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In hydrolysis/nucleophilic substitution of haloalkanes, the bond enthalpy indicates the rate of hydrolysis (e.g. the $\ce {C-I}$ bond is weaker than the $\ce {C-Cl}$ bond so in hydrolysis the alkane with the $\ce {C-I}$ bond will break more quickly and react with the $\ce {OH-}$ ions)

However, in free radical substitution reactions, if you have a molecule like $\ce {CCl2F2}$ , then is the $\ce {C-F}$ bond stronger or weaker than the $\ce {C-Cl}$ bond (so which will require less energy to break)?

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For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker.

This can be rationalized by considering the poorer orbital overlap between the $\ce {Cl}$ bonding orbital (3 s/p character, more diffuse than 2s/p) and the $\ce {C}$ $\mathrm{sp^3}$ hybrid (predominantly $\text{2p}$ character) than that between the $\ce {F}$ bonding orbital (2 s/p character) and the $\ce {C}\ \mathrm{sp^3}$ hybrid.

Some data on typical BDE values in $\ce {F/Cl}$ compounds can be found here.

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  • $\begingroup$ But why would polar bonds have a higher bond enthalpy? $\endgroup$ – user58953 Dec 11 '14 at 17:28
  • $\begingroup$ This is often the case; bonds with poorer orbital overlap are typically weaker even with a better electronegativity match. $\endgroup$ – J. LS Dec 11 '14 at 21:16
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Rather than looking at different bond strengths, when determining which bond reacts favourably one should look at reactants and products.

In nucleophilic substitutions or eliminations, this means considering the more stable halide anion. The larger the halide, the more diffuse and thus the better stabilised the negative charge is.

In radical reactions, the stability of the product is also more important. Here again, the lower an element is in the periodic table, the less unstable the radical is.

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I would like to offer an alternative explanation. The answer provided by J. LS does not provide the main reason for why the $\ce {C-F}$ bond is stronger than the $\ce {C-Cl}$ bond. Yes, it is true that fluorine's $\ce {2p}$ orbital is less diffuse than chlorine's $\ce {3p}$ orbital. However, it is also important to note that orbital energies need to be accounted out in determining the effectiveness of the covalent interaction between two orbitals. In order for two orbitals to interact strongly, they must be of similar sizes and energies. The Pauling electronegativity values of fluorine and chlorine are $\ce {3.98}$ and $\ce {3.16}$ respectively while the value for carbon is $\ce {2.55}$. Clearly, the orbital energies of chlorine and carbon would be more similar than the orbital energies of fluorine and carbon. Thus, we cannot conclude that the covalent interaction between carbon's and fluorine's atomic orbitals is stronger than that between carbon's and chlorine's atomic orbitals.

What then can we use to rationalise the bond strengths?

We can actually explain this using the idea of bond polarity and ionic contributions to the covalent bond. Lemal (2004) reports that the partial charge on carbon in the molecule $\ce {CF4}$ is $\ce {+0.76}$. This illustrates that partial charges are actually very significant when looking at the $\ce {C-F}$ bond. Thus, it is due to this ionic character of the $\ce {C-F}$ bond, that is, the attraction between the partial positive charge on carbon and the partial negative charge on fluorine, that results in the $\ce {C-F}$ bond being significantly stronger than the $\ce {C-Cl}$ bond.

Reference

Lemal, D. M. (2004). Perspective on Fluorocarbon Chemistry. The Journal of Organic Chemistry,69(1), 1-11. doi:10.1021/jo0302556

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