Chemistry - What is the strongest oxidising agent?

Solution 1:

Ivan's answer is indeed thought-provoking. But let's have some fun.

IUPAC defines oxidation as:

The complete, net removal of one or more electrons from a molecular entity.

My humble query is thus - what better way is there to remove an electron than combining it with a literal anti-electron? Yes, my friends, we shall seek to transcend the problem entirely and swat the fly with a thermonuclear bomb. I submit as the most powerful entry, the positron.

Since 1932, we've known that ordinary matter has a mirror image, which we now call antimatter. The antimatter counterpart of the electron ($\ce{e-}$) is the positron ($\ce{e+}$). To the best of our knowledge, they behave exactly alike, except for their opposite electric charges. I stress that the positron has nothing to do with the proton ($\ce{p+}$), another class of particle entirely.

As you may know, when matter and antimatter meet, they release tremendous amounts of energy, thanks to $E=mc^2$. For an electron and positron with no initial energy other than their individual rest masses of $\pu{511 keV c^-2}$ each, the most common annihilation outcome is:

$$ \ce{e- +\ e+ -> 2\gamma}$$

However, this process is fully reversible in quantum electrodynamics; it is time-symmetric. The opposite reaction is pair production:

$$ \ce{2\gamma -> e- +\ e+ }$$

A reversible reaction? Then there is nothing stopping us from imagining the following chemical equilibrium:

\begin{align} \ce{e- +\ e+ &<=> 2\gamma} & \Delta_r G^\circ &= \pu{-1.022 MeV} =\pu{-98 607 810 kJ mol^-1} \end{align}

The distinction between enthalpy and Gibbs free energy in such subatomic reactions is completely negligible, as the entropic factor is laughably small in comparison, in any reasonable conditions. I am just going to brashly consider the above value as the standard Gibbs free energy change of reaction. This enormous $\Delta_r G^\circ$ corresponds to an equilibrium constant $K_\mathrm{eq} = 3 \times 10^{17276234}$, representing a somewhat product-favoured reaction. Plugging the Nernst equation, the standard electrode potential for the "reduction of a positron" is then $\pu{+2 116 413 V}$.

Ivan mentions in his answer using an alpha particle as an oxidiser. Let's take that further. According to NIST, a rough estimate for the electron affinity of a completely bare darmstadtium nucleus ($\ce{Ds^{110+}}$) is $\pu{-204.4 keV}$, so even a stripped superheavy atom can't match the oxidising power of a positron!

... that is, until you get to $\ce{Ust^{173+}}$ ...

Solution 2:

There is no definitive answer; if you think you have one, you are wrong.

See, this is much like asking "what is the northernmost big city". Depending on where you draw the line for being "big", the answer may be Moscow (latitude $55^\circ$N, population 13M), St. Petersburg ($60^\circ$N, 5M), Murmansk ($68^\circ$N, 300K), and quite a few others. There is no natural and universally accepted way to draw that line; it is inherently arbitrary.

How's that similar, might you ask, as you don't have any such line in your question? Yes you do, and here is it: you want the compounds which exist. Really, you don't want any compounds which don't exist, do you?

Now there's a catch: we have quite a few different subtle grades to "exist", and no natural way to draw the line that would make it clearcut black-and-white. In fact, in absence of further context it is about as ill-defined and vague as "being big" for a city. In my personal taste, $\ce{HArF}$ does not exist; if you point me to the papers that claim otherwise, I'll throw at you (figuratively) $\ce{He^2+}$, AKA an $\alpha$ particle, which was known for a good century longer, surely does exist, and will easily oxidize $\ce{HArF}$, I'm pretty confident on that.

There is another dimension to the problem. Oxidative ability of any compound is not measured by one number so that you could compare them. True, there is redox potential, but it is measured in standard conditions, and in different conditions things may turn out other way around. So there is not going to be an answer even if we would unanimously agree on the definition of "exists" (which we wouldn't).

So it goes.