An oxidizing agent pulls the electron cloud of the substance being oxidized towards itself, for example:

$$\ce{F2 + 2 X- -> 2F- + X2}\qquad (\ce{X} = \ce{Cl}, \ce{Br}, \ce{I})$$

We also know that fluorine has very high charge/mass ratio as it is very small in size and its electron gain enthalpy is know to be very high, i.e gain of electron is favourable thermodynamically.

Also, many sources state that:

  • Fluorine has low bond dissociation enthalpy and high hydration enthalpy due to which fluorine is a good oxidizing agent.
  • Fluorine is achieving octet, so electron gain is favorable.
  • I found another source stating that bond dissociation enthalpy is the major driving factor.

My questions(along with reasons why fluorine shouldn't be oxidising agent)

  1. What has hydration enthalpy and bond dissociation enthalpy got to do with oxidizing power? (As we know that lithium is the best reducing agent, and it also has high hydration enthalpy but we can’t say it is an oxidising agent.)
  2. What am I missing here? What is the reason for such high standard electrode potential for fluorine?
  3. Why wouldn't the electron cloud of fluorine atom repel the electron cloud of substance being oxidized? It seems paradoxical that the fluorine pulls electron cloud towards itself so easily.

Even in the post Why does chlorine have a higher electron affinity than fluorine? it is mentioned that electron cloud is repelled.

Is there any better explanation other than "octet is being achieved", such that it resolves the repulsion issue, to explain why fluorine is a good oxidising agent?

  • $\begingroup$ look up Born-Haber cycle, hydration energy effects will make sense. $\endgroup$ Commented Mar 10, 2019 at 17:25
  • $\begingroup$ Maybe write it this way to gain some insight: $$\ce{F-F(g) + 2Cl-(aq) -> 2 F-(aq) + Cl-Cl(g)}$$ $\endgroup$
    – Karsten
    Commented Apr 15, 2020 at 22:18

2 Answers 2


If we consider

$\ce{F2(g) -> 2 F(g)}$ with $\Delta H_\mathrm{F, bond}$

$\ce{2 F(g) + 2 e- -> 2 F-(g)}$ with $\Delta H_\mathrm{F, aff}$

$\ce{2 F-(g) ->[H2O] 2 F-(aq)}$ with $\Delta H_\mathrm{F, hydr}$

and respectively

$\ce{Cl2(g) -> 2 Cl(g)}$ with $\Delta H_\mathrm{Cl, bond}$

$\ce{2 Cl(g) + 2 e- -> 2 Cl-(g)}$ with $\Delta H_\mathrm{Cl, aff}$

$\ce{2 Cl-(g) ->[H2O] Cl-(aq)}$ with $\Delta H_\mathrm{Cl, hydr}$

Then overall reaction enthalpy for $\ce{ F2(g) + 2 Cl-(aq) -> Cl2(g) + 2 F-(aq)}$ is

$$\Delta H_\mathrm{r} = \left({\Delta H_\mathrm{F, bond} - \Delta H_\mathrm{Cl, bond}}\right) + \left({\Delta H_\mathrm{F, aff} - \Delta H_\mathrm{Cl, aff}}\right) + \left({\Delta H_\mathrm{F, hydr} - \Delta H_\mathrm{Cl, hydr}}\right)$$

The differences between fluorine and chlorine enthalpies of atomization(bond breaking ), electron affinity and ion hydration lead to the overall reaction enthalpy, that is pushing reaction toward chloride oxidation.

The truth is, that the equilibrium is driven by Gibbs energy, not enthalpy,

$$\Delta G^{\circ}_\mathrm{r}=\Delta H^{\circ}_\mathrm{r} - T \cdot \Delta S^{\circ}_\mathrm{r}=-RT\ln{K}$$

but for the rough reasoning you get the picture. Futhermore, the entropy change will not be major factor, due analogical structures.


Electron Affinity

Using Slater's rules we can calculate the effective Charge that an electron added to $\ce{F*}$ would encounter. $$Z_{eff} = 9 - (0.35 \cdot 7 + 0.85 \cdot 2) = 2.4$$ This is higher than the effective charge of any other element before fluorine, which makes it more favorable to have an electron added to fluorine, rather than leaving it where it was.
For any element after fluorine, the lowest-energy orbital available is still a lot higher in energy than the 2p-orbital in flourine, because of the higher quantum number and the greater distance to the core. Which again makes it more favorable to add an electron to fluorine rather than leaving it in another atom. (Disregarding noble gases here)


But wait, there is the effect of flourines small size. And it is true that electron-clouds do repel each other through Coulomb-forces and, when very close, through Pauli-repulsion. And this effect does lead to the situation of fluorine having a slightly less (but similar and still impressive) negative electron affinity than chlorine.
So we can conclude that if we had a single atom each of fluorine and chlorine and a spare electron, it would rather be added to the chlorine atom.

Dissociation energy

But if we talk about oxidising agents, we do not talk about single atoms. Fluorine is naturally a molecule and as you mentioned, its dissociation energy is extraordinarily low for a single covalent bond. This anomaly is another result of its small size and the resulting repulsion of the electron clouds of the two fluorine atoms. (Actually, it has to do with the overlap of fluorines orbitals, which is more complicated but still a result of size)

For an oxidation to take place, the molecular fluorine has to undergo homolytic dissociation. $$\ce{F_2 -> 2F*}$$ This means that the dissociation energy needs to be provided for the reaction to take place. With a lower dissociation energy, the kinetic barrier is lower as well as the process being thermodynamically more favorable.


If there were no anomalistic effects resulting from its small size, fluorine would be the best elemental oxidising agent, resulting from its position in the periodic table. However there are effects. But luckily for fluorine the effect of repulsion counters the effect of the lower dissociation energy, which leads to it still being the best elemental oxidising agent.

The hydration enthalpy is relevant in liquids only, which is not necessarily the case for oxidations, so it should not be regarded when considering fluorines oxidising potential.

  • 3
    $\begingroup$ Welcome to Chemistry.se! This does unfortunately not answer the main question of the post. It also reads more like speculation and conjecture. It would be nice, if you back up your claims with references. $\endgroup$ Commented Jan 29, 2020 at 13:51

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