As we go down the periodic table, Group 1 alkali metals hold their single outer valence electron more loosely, and so reactivity increases. Below potassium (Na, Li), we can store the metal in oil with no difficulty. Above potassium (Rb, Cs), we need to exclude all oxygen to prevent reaction in air. Potassium (K) is the unique element where we can technically store it in oil, but it is highly likely to react with any available oxygen to form explosive peroxides. I'd like an explanation of why this is; i.e. why is the reactivity favorable for the formation of peroxides. Thanks!


If you look at the standard Gibbs free energies for the alkali metal peroxides, actually, lithium should form it most easily. The standard Gibbs free energies of formation are -564.8, -430.1, -418.4, -351.5, and -359.8 kJ/mol for $\rm Li_2O_2, Na_2O_2, K_2O_2, Rb_2O_2,$ and $\rm Cs_2O_2$, respectively.

If we're looking at reaction of an alkali metal with oxygen to form explosive compounds, perhaps we should look at superoxides. The superoxides are powerful oxidizing agents, and they can explode when mixed with water, acids, organics, or powdered graphite.

At room temperature, oxygen readily reacts with K, Rb, and Cs to form superoxides ($\rm MO_2$), while Li and Na don't. There is a direct correlation between superoxide stability and electropositivity of the metal concerned (source: Cotton & Wilkinson's Advanced Inorganic Chemistry). $\rm NaO_2$ can be obtained by high pressure and temperature reaction of $\rm Na_2O_2$ with $\rm O_2$, and $\rm LiO_2$ won't form at room temperature.

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    $\begingroup$ Nice answer -- here's a vid of exactly how high-temperature you need to get to make $\ce{Na2O2}$: youtube.com/watch?v=ZqEWUw6sgpA $\endgroup$ – Curt F. Jan 28 '16 at 20:03

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