# Why is difluoride a stronger oxidizing agent than dichloride? [closed]

I am preparing for a chemistry olympiad and I am stuck with the question that why $$\ce{F_{2}}$$ is stronger oxidizing agent than $$\ce{Cl_{2}}$$.

## closed as off-topic by orthocresol♦, Jon Custer, Todd Minehardt, Wildcat, A.K.Jul 30 '16 at 23:19

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• Fluorine has a more positive redox potential (probably the most positive) at +2.87 V thus it spontaneously oxidises most everything else as they have a lower potential. Cl has a potential 1.35 V. Note also that Oxidation is loss of electrons, Reduction gain, thus F is reduced in oxidising Cl. – porphyrin Jul 30 '16 at 9:39

If you think about it, strong oxidizing agents must have a high "desire" or propensity to be reduced themselves (as the nature of redox reactions is such). We'll build up on the explanation from the ground up.

Comparing both fluorine and chlorine gas, both are halogens and both only need two electrons each to be reduced to fluoride and chloride ions respectively:

$$\ce{X2 + 2e- <=> 2 X-}$$

Since it appears that the two are similar in most respects, we should look at the clearest differences between Fluorine and Chlorine instead: the number of protons and the number of electrons.

Fluorine is a row 2 element with 9 protons an electronic configuration 1s2 2s2 2p5, whereas chlorine is a row 3 element with 17 protons and electronic configuration 1s2 2s2 2p6 3s2 3p5. Although chlorine has more protons to attract electrons (and therefore be reduced / act as an oxidizing agent), it also has two inner shells worth of electrons that create a shielding effect to reduce the attraction. Fluorine, however, only has one inner shell to do this.

From a simplistic physics point of view, electronic attraction is governed by the charge and the inverse square of the distance. You can approximate that the shielding effect will make the effective charge of the atomic nucleus the same in both halogens, so distance becomes the key deciding factor. Since chlorine is larger, and therefore the attracted electron further than it would be if it was instead bound to fluorine, the attraction (and therefore oxidizing power) is greater in fluorine than in chlorine.

However, looking at the electron affinity (EA) values you'd find that fluorine doesn't follow the group trend of decreasing EA and there's a very good discussion on that over at ChemLibre. Long story short, it's because fluorine is so small that you have to account for the electronic repulsion as well which reduces the bond strength of fluorine vs chlorine (36.6 vs 58 kcal/mol)

There is another measurement which seems to be consistent with this whole fluorine being more oxidizing than chlorine though - and that is reduction potential. The reason that there is a difference between the values for reduction potential and electron affinity even though they are both supposed to be measuring the same thing (that is the energy change related to accepting an electron), the key is the environment in which they are measured. Electron affinities are measured in the gas phase where it is purely the energy related to the electron transfer whereas reduction potentials are measured in aqueous solution and so factors such as hydration energy start to play a role.

As your textbook correctly points out, the high hydration energy is due to fluorine's higher charge density compared to that of chlorine.