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(Please note that all comparisons I have made below are with respect to halide ions only.)

According to what I have learnt, the fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density. This implies that it has the highest tendency of sharing/giving away its electron density.

Why isn't the same logic applicable when reducing strength is compared? Fluorine should have the highest tendency to lose electrons, and hence should be the best reducing agent.

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  • $\begingroup$ F2 molecules have quite weak and rather unusual bonds. F makes strong bonds this other elements, on the other hand. $\endgroup$ – Mithoron May 8 '18 at 18:16
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Fluorine is the most electronegative element. But due to its high effective nuclear charge it holds up its electron density very tightly, moreover its a non-metal so preferably accepts electron to form stable (F–) ion which has configuration of Ne. F2 ------> F- + e- is very favorable therefore it gets reduced very easily (reduced form F- is very stable than F2) and so it oxidises other elements and self gets reduced.

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  • $\begingroup$ Then why is it that F- is the most basic halide ion? If it is stable, it should be less basic. $\endgroup$ – PhysicsStudent May 9 '18 at 11:27
  • $\begingroup$ Let's take it in different sense, atom achieves stability by the octet rule so F- has a complete octet but there's a better chance of getting octet complete WITHOUT BEARING ANY CHARGE that's by forming a bond. $\endgroup$ – Aman Chande May 12 '18 at 17:25
  • $\begingroup$ So by accepting a proton it forms HF which is more stable than F- as it HF is neutral. Also that's not with F2 because of high repulsion ( small size and high electron density) the molecule is unstable. $\endgroup$ – Aman Chande May 12 '18 at 17:27
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You're looking at two different changes. The end state is different so it makes no sense to compare these two scenarios.

If fluoride is oxidized, it loses an single electron.

If fluoride acts as a base and donates a pair of electrons, it shares a pair of electrons.

In the first case, you now have fewer electrons around a very small charge. The effective nuclear charge on the valence electrons is quite high. The extra electron was attracted to this positive charge, and if you want the process to be favorable, you needed to stabilize that electron more than destabilization from removing it from the fluoride. But you're going to find something that's going to want an electron electron more than a fluorine atom...

In the second case, you have a high density of electrons fairly close to the nucleus. But you can stabilize the overall system by using some of those electrons to stabilize another positive charge. So, you're reducing some electron repulsion at the expense of attraction within the fluoride, but you're stabilizing another positive charge, which is overall favorable.

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  • $\begingroup$ Zhe, could you add some examples for both cases with Fluorine, please? $\endgroup$ – HolgerFiedler May 10 '18 at 16:20
  • $\begingroup$ @HolgerFiedler I don't understand the request... $\endgroup$ – Zhe May 10 '18 at 17:10
  • $\begingroup$ „If fluoride is oxidizes, it loses an single electron.” - give an example of such a compound “if fluoride acts as a base and donates a pair of electrons, it shares a pair of electrons.” - give an example of such a compound $\endgroup$ – HolgerFiedler May 10 '18 at 17:47
  • $\begingroup$ I can do that, but I fail to see how that helps at all with understanding the OP's question... $\endgroup$ – Zhe May 10 '18 at 19:48
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A flourine atom is very electronegative, fluorine is the most electronegative element it has the highest effective nuclear charge.

If a F atom accepts an electron it then has the same electronic configeration as neon. This gives fluorine a great driving force to act as an oxidant.

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  • $\begingroup$ Appropriate answer $\endgroup$ – theenigma017 May 9 '18 at 11:30
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... fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density.

A look at the Periodic table and the ionization energies is always a good idea to see what’s going on.

  1. The highest energies, required to remove an electron from the atom, are needed for the Noble gases. That means they are most aggressively reclaiming a removed electron back. First ionization energies of elements

  2. The less the number of electrons around the nucleus of Noble gases, the higher the aggressiveness to retrieval a lost electron.

  3. The elements in the column before the Noble gases ($\ce {Fl}, {CL}, {Br}, ...$) are doing the same in an attenuated way. And more than that, in bounds these elements try to complete their electron shell to the Noble gas configuration.

Take the Noble gas compounds. Fluorine is able to remove electrons from Xenon:

All images from WP

The tendency to fulfill the electron configuration is higher for Fluorine when the tendency for Xenon to preserve its electron configuration. The Octet rule works better for elements with high ionization energy when for elements with less ionization energies. (Perhaps instead of ionization energy in this case we have to remember the electron affinity, but this seems to me to be more a scholastic difference in definition.) According the above image Fluorine has a ionization energy of roughly 17 eV, Xenon has about 13 eV. That is why the octet rule in this case “works” for Fluorine and against Xenon.

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  • $\begingroup$ Noble gas, noble gas, noble gas. Question about halogens. $\endgroup$ – ParaH2 May 11 '18 at 16:25

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