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#1 2010-08-06 17:39:45


Homolytic, heterolytic bond strengths and lattice energy

I saw the following question posted on Yahoo!Answers

Yahoo!Answers in chemistry wrote:

Bond Strength, Energies...and a bonus question about surface interactions?
First off, if you decide to explain these things for me try to give a practical, qualitative explanation along with any mathematical descriptions/crazy jargon you use. Also, please don't answer these questions with vague explanations like "oh, well ionic bonds don't share electrons like covalent bonds do" because this doesn't provide an in depth explanation of how that effects things like bond dissociation energies

1) Is it accurate to say that "bond energy" is essentially the average of the "bond dissociation energy" within a molecule. Also, I want to confirm that bond dissociation energy is always referring to the homolytic cleavage of a bond while lattice is referring to heterolytic?

2) When people say ionic bonds are weaker than covalent, are they really saying the "Bond dissociation energy" for a covalent compound is generally higher than the "lattice energy" for an ionic compound. IE - (read this carefully) wouldn't homolytic cleavage of ionic compounds be more energy intensive than covalent ones? Analogously, wouldn't the lattice energy for a covalent compound be higher than an ionic one? I'm confused about the interplay between homo and heterolytic cleavage and how that impacts bond strengths in ionic and covalent compounds.

3) This question arises from my confusion in #2.It is often said that the strength of C-F bond is due to a powerful dipole moment between the two atoms that creates a somewhat ionic character. This is also in line with the idea that the flouride ion makes a poor leaving group in organic mechanisms due to its high electron density and small bond length. But how does the ionic character strengthen if ionic bonds are supposedly weaker than covalent?? Wouldn't the fact that flourine likes to hold the negative charge make a C-F weaker not stronger???? As shown below, can you explain why C-F is stronger than C-C?

I feel the strong electronegativity of flourine counteracts it's high electron density...the two characteristics seem to go against each other in providing bond strength. And if indeed it is the small size of the flourine atom, why is the C-H bond still weaker than C-F if H is clearly smaller than F.

C-H 98 kj/mol

C-F 117 kj/mol

C-C 85 kj/mol

H-H 104 jk/mol

F-F 158 kj/mol

It is shown that C-F has a higher bond dissociation energy, which to me says that homolytic cleavage is costly. But in organic mechanisms, Flourine ALSO makes a poor leaving group which would be a heterolytic cleavage. Does that mean that the C-F bond has high "lattice" energy too?
This also begs the question, people say the iodine makes a good leaving group due to it's large size and lack of density. The bond dissociation energy is, appropriately, very low compared to C-F which is charge dense. BUT if BDE refers to homolytic cleavage, how does it apply to leaving groups in organic mechanisms when leaving groups break by heterolytic cleavage.

This is a $64,000 question (old tv game show). I have been trying to answer this question also (and many chemists have also tried to provide an answer). I don't think you will find an good answer or if you do, please contact me.

Homolytic bond cleavage yields radicals, R-R -> R  +  R
Heterolytic bond cleavage yields ions, R-X -> R(+) + X(-)

Bond dissociation energies measure homolytic bond strengths. Lattice energy is used to calculate the energy to separate a mole of a solid into a gas of its ions.

So, the poster is paying attention to what is being said, therefore he arrives at his questions. How can this be?
Herein lies the problem, "How can you measure what you think you want to measure?" Is heating sodium chloride the best way to determine how much energy to takes to separate NaCl into a sodium and chloride ions?  Heating may measure lattice energy, but it doesn't measure bond energy. How do they differ? If you make a bridge out of toothpicks, the bridge can be much stronger than an individual toothpick. The lattice energy of ice is relatively low and the mp is low. If you used organic compounds, the lattice energies would be lower still. However, in each of these cases, breaking the lattice would not require breaking any covalent bonds. These lattices are made up of covalent and other weaker bonds, hydrogen bonds or London forces.

When an object is broken, what you measure is the weakest force. That is the analogy with a chain. You will find the weakest link. In a lattice of organic compounds, the weakest link are weak forces, like hydrogen bond, London forces, dipole-dipole interactions, etc. Furthermore, the organic lattice, once broken, becomes a solvent to further dissolve the components making up the lattice. That is, what happens upon melting. The solution becomes a good medium to further dissolve the solid.

On the other hand, compare this with melting of sodium chloride crystals. Arguably, the "ionic bonds" making up the crystal are stronger bonds than those making up organic solids. Furthermore, sodium chloride probably is not a good solvent for dissolving sodium chloride crystals. If you added a small amount of water, you would find water is much more effective in breaking the lattice.

If one wished to use use lattices to compare the strength of ionic and covalent bonds, then one should compare NaCl and diamond. The bonds in NaCl are ionic and the bonds in diamond are covalent. In both cases, the lattice is made up the the same types of bonds. This is different from other organic compounds in which a relatively small number of weak bonds may be contributing to the crystal lattice, such as cholesterol, benzoic acid, acetanilide, etc.

Let's try to make simple comparisons. If we wished to compare lattice energies, then we should compare NaCl to diamond. In melting NaCl, we are breaking the lattice. If we attempt to do the same with diamond, we must also break the lattice. Because four covalent bonds are in diamond, it is very difficult to break and diamond is one of the hardest compounds known and much harder than salt crystals.

If we wished to break NaCl into its ions, the easiest way to do so would be with water. That is a very easy reaction. If we attempted the same reaction with chloromethane, nothing would happen because the bonds are too strong and the ions too unstable. If we used t-butyl chloride, then a slow ionization would take place. This should indicate that ionic bonds are weak and covalent bonds are strong (to use the chemistry vernacular). It should also indicate that homolytic cleavage is more difficult than heterolytic cleavage (into ions). However, because this ionization involves participation of solvent, no clear and predictable method of calculating bond strengths has arisen from solubility or other methods to detemine ionic or heterolytic bond strengths.

The bond dissociation energies measure the homolytic cleavage of the bonds and not the heterolytic strength. The values are important from a conservation of energy point of view, but not necessarily for estimating heterolytic bond strengths. Again, let us return to NaCl. If you calculate the bond energies, it is an exothermic reaction. Sodium donates electrons to chlorine. Think about it. If the affinity of sodium for its valence electrons is so weak, then it readily gives them up and chlorine has such a high affinity for an additional electron, that it readily accepts one, then how great should the attraction of a sodium cation be for the electrons of a chloride anion?

Note that I used the term "chemistry vernacular" in referring to ionic and covalent bonds. I posit those are not good terms, but simply describe weak versus strong bonds. Ionic bonds are generally weak and covalent bonds are strong. If we take that definition, then some bonds that break heterolytically could be weak or strong. Thus, silver chloride may form strong bonds that break into ions and trityl chloride may form weak bonds that also break into ions. We may begin to recognize that atoms that have a higher affinity for electrons may hold them more strongly than those with a weak affinity. The metals as a class have a low affinity for electrons and as we increase the number of electron around atoms, we see a trend toward metallic character.

The other bond strength we should consider is homolytic bond strengths. Generally, when we break bonds using the least amount of energy possible, we find heterolytic bond cleavage. However, we do find some molecules that can also break bonds homolytically at relatively low energy levels. The halogens and peroxides are examples. In characterizing the reaction path of these low energy radicals, they generally engage in reactions to form stable two electron bonds. That is, unpaired electrons are generally less stable than paired electrons.

Fluorine or how is fluorine different? Fluorine is more like oxygen than it is like chlorine. Although fluorine holds its electrons tightly, hence strong bonds, it doesn't hold them as tightly as chlorine, hence less likely to form ions. Fluoride is more basic, HF is a weaker acid than chloride. Fluorine is less dense than chlorine. The electron density of fluorine is the least of the halogens. The volume of atoms is proportional to the volume of its electron clouds. Iodine is more dense than bromine>chlorine>fluorine. Iodine holds its electrons most tightly. It has the largest number of protons to increase the Coulombic attraction. Since iodine holds its electrons the most tightly, the attraction to a proton is weak and HI is a strong acid.

How can iodine be a good nucleophile and a good leaving group? In measuring the acidity of HI, this is a thermodynamic property. It is affected by the inherent ability of iodine to withdraw its electrons. Nucleophilicity is a kinetic property. It is a function of which reaction can be faster. Two explanations have been suggested for the nucleophilicity of iodine. The first is a hydrogen bonding effect. The greater the hydrogen bonding, the less available will its electrons be for reaction. Since the electrons of iodine are held more tightly than the other halogens, it would be less subject to the retardation of hydrogen bonding. I think this is correct, but not sufficient for all of the properties of iodine. The other is that because the outer electrons of iodine are subjected to a Coulombic attraction to the nucleus and a Coulombic repulsion to the inner electrons. Because we don't know the properties of electrons completely, we may think that the electrons may extend further beyond its nucleus but with only a weak force. This weak force may kinetically be sufficient to form a bond. Alternately, we may think that local ionizations take place and that one of these electronic peaks forms a bond.

Even though iodine can kinetically form a bond at a higher rate than other halogens, the bond strengths may not be large. That is, if iodine pulls its electrons in more than the other halogens, then the thermodynamic bond strength may be weaker. When we measure the kinetic rate of an SN1 reaction, we are determining the slowest or rate limiting step. If we compare iodine to the other halogens, the rate limiting bond cleavage of iodine is much faster than the other halogens. This is because iodine pulls its electrons toward its nucleus more than the other halogens. Check the densities of the halogens.


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