline \end{array}$$

Why is the enthalpy of formation of $\ce{O^2-}$ so large? The answer is that, to get from $\ce{1/2 O2}$ to $\ce{O^2-}$, you need to first break the $\ce{O=O}$ bond, then add two electrons to the oxygen. Furthermore, the second electron affinity is often an unfavourable process. For the other two anions, you don't need to break the $\ce{O=O}$ bond.

So now, we can see a trend at work already. Going from the superoxides to peroxides to oxides, the more negative lattice enthalpy favours the formation of the oxide. However, the increasing heat of formation of the anion favours the formation of the superoxide.

When do each of these two factors win out? Well, when lattice enthalpies are comparatively large, we would expect the lattice enthalpy factor to outweigh the heat of formation of the anion. Lattice enthalpies are large precisely when the cation is small, and therefore lithium forms the oxide when heated in oxygen. However, with caesium, lattice enthalpies are smaller, less significant, and the heat of formation of the anion wins out; caesium therefore forms the superoxide.

The trend is of course not black and white. Going down the group from lithium to caesium, we might guess that perhaps there are one or two elements that form the intermediate peroxide. That element is sodium. You could say that the larger lattice energies of sodium salts sufficiently compensate for the formation of the peroxide ion, but aren't enough to compensate for the formation of the oxide ion.

I leave you with the last bunch of numbers, which tabulate the values of $\Delta H_1$ through $\Delta H_3$ for all the elements (all values in $\mathrm{kJ~mol^{-1}}$). You can actually calculate these yourself by plugging the data above into the Hess cycles. It seems that the data is a little different from that given in the Chemguide screenshot you have, but the conclusion is the same, so I'll ignore that:

$$\begin{array}{cccc} \hline \ce{M} & \Delta H_1\text{ (superoxide)} & \Delta H_2\text{ (peroxide)} & \Delta H_3\text{ (oxide)} \\ \hline \ce{Li} & -386 & -435 & \mathbf{-459} \\ \ce{Na} & -362 & \mathbf{-375} & -342 \\ \ce{K} & \mathbf{-349} & -320 & -234 \\ \ce{Rb} & \mathbf{-338} & -295 & -194 \\ \ce{Cs} & \mathbf{-316} & -259 & -143 \\ \hline \end{array}$$

As described earlier, the salt with the most negative enthalpy of formation will be preferentially formed. These are bolded in the table.


I believe the answer given by @orthocresol is far too complicated. There is a very simple, yet understandable answer to your question. Apart from energy factors, ther is something called coordination in formation of a compound. COORDINATION is simply the number of atoms the central atom in a compound can accomodate. As we go down the group, size increases for alkali metals (rather for any group, except maybe for transition metals, but that is another story). Due to this increase in size, larger ionscan form bonds with larger negative ions. So Li being small can be stable with small oxide ion only.

Na ion, being larger, can remain stable with larger peroxide and superoxide ions, though peroxide is more stable. Finally Potassium is most stable with the superoxide ion.

note : the size of the negative ions decreases as follows- Superoxide>peroxide>oxide

I havent covered the elements further down in group 1 as they are very reactive and react with all the ions of oxygen to form stable species.


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