What is the strongest oxidising agent?

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?

• "not much of it is available" is like understatement of century. HArF wasn't isolated at all, only "detected". Highest std. potential known goes to KrF2 with impressive 3.5 value. FOOF doesn't have one because it decomposes in cryogenic temp. Oct 4 '18 at 14:28
• Colleagues, please refrain from close votes. This is a very natural question, which is likely to come up often in the future. Also, it does have a very certain answer, and there is nothing opinion-based about it. That the answer kinda defeats the purpose of the question is another story. After all, an answer "there is no answer" can be as final and as important as any other possible answer. Oct 4 '18 at 20:27
• @IvanNeretin Well, you two didn't do it much favor by nitpicking and stretching to antimatter and beyond. You should better act as if OP posed it properly, by putting a criterion like std. potential and agent as compound which is actually used to oxidise other compounds. Oct 4 '18 at 23:42
• @user17915 Yes, at very low temperatures and with a healthy dose of ultraviolet light (which is very good at forcing energetically unfavourable bonds) or in plasma. The bond is very weak, needless to say. Oct 5 '18 at 6:56
• I edited the question to be more precise. I suggest OP to check it out. Oct 6 '18 at 20:30

3 Answers

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

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

• My black hole will oxidize your vacuum and spit out the positron. Oct 4 '18 at 13:46
• @IvanNeretin I do enjoy an escalation which reaches the most fundamental limits of physics! Oct 4 '18 at 13:51
• Of course the corollary to this is that a sufficiently at sufficiently high concentrations gamma photons can reduce Fe3+ to Fe2+ (and a positron). The experimental conditions here might be a little harsh, though. Oct 5 '18 at 7:01
• @NicolauSakerNeto is it possible for you to edit in how you did your calculations? I have been trying to replicate your results but I cannot. Additionally, what do you mean by entropy being negligible? And how did you calculate the equilibrium constant? Jul 23 at 3:22
• Well, there is not much to calculate because I made some coarse approximations to get the point across. The electron and positron both have a rest mass of $\mathrm{511 keV\ c^2}$, and due to conservation of energy all this rest mass is converted into the energy of the combined photons ($\mathrm{E=mc^2}$ on the left side). I then call this the "enthalpy change of the reaction" (exothermic by $\mathrm{2 \times 511 keV = 1.022\ MeV })$. Jul 23 at 3:29

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.

• HArF exist , its mentioned on wikipedia with all information with melting point and bond energies . What about F2 can protonate From F+ . I don't think so Oct 4 '18 at 8:47
• Yeah, I've seen that page, and that's what I call "don't exist"; also, the page doesn't claim it has a melting point, even if a cursory glance might have convinced you otherwise. The rest is puzzling to me; I never said a thing about anything protonating anything. Oct 4 '18 at 8:56
• I'm with @IvanNeretin - HArF does not exist in any usable or useful form. When you show me a reaction with it acting as an oxidant I will acknowledge its place in the ranks of powerful oxidising agents. Oct 4 '18 at 13:29
• @Waylander Besides, even if we accept it as existing, why would it be any remarkable as an oxidizer? It's just Ar in oxidation state 0, after all. Oct 4 '18 at 13:48
• There is a subtle difference between exists and has been made in bulk and studied. I'm with those who think candidates should have been made in quantity and studied not just shown to exist in very cold matrices of inert gases. Oct 5 '18 at 10:39

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!

• The no-hair theorem for black holes suggests that a black hole "wouldn't know" it was made of positrons, and would behave as any black hole with the same mass and charge, so I don't think you get any annihilation benefits. Also, a black hole can't be generated with just positrons, as the electromagnetic repulsion would vastly outstrip the gravitational attraction. There would have to be a lot of electrically neutral "filler" mass to keep things together. Lastly, all black holes are thought to be overall very close to electrically neutral. May 22 '19 at 5:15
• All in all, it's an interesting idea, but you may not get qualitatively greater oxidation power even though it is a much more exotic system (even compared to positrons and regular black holes!). May 22 '19 at 5:17
• Yeah! But it isn't completely proved right? I did doubt that the repulsions might not let it happen. But what I am saying is if it did exist, it would be powerfull. By the way wouldn't the no-hair theorem lead to asymmetry in matter and anti-matter? But I must confess I am not knowledgeable enough to make these statements. Probably only after a few years (just about to start my UG course) I can make legible statements. May 22 '19 at 12:42
• The no-hair theorem is at least a good approximation of reality which meshes well with current astrophysical theory, so without a very good reason otherwise (i.e. degrees in astrophysics!) you and I do well to regard it as robust. I'm not sure this analysis has much to say with regards to matter/anti-matter asymmetry. I see you're an eager young student! The fun thing in science is putting your imagination in a straitjacket; learn the rules of the Universe as well as you can, then find elegant ways to exploit them to do weird and beautiful things! May 22 '19 at 13:05