[orgopete]

I want to resurrect this topic which has been discussed before:
http://www.chemicalforums.com/index.php?topic=19788.0
http://www.chemicalforums.com/index.php?topic=16293.0

I am aware of some of the past thoughts but I want to learn more. First of all, I do not wish to invoke electronegativity arguments as I believe this is a scale greatly biased by the redox reactions. Thus the electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. This is simply different than the electron withdrawing power. Thus HF is a weak acid and HI is strong. That simply restates the question, but it doesn't explain why.

It is easy to understand why HI is a strong acid. Its nuclear charge is large (+53) and thus it can pull its electrons inward. Compared to the other halogens, the repulsive field of an iodine nucleus grows faster than bond length. That is, if we set the repulsive force of HI to be equal to that if HF, then the bond length of HI would be 223 pm versus 161 pm actual, x^2=(53*92^2)/9. Granted, this is simplistic reasoning, but I think it is consistent with a general trend in bond lengths, e.g., CH>NH>OH>HF due to nuclear charge. For this reason, we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. This should be a general principle of physics that matches the acidity trends of simple acids, HI>HBr>HCl>>HF; HF>H2O>NH3>CH4.

You may see arguments that HF is a weak acid because of low entropy. While I think this is true, I don't agree with the interpretation of it. The entropy is lower because HF doesn't ionize, not that it doesn't ionize because its entropy is lower. I don't think of entropy as a normal force. Entropy can be small if a bonding force is present, otherwise molecules will behave as ideal substances, that is without interaction.

The issue I am trying to resolve is gas phase v solution phase reactions. I think they are simply different. McMurray states, “… bond dissociation energies refer to molecules in the gas phase and aren’t directly relevant to chemistry in solution.” The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase. I have used this argument for a solution phase Born-Haber cycle in a formation of NaCl(aq), -386 kJ/mol v -411 kJ/mol (s). In a solution phase reaction, no bond is formed between Na(+) and Cl(-). The energy is largely the energy of the redox reaction of Na->Na(+) and Cl->Cl(-). So, if you perform a reaction, it will be guided by Hess's Law.

I saw a post for the reaction of LiBr + MeI  LiI + MeBr, the net enthalpy of bonds formed versus broken would be +18 kJ/mol. However, I began to think the bonds of LiBr and LiI are actually not being broken and thus should not be part of the calculation. Now, the reaction enthalpy is a difference in the bond energy of methyl bromide and methyl iodide, -72 kJ/mol. However, as I understand this calculation, this is a measure of the homolytic bond energy. The homolytic bond energy is being used to calculate the thermodynamics of a heterolytic reaction.
 
Further, the commenter posted:

In another paper, they gave the following equilibrium constants for a (reversed Finkelstein) reaction LiBr + RI = LiI + RBr: MeI (K = 1.7), n-alkyl (K = 7), iPr (K = 30), t-Bu (K ca 100). They furthermore stated that for the reaction LiCl + RBr = LiBr + RCl, the equilibrium should even be further on the RCl side (K larger). (See: DOI: 10.1039/JR9550003173)

Now, I realized I could calculate a Gibbs free energy from the equilibrium. The methyl bromide reaction has a rather modest enthalpy of -1.32 kJ/mol and isopropyl -8.4 kJ/mol by my calculations. As an organic chemist, these values seem about right. I am finding these calculations daunting. The enthalpy values should include all of the factors in a reaction, not just the bond energy. I don't know how to adjust for an entropy change, though I would have thought it to be small and toward equilibrium in this substitution reaction.

Now I've seen the heat of neutralization of HF is -68 kJ/mol while HCl, HBr, and HI are -57 kJ/mol. The simplicity of HF makes this an ideal molecule to understand these energy differences. I can understand HCl, HBr, and HI as they simply form the same H3O(+). The enthalpy of neutralization is the neutralization of hydronium ion. This is analogous to my ignoring the bonds of LiI and LiBr as they are broken upon dissolving in acetone.

One analysis that seems to be in line as to how this problem may be solved has been contributed by Jim Clark. This example illustrates the problem I have with understanding the data. It illustrates a Born-Haber cycle with enthalpies of vaporization, bond energy, ionization, electron affinity, and hydration. I find it difficult to verify any one of these data points and if I can, they may differ significantly from this source.

I don't know if I am trying to understand that which cannot be readily understood. I am dubious of gas phase acidity as it can often differ from solution phase acidity. For me, a clear warning to using values from one phase to predict reactions in another is the endothermic gas phase ionization of sodium. If sodium preferred to hold its electron, I would expect to find sodium in its ground state, yet metallic sodium reacts exothermally in solution.

I hope someone can clarify this problem.

[AWK]

http://www.ch.imperial.ac.uk/rzepa/blog/?p=16320

[orgopete]

http://www.ch.imperial.ac.uk/rzepa/blog/?p=16320

I had seen that, but I didn't find it helpful. I guess I should have specified I was interested in being able to calculate the energy differences. Otherwise, I find the acidity of HF to be as I would have expected, greater than H2O and less than HCl.

[AWK]

Liquid HF (anhydrous) is a superacid

[Irlanur]

The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase.

Wut?

I find the post not very well structured and honestly have a hard time what your actual question is...

[Enthalpy]

A part of the answer is how water eases the formation of the halide ion, because without the solvent, there would be extremely little ionization.

And then, one simple reason is the diameter of the halide ion. The bigger iodide concentrates less the charge of the excess electron than the smaller fluoride does, and the less concentrated charge needs less electric energy.

Modelling water as a continuous insulator with permittivity already tells it. q (puts signs as you like) concentrated at the surface of a radius r sphere means an energy varying like 1/(r×ε) where both the big r and water's big ε help the ionization.

Or if you prefer to imagine the discrete water molecules, there are more of them near the bigger ion, so they mitigate more easily the effect of the excess electron.

[Enthalpy]

[...] The electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. [...]

If you check the bond strength of the halogens with other atoms like C and H, it's not very different from C-C or C-H bond strength. C-H and C-Cl are almost equally strong bonds despite Cl being more electronegative, and the electronegative Br makes a weaker bond than C-H. What does make the light halogens reactive is rather that the X-X bond is weak, and this acts both on the reaction enthalpy and on the activation energy.

[...] we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. [...]

I don't see a difference (except in reaction direction and energy sign) between the addition of an electron to a halogen atom (the electronegativity) and the withdrawal of an electron from a halide ion.

[orgopete]

The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase.

Wut?

I find the post not very well structured and honestly have a hard time what your actual question is...

I agree. I should have kept it simple at why is the heat of neutralization of HF is -68 kJ/mol while HCl, HBr, and HI are -57 kJ/mol, but ...

[orgopete]

A part of the answer is how water eases the formation of the halide ion, because without the solvent, there would be extremely little ionization.[/unquote]
This is little ionization in water

And then, one simple reason is the diameter of the halide ion. The bigger iodide concentrates less the charge of the excess electron than the smaller fluoride does, and the less concentrated charge needs less electric energy.

A lot of people like to make that argument. I am thinking electrons are negative, in the case of iodide, there are 54 of them. How do you figure the charge is less than fluoride?

Modelling water as a continuous insulator with permittivity already tells it. q (puts signs as you like) concentrated at the surface of a radius r sphere means an energy varying like 1/(r×ε) where both the big r and water's big ε help the ionization.

Or if you prefer to imagine the discrete water molecules, there are more of them near the bigger ion, so they mitigate more easily the effect of the excess electron.

If I understand this correctly, the big "r" means the ion is being treated as a Gaussian shell. I think that is only true at a distance further from an anion. If electrons have a negative charge and we invoke the inverse square law, then at close distances, electrons will behave as local charge. I reason that is why you can see the tetrahedral structure of fluorine at close distances.

Am I construing this correctly, you do not think the repulsive effect of an iodine's nucleus is a factor?

[orgopete]

[...] The electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. [...]

If you check the bond strength of the halogens with other atoms like C and H, it's not very different from C-C or C-H bond strength. C-H and C-Cl are almost equally strong bonds despite Cl being more electronegative, and the electronegative Br makes a weaker bond than C-H. What does make the light halogens reactive is rather that the X-X bond is weak, and this acts both on the reaction enthalpy and on the activation energy.

Enthalpy of formation:
HF (-272), HCl (-92.3), HBr (-36), HI (+26)

Bond energy
HF (570), HCl (432), HBr (366), HI (298)
CF (452), CCl (352), CBr (293), CI (236)

I know these are not exactly the same thing, but really, don't you think the enthalpy of formation for HF is high because fluorine is more reactive than the other halogens? The bond energy is a kind of reverse of the formation. If we are using Pauling electronegativity, he assumed the bonds were stronger (which I think may be partly true), but I also think fluorine is a lot more reactive. I believe it is this assumption that leads to the awkward result of HF being more ionic than HI, but does not ionize. HI is a stronger acid and iodine is more electron withdrawing.

[...] we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. [...]

I don't see a difference (except in reaction direction and energy sign) between the addition of an electron to a halogen atom (the electronegativity) and the withdrawal of an electron from a halide ion.

This is two different things. I'm arguing that in the case of the electrons of fluoride, if we likened them to the electrons of a carbon (anion), nitrogen, and oxygen, there are two factors in their basicity. One, the smaller the nuclear charge, the further the electrons are to the nucleus. As an inverse square force, the nuclear field diminishes exponentially. The force of a pair of electrons would create an inverse square force. Comparing fluoride to the other electron pairs, it would be closest to the nucleus, the larger nuclear charge and the least basic. If compared to iodide, it would be the most basic.

I argue that this should be a general observation. For carbon, the shorter the CH bond, the more acidic. I use this to argue why ammonia is a stronger base than fluoride. Even though ammonia has one more proton, they are much further from the basic electrons.

The other point about adding an electron to a halogen versus knocking one off. I agree they may be quite similar. That could be reassuring, but the gas phase loss of an electron from sodium is +496 kJ/mol. So we should assume sodium will release 496 kJ/mol to reform sodium metal. What happens if you add sodium to water? It is a different reaction. The bonds to sodium ions are long and weak. A sodium cation is surrounded by 10 electrons. They are negative. It is only at a greater distance than the greater nuclear charge is felt. Up close, the electron are repelling. Sodium does not hold its electrons. I am imagining a different trend here.

[orgopete]

I'm not sure I have explained my bond length argument well. I reason that in a Bronsted acid, the attractive force should be an inverse square force to the nearest electron pair. The closer a proton to an electron pair, the greater the force. If you compare two acids, the stronger acid should correspond with the weaker force and via the inverse square law, a greater distance between the proton and its electron pair. If we compare HF and H2O, fluorine with a larger nuclear charge exerts a greater pull on its electron pairs. The electron pairs are normally repelling to one another so they resist being pulled in. This is a continuing effect for all atoms. While fluorine pulls its electrons in more than oxygen, the proton follows and a shorter bond results. However, the gap between the electron pair and a proton increases due to the repulsive effect of the fluorine nucleus. The larger gap results in a heterolytically weaker bond, and stronger acid.

I expect a similar trend should exist for the other haloacids, except the position of the electrons is masked by an increase in the number of electrons in its valence shell. I would expect iodide to exert a greater pull on its electrons than fluoride. It would also have a greater repulsive field to a proton. The net effect is the gap between an electron pair and a proton would be the greatest for iodide.

I argue it isn't the bond length that matters, but the proton-electron pair distance. Thus, HF has the shortest bond and is the strongest acid of C, N, O, and F. HI is the longest bond of F, Cl, Br, and I, but it has the longest bond because we have added an additional layer of valence electrons to which a proton is attracted. It isn't the bond length that matters, it is the gap.

In the case of carbon, the bond of an acetylene is the shortest. I interpret this to indicate that with a triple bond, the electrons are being pulled away from the HC bond and in toward the nucleus. This shift decreases the bond length and increases the proton-nuclear repulsion. The result is a larger gap between the proton and electron pair, weaker force, and greater acidity. With a tertiary hydrogen, the electron donation of carbon reverses this effect. The electrons directed to the proton are pushed out further, the bond is longer, but the proton-electron pair gap is smaller. The result is a greater force, a heterolytically stronger bond, and weaker acid.

[mjc123]

How about this from Wikipedia https://en.wikipedia.org/wiki/Hydrogen_fluoride:
"The weak acidity in dilute solution is sometimes attributed to the high H—F bond strength, which combines with the high dissolution enthalpy of HF to outweigh the more negative enthalpy of hydration of the fluoride ion.[19] However, Giguère and Turrell[20][21] have shown by infrared spectroscopy that the predominant solute species is the hydrogen-bonded ion-pair [H₃O⁺·F⁻], which suggests that the ionization can be described as a pair of successive equilibria:
H₂O + HF  [H₃O⁺·F⁻]
[H₃O⁺·F⁻]  H₃O⁺ + F⁻
The first equilibrium lies well to the right (K ≫ 1) and the second to the left (K ≪ 1), meaning that HF is extensively dissociated, but that the tight ion pairs reduce the thermodynamic activity coefficient of H₃O⁺, so that the solution is effectively less acidic.[22]"

[orgopete]

How about this from Wikipedia https://en.wikipedia.org/wiki/Hydrogen_fluoride:
"The weak acidity in dilute solution is sometimes attributed to the high H—F bond strength, which combines with the high dissolution enthalpy of HF to outweigh the more negative enthalpy of hydration of the fluoride ion.[19] However, Giguère and Turrell[20][21] have shown by infrared spectroscopy that the predominant solute species is the hydrogen-bonded ion-pair [H₃O⁺·F⁻], which suggests that the ionization can be described as a pair of successive equilibria:
H₂O + HF  [H₃O⁺·F⁻]
[H₃O⁺·F⁻]  H₃O⁺ + F⁻
The first equilibrium lies well to the right (K ≫ 1) and the second to the left (K ≪ 1), meaning that HF is extensively dissociated, but that the tight ion pairs reduce the thermodynamic activity coefficient of H₃O⁺, so that the solution is effectively less acidic.[22]"

Here is my take. A lot of what is being offered is an attempt to counter the implications of Pauling's electronegativity theory. Even Pauling seemed surprised at its failure when he states, “…the bond type has no direct connection with ease of electrolytic dissociation in aqueous solution.” If we were discussing H₂O, I doubt anyone would be surprised at its low acidity. HF should have followed the trends of its row. If electronegativity theory had not been advanced, I doubt its acidity would have been such an issue. Although I tried to discount electronegativity theory, but I cannot prevent others from bringing it up.

I might think the Wikipedia article might be written as:
 HF + H₂O    H₂O·HF    H₃O⁺ + F^-

That would seem reasonable. HF is a good proton donor for hydrogen bonding. I'd guess even better than water. Even though it is hydrogen bonded, it still doesn't ionize. The difference is I have failed to write a negative charge on the fluorine atom. My perspective is if HO^-, H₂O, and H₃O⁺ are compared, the formal charges reflect the net number of hydrogens added (irrespective of hydrogen bonding to other water molecules). The charge of the oxygen remains unchanged at +8. Adding a negative charge to the hydrogen bonded structure doesn't change the charge of the fluorine. The difference between the Wikipedia article and what I have written is the Wikipedia article expects HF to ionize hence it is written with an negative charge, yet it doesn't ionize.

[Irlanur]

I reason that in a Bronsted acid, the attractive force should be an inverse square force to the nearest electron pair. The closer a proton to an electron pair, the greater the force.

This is simply an extremly unhelpful language in the context of chemical bonds. The forces on the atoms at the equilibrium distance are simply 0. And the force constant in a harmonic approximation doesn't say too much about the dissociation energy. It becomes even more useless if a proton is hoping back and forth between different bases, which is quite often the case with acidic protons in "normal" pH solutions.

[Enthalpy]

I am thinking electrons are negative, in the case of iodide, there are 54 of them. How do you figure the charge is less than fluoride?

You wrote that. I didn't. I wrote "the excess electron". The net charge of F^-, Cl^-, Br^- and I^- is one excess electron.

About any angular structure in the electron shell: at least F^- has spherical symmetry. Adding the probability densities for three 2p orbitals removes all the angular dependencies. I suppose, but didn't check, that this holds for halides with bigger orbitals too.