[antelope]

The atomic size has a greater stabilising effect for charged atoms than electronegativity. For example, an iodide ion is more stable than a fluoride ion. "This is because the electron is distributed/orbits over a larger area."

I'm finding it hard to understand why this is the case. If the electron is further away from the protons, doesn't it make it more loose and the electron is more easily donated (reactive)? If it is more reactive, than it follows that it is less stable. So how is it that a less electronegative and bulkier atom, which doesn't hold electrons as tightly, is more stable than a smaller but tightly electron-bound atom?

[mjc123]

This doesn't seem to make sense. For a start, more stable relative to what?
Iodide is much more easily oxidised than fluoride: E°(F₂/F^-) = 2.84 V; E°(I₂/I^-) = 0.54 V
The electron affinity of the F atom is greater than that of I.
If you have a fluoride, there's really not a lot you can do with it, apart from electrolysis or forming a complex with a Lewis acid, e.g. F^- + BF₃  BF₄^-
I think your reasoning makes sense.

[antelope]

I see your point and agree with what you said. But I was just using those two as a specific example to make a more general point. To put it in context, I was refering to the subject on acids and bases.

You can estimate the strength of an acid based on the stability of its conjugate base. One factor used to determine stability is the electronegativity and atomic size of the atom that has lost a proton.

What got me confused was that HI is more acidic than HF i.e. I- is more stable than F-. This was based on the reason that iodine is a larger molecule than flourine, even though flourine is more electronegative.

[Enthalpy]

Elements of thought:

The strength of a neutral atom to gain an electron isn't the same as the strength of the positive ion to gain one and get neutral (the latter being, possibly with a sign change, the difficulty to abstract an electron from an atom).

Electronegativity is defined for one atom alone in vacuum. What happens in a solvent is different.

Electronegativity is a huge value like 2eV. Equilibria at room temperature, or 26meV, depend on much smaller energies, between sets of compounds, ions, radicals whose transformation has a nearly energy-neutral balance.

I put energy balance because individual transformation steps can overcome a higher barrier, only improbably hence less often. So while molecules may collide every ns (gas) or ps (liquid), a reaction can take hours.

Electronegativity doesn't determine simply the strength of a bond. For instance, H-H has 436kJ/mol and H-Cl 432kJ/mol, but the reaction proceeds because Cl-Cl has only 243kJ/mol.

[mjc123]

This illustrates why you always need to specify stable relative to what?. Evidently you meant relative to HX, but nothing you said indicated that.
To say that HI is more acidic than HF because I^- is more stable (to protonation) than F^- is merely tautologous; it is not an explanation, but just saying the same thing in a different way. This concept is more useful when you are comparing similar compounds, e.g. alcohols and phenols. In both cases you are breaking an O-H bond, but the phenoxide ion is stabilised by conjugation in a way not possible to the alkoxide (nor - and this is important - to the neutral phenol, at least to the same extent). Hence phenols are more acidic than alcohols.
It is less useful when comparing such different things as HI and HF. What "stabilises" I^-?
It may be more helpful to break it down into steps. You break an H-X bond and hydrate the resulting ions. The key things are the H-X bond strength and the X^- hydration energy. Now the H-F bond is very strong, about 300 kJ/mol stronger than H-I. The heat of hydration of F^- is about 200 kJ/mol more negative than I^-. So the ionisation of HI is energetically more favourable than HF. (That is an oversimplification; you are considering heterolytic cleavage rather than homolytic [which is what reported "bond energies" refer to], and you need to think about entropy as well. But I hope you get the drift.)

[Corribus]

To say that HI is more acidic than HF because I^- is more stable (to protonation) than F^- is merely tautologous;

If you write it like that, I would agree. But I don't think there's anything wrong about pointing out that what we define as acidity is related to the relative concentration of charge in fluoride vis-a-vis iodide.