What is the strongest oxidising agent?

Question

I searched for the strongest oxidising agent and I found different results: $\ce{ClF3}$, $\ce{HArF}$, $\ce{F2}$ were among them.

Many said $\ce{ClF3}$ is the most powerful as it oxidises everything, even asbestos, sand, concrete, and can set easily fire to anything which can't be stopped; it can only be stored in Teflon.

And $\ce{HArF}$ could be a very powerful oxidant due to high instability as a compound of argon with fluorine, but was it even used as such?

What compound is actually used as oxidising agent and was proven to be stronger then others, by, for example, standard reduction potential?

Answer 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 $\mathrm{\frac{98\ 607\ 810\ kJ\ mol^{-1}}{96\ 485.33212\ C\ mol^{-1}} = +1\ 021\ 998\ V}$.

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 $\ce{Ust^{173+}}$ ...

Answer 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.

Answer 3

I think the answer would be a black hole made up of antimatter (specifically positrons) . Here you have the annihilation energy along with immense gravitational energy and immense electromagnetic energy. The gravitational tidal forces are strong coupled with the fact that the electromagnetic force repels the nucleus and attracts the electrons it is even more. I can't think of anything better!

Answer 4

The most powerful oxidising particle would be positrion. In term of chemistry the strongest oxidiser would not be HArF and definitely not ClF3. ClF3 is more powerful compared to ClF3, but not as powerful as oxygen fluoride. O2F2 is far more reactive than chlorine fluoride or fluorine, in fact it will decompose into fluorine and oxygen at -163 celsius. Oxygen fluoride have a trend of being more reactive the more oxygen and flurine are in the compound. There are compound that are even more reactive than argon fluorohydride, as mentioned by other users dihelium will oxidse HArF.

Trihelium may be even more reactive compared to dihelium as the Van der Waals bond is even longer. There is an acid that is discovered in 1925 called Helium Hydride which is more adicidc than any man-made superacid, it has a lower porton affinity when compared to helium. Neon compound may be the most more oxidising compound but there is no stable neon compound (this is due to how neon is more inert than helium).