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Topic: Why is HF a weak acid  (Read 10964 times)

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Offline Enthalpy

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Re: Why is HF a weak acid
« Reply #15 on: November 26, 2016, 04:27:59 AM »
Bond strength compared with electronegativity and reactivity:

439kJ/mol CH3−H
432kJ/mol H-Cl
350kJ/mol CH3−Cl
218kJ/mol H-H
121kJ/mol Cl-Cl

On this example you see that bond strength with carbon doesn't result from electronegativity nor from the difference of electronegativity between the atoms, and that chlorination happens because the Cl-Cl bond is weak.

Also, the stronger H-H bond is what distinguishes hydrogen from the halogens.

Offline orgopete

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Re: Why is HF a weak acid
« Reply #16 on: November 28, 2016, 09:50:10 PM »
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.

... 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.
I was referring to the notion that the iodide concentrates less charge. I thought it was actually the opposite. There are 54 electrons per atom compared to 10.

I think perhaps my thought that the nuclear charge is important is not being given enough credit. If protons are being repelled by the nuclear charge, then it seems plausible that one should compare the charges and distance for HF and HI. HF, 9 protons, has a bond length of 92 pm while HI, 53 protons, has a bond length of 161 pm. By my calculations, the proton of HI has a 1.9 times greater repulsive force from the nucleus. I find this a compelling argument. I use a similar argument to explain why ammonia should be more basic than fluoride. It isn't the net charge per se, but the collection of forces of independent particles. That is, I use the charge and distance of a nitrogen nucleus plus the charge and distance of the three more remote protons to calculate the repulsive force a proton would feel at the distance of an ammonium ion.

I further find this plausible that a greater charge from a larger nucleus can compress the inner electrons to a greater extent than lighter atoms. If the electrons are being pulled in further, a proton must approach closer to the nucleus to be attracted to the electrons. I am prone to believe this nuclear charge may accommodate nuclear attacks on iodine in net reductions by iodide and iodo compounds forming halogen bonds as electron acceptors.
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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.

Poly HF has a zig-zag structure. This makes me think it would have a similar tetrahedral geometry as borohydride, methane, ammonia, and water. How do you know it is spherical?
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Offline orgopete

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Re: Why is HF a weak acid
« Reply #17 on: November 28, 2016, 10:02:40 PM »
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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.

In the model I was imagining, I was proposing something like that shown below. In that model, a pair of sp3 electrons can both attract a proton and to be attracted to a nucleus. If the proton were an acid, then I'm further imaging the attraction of the proton should be an inverse square force to the electron pair. If I compare the second row elements, LiH, BeH2, and BH4(-), the electrons remain with the proton and are hydride donors. Carbon is in between. As the nuclear charge increases, the electrons are pulled closer and I reason the force between the sp3 electrons and a nucleus is greater than to a proton and so they are acids. The greater the nuclear charge, the stronger the acid. I reason it is because the proton-electron pair distance is larger and its force is smaller.
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Offline orgopete

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Re: Why is HF a weak acid
« Reply #18 on: November 28, 2016, 10:15:58 PM »
Bond strength compared with electronegativity and reactivity:

439kJ/mol CH3−H
432kJ/mol H-Cl
350kJ/mol CH3−Cl
218kJ/mol H-H
121kJ/mol Cl-Cl

On this example you see that bond strength with carbon doesn't result from electronegativity nor from the difference of electronegativity between the atoms, and that chlorination happens because the Cl-Cl bond is weak.

Also, the stronger H-H bond is what distinguishes hydrogen from the halogens.

I'm not following the argument (nor the bond strength data posted). Are you agreeing with me or disagreeing? I am of the opinion that gas phase and solution phase data are simply different. I would conclude that even though the gas phase data is similar, HCl has a heterolytically weaker bond than chloromethane. If that is what you are arguing, then I agree. If you are saying they are similar, then I disagree.
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Offline Irlanur

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Re: Why is HF a weak acid
« Reply #19 on: November 29, 2016, 04:50:21 AM »
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I reason it is because the proton-electron pair distance is larger and its force is smaller.

doesn't change anything about the fact that the force is 0 at Equilibrium. If it the force wouldn't be zero, then the proton would move closer.

The only possibility I see to include forces in any way is by discussing the force constants, which come from a taylor approximation around the equilibrium position. But again this has rather little to do with acidity.

Offline orgopete

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Re: Why is HF a weak acid
« Reply #20 on: November 29, 2016, 09:56:56 AM »
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I reason it is because the proton-electron pair distance is larger and its force is smaller.

doesn't change anything about the fact that the force is 0 at Equilibrium. If it the force wouldn't be zero, then the proton would move closer.
I think I get this argument. The forces on any subatomic particle will be found in the valley of a Morse curve, zero. I agree that will describe the equilibrium condition, I'm not sure I agree with the implications. It suggests an equal energy level (forces) for two different acids. I think even though the protons may be found in a valley, I don't think the valleys have the same energy levels (forces) nor the same shapes, but I think I'll let you explain this to us.

If the forces acting on a proton in HF and HI are 0 at equilibrium, why is HI a stronger acid?
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Offline Enthalpy

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Re: Why is HF a weak acid
« Reply #21 on: November 30, 2016, 02:34:15 PM »
I was referring to the notion that the iodide concentrates less charge. I thought it was actually the opposite. There are 54 electrons per atom compared to 10.

I stop here. Good luck.

Offline Irlanur

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Re: Why is HF a weak acid
« Reply #22 on: December 01, 2016, 03:35:10 AM »
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I think I get this argument. The forces on any subatomic particle will be found in the valley of a Morse curve, zero. I agree that will describe the equilibrium condition, I'm not sure I agree with the implications. It suggests an equal energy level (forces) for two different acids. I think even though the protons may be found in a valley, I don't think the valleys have the same energy levels (forces) nor the same shapes, but I think I'll let you explain this to us.

If the forces acting on a proton in HF and HI are 0 at equilibrium, why is HI a stronger acid?


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energy levels (forces)

Why do you put the (forces) in brackets? The force is the negative gradient of the potential energy. You can deduce nothing about the energy from the fact that the force is 0 at equilibrium.
The acidity is determined by the Gibbs free energy difference between the protonated and the deprotonated form. you cannot make any (rigorous) conclusions whatsoever about this energy difference from the knowledge about the forces. 

Offline orgopete

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Re: Why is HF a weak acid
« Reply #23 on: December 01, 2016, 10:00:58 AM »
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Why do you put the (forces) in brackets?

It is because I perceive a difference between chemistry and physics over force v energy. Chemists refer to and calculate energy or energy differences but without referring to the force that creates it. I argue you can only have an energy difference as the result of a force. I think this is the source of why you may find, "you cannot calculate the energy difference with classical physics, you must use quantum theory". I perceive this to mean, we know there is an energy difference, but we haven't been able to describe the forces that actually cause the energy difference. For example, an aromatic ring has a lower energy level than predicted for a cyclohexatriene. What is the force leading to the lower energy level? (This is a very old idea. Gilbert Lewis suggested the discredited Parson's magnetron model. I think anyone able to elucidate an equation for the force between electrons at subatomic distances would receive a Nobel prize.)

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The force is the negative gradient of the potential energy. You can deduce nothing about the energy from the fact that the force is 0 at equilibrium.
The acidity is determined by the Gibbs free energy difference between the protonated and the deprotonated form. you cannot make any (rigorous) conclusions whatsoever about this energy difference from the knowledge about the forces.

I think we may be in agreement here. At equilibrium, the difference in force acting on two atoms would be zero and this does not predict a result of a collision with another atom. My analogy here is that in the gas phase, molecules can experience no difference in the forces acting on its bonds. If a collision were to occur, the bonds may break as a result. The energy required to break a bond in the gas phase can be quite different from the energy to break a bond in solution.

It should be obvious that I cannot explain my thinking on this subject. I am interested in the implications of this difference. I'll wait to see whether you even agree with this point of view or how I might understand yours.
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Offline Irlanur

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Re: Why is HF a weak acid
« Reply #24 on: December 01, 2016, 10:44:04 AM »
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It is because I perceive a difference between chemistry and physics over force v energy. Chemists refer to and calculate energy or energy differences but without referring to the force that creates it. I argue you can only have an energy difference as the result of a force. I think this is the source of why you may find, "you cannot calculate the energy difference with classical physics, you must use quantum theory". I perceive this to mean, we know there is an energy difference, but we haven't been able to describe the forces that actually cause the energy difference.

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I argue you can only have an energy difference as the result of a force

I don't think it is very useful to think this way. You say the the force is the cause for an energy difference. You can also say it the other way around: an energy difference leads to a force. It's just a functional relationship, which one can express in differential or integral form. The question here is not which one is "truer", but which formalism is easier to use. You can of course neglect the last roughly 100 years of research in theoretical chemistry, but let's not do this for now. What we can do (with various accuracy) quite well at the moment is calculate the energy of a chemical system where we set the positions of the nuclei as fixed (Born-Oppenheimer). We can then parametrically scan what energy we get for different nuclear positions. This gives us what we call an energy hyper-surface. The fixed nuclei are also needed for what we usually called a "structure". Usually we don't need to calculate the force to predict anything. In case we do (e.g. for IR-spectra), we can simply differentiate the energy.

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My analogy here is that in the gas phase, molecules can experience no difference in the forces acting on its bonds.
I don't understand the sentence. A force cannot really act on a bond, only on the elementary particles. or do you mean the "bonding electrons"?

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The energy required to break a bond in the gas phase can be quite different from the energy to break a bond in solution.

Yes of course, but this is simply because these are two different reactions. Let me clarify this:

Let's say we have an acid in the gas phase at very low pressure, we can then approximate the system as an ensemble of non-interacting subsystems. One subsystem is just a single acid molecule A-H. Now there is an energy difference between the states A-H and A(-) + H(+). The transition between the two is continuous. the second state in principle only exists in our minds, and what we usually calculate is the energy of a state where the proton and the base are infinitely far away from each other, so that E(A(-)+H(+))=E(A(-))+E(H(+)).

The approximation of isolated centers is pretty good in the gas phase at low pressures. It is absolutely terrible in solution. What we then usually write as A-+H+ or sometimes as A-+H3O+ is extremely far away from reality. The energy of the system and the energy differences can vary substantially if there is a solvent. One can probably say that this is one of the main reasons why Compuational Chemistry is not what we would like it to be.

Offline orgopete

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Re: Why is HF a weak acid
« Reply #25 on: December 08, 2016, 09:57:32 AM »

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I argue you can only have an energy difference as the result of a force

I don't think it is very useful to think this way. You say the the force is the cause for an energy difference. You can also say it the other way around: an energy difference leads to a force. It's just a functional relationship, which one can express in differential or integral form. The question here is not which one is "truer", but which formalism is easier to use. You can of course neglect the last roughly 100 years of research in theoretical chemistry, but let's not do this for now. What we can do (with various accuracy) quite well at the moment is calculate the energy of a chemical system where we set the positions of the nuclei as fixed (Born-Oppenheimer). We can then parametrically scan what energy we get for different nuclear positions. This gives us what we call an energy hyper-surface. The fixed nuclei are also needed for what we usually called a "structure". Usually we don't need to calculate the force to predict anything. In case we do (e.g. for IR-spectra), we can simply differentiate the energy.

I rather liked my model of an acid. I thought it was easy to understand the forces holding a proton to the electrons of an anion and I think it is actually "easier to use". I argue it is a simple inverse square force, the further a proton is from an electron pair, the weaker the force. I think it explains why ammonia is a stronger base than fluoride. I think it generally holds, but I am interested in knowing where it fails also.

If I didn't know what the force actually was or how to calculate it, then I too would say it is unimportant. I'd consider Newton's connection of gravity and mass together to be more important than anyone who may have built a scale to weigh things. Just because you can weigh something or calculate it doesn't explain what the force was that caused it.

Anytime you determine there is an energy well, don't you just wonder what caused it? Can't you make a guess? Shouldn't everyone at least make a guess? We may not succeed as Newton had, but even simple logic tells me there must be a force for an energy well to exist. This is what Gilbert Lewis thought over 100 years ago. I think there is a lot of merit to this kind of thinking. I cannot imagine anyone telling Newton they have a scale for measuring weight (mass), so they don't need a formula to calculate it.
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The energy required to break a bond in the gas phase can be quite different from the energy to break a bond in solution.

Yes of course, but this is simply because these are two different reactions. Let me clarify this:

Let's say we have an acid in the gas phase at very low pressure, we can then approximate the system as an ensemble of non-interacting subsystems. One subsystem is just a single acid molecule A-H. Now there is an energy difference between the states A-H and A(-) + H(+). The transition between the two is continuous. the second state in principle only exists in our minds, and what we usually calculate is the energy of a state where the proton and the base are infinitely far away from each other, so that E(A(-)+H(+))=E(A(-))+E(H(+)).

The approximation of isolated centers is pretty good in the gas phase at low pressures. It is absolutely terrible in solution. What we then usually write as A-+H+ or sometimes as A-+H3O+ is extremely far away from reality. The energy of the system and the energy differences can vary substantially if there is a solvent. One can probably say that this is one of the main reasons why Compuational Chemistry is not what we would like it to be.
Two things. Yes, the difference between solution and gas phase energy is what I am interested in. Pauling's electronegativity theory is based upon bond energy as a sum of covalent and ionic portions. I agree there is a difference between gas and solution phase. By a conservation of energy principle, one should be able to make a calculation. I want to know how to do it. HF seems like an ideal example. It is small, I am pretty familiar with the properties of the rows and columns. It is at an end of an extreme which can be helpful to understanding or discovering any principles. So why is the enthalpy of neutralization of HF -68 kJ/mol while H3O+ is -57 kJ/mol? Who determined -68 kJ/mol? If HF is a weaker acid, than I don't see why more energy should be released. However, I find these calculations complicated.

The second thing is that it seems paradoxical that I should rely upon computational methods and that I should also realize that computational chemistry is not what we would like it to be.
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Offline orgopete

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Re: Why is HF a weak acid
« Reply #26 on: December 08, 2016, 10:47:43 AM »
I think I understand the Born-Haber cycle for NaCl. You can also find a value for a solution phase Born-Haber cycle and it is slightly lower. I reason that in a solution phase Born-Haber cycle, no bond is formed between NaCl so the net energy is mostly the energy of the redox reaction. By a conservation of energy principle, there shouldn't be any difference in the net energy of a reaction even if a different mechanism. I think I'm still running up against the  “… bond dissociation energies refer to molecules in the gas phase and aren’t directly relevant to chemistry in solution.” McMurry, 8th edition of organic chemistry. McMurry still has some examples. The ones that are gas phase an analogous to the measurement of the heats of formation, I expect are reasonable. Those that are entirely ionic is what I'm expecting McMurry is referring to. It doesn't seem to me that the values shouldn't be reliable from a conservation of energy point of view, it is simply a matter of how.
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Offline Irlanur

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Re: Why is HF a weak acid
« Reply #27 on: December 09, 2016, 09:11:14 AM »
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I rather liked my model of an acid.
Well. I don't find this very fruitful and will also leave the thread at this point.

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Re: Why is HF a weak acid
« Reply #28 on: December 10, 2016, 04:51:23 PM »
Great question. Here's my shot at it. Fluorine is very electronegative which makes it the ideal halogen for having a very acidic hydrohalide( HX). However, fluorine has a little problem when it comes to acidity...and I do mean "little"..its size doesnt allow for stabilization of the added electron from hydrogen. After dissociation in a aqueous solution, the H+ and -F are on their own. -F is too unstable because its not successfully delocalizing that electron over its relatively small nucleus. In order for reactions to make sense, they end game is stability. Just remember stability governs reactions even LIFE.  :)
« Last Edit: December 10, 2016, 05:07:51 PM by Scitalk »

Offline orgopete

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Re: Why is HF a weak acid
« Reply #29 on: December 11, 2016, 02:49:28 AM »
Great question. Here's my shot at it. Fluorine is very electronegative which makes it the ideal halogen for having a very acidic hydrohalide( HX). However, fluorine has a little problem when it comes to acidity...and I do mean "little"..its size doesnt allow for stabilization of the added electron from hydrogen. After dissociation in a aqueous solution, the H+ and -F are on their own. -F is too unstable because its not successfully delocalizing that electron over its relatively small nucleus. In order for reactions to make sense, they end game is stability. Just remember stability governs reactions even LIFE.  :)
Although this is a little different than the kind of data I'm seeking, I'll try to explain. 10 electrons is the most stable state of fluorine. If comparing HF and F(-), each has 10 electrons in each so I don't imagine any instability. Boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, and magnesium are all in their most stable states with 10 electrons. Except for boron, these atoms are all less stable if they have less than 10 electrons. Adding an electron to fluorine increases its stability. Adding an electron is a redox reaction and is a different reaction with different energies than the forces holding a proton to an electron pair (my question). I further argue that for methane, ammonia, and water, the electrons are arranged in pairs in a tetrahedral arrangement. By analogy, HF and F(-) should have the same structure. If this is so, then all electron pairs may be treated equally and delocalization shouldn't change the structure of the atom or the basicity of fluoride (or any atom). Basicity is a property of an electron pair. I argue basicity should be an inverse square force, the closer, the stronger. If you compare the bond lengths, CH>NH>OH>HF. The nuclear charge of carbon is less so its electron pair extends further from its nucleus than the others, hence its electron pair should be more basic to a proton. By analogy, HF would be the least basic of this row.

If this model holds (and I think it does), then it should hold for all atoms. For the series BH4(-), CH4, NH3, H2O, and HF, they all contain 10 electrons so the difference in forces should be found in nuclear-electron pair, nucleus-proton, and electron pair-proton distances. We do not know the locus of electrons, but I argue they should still follow the laws of physics, namely inverse square forces. Anyone can do the calculations as I have and discover HF should be the weakest acid. It should be weaker than HI because of the greater charge of iodine pulls its valence electrons proportionately closer to its nucleus increasing its nucleus-proton repulsion and reducing the electron pair-proton attraction or acidity.

I know how to calculate the energy differences in reactions. What I do not know how to do is to adjust for the various changes that must be made to accommodate for the difference between gas phase and solution phase energies. For example, HF is a strong bond in the gas phase but weaker in solution. HCl, HBr, and HI are similar. The loss of an electron by sodium is endothermic in the gas phase but exothermic in solution. We should still be able to calculate the energy differences in reactions. Acidity is a solution based force and is different from gas phase forces. How do we adjust for the values to be used?
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