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The reactivity of halogenation in sunlight is as follows:

$$\ce{F2}>\ce{Cl2}>\ce{Br2}>\ce{I2}$$

So why does we often label chlorine as most reactive halogen towards halogenation reaction?

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    $\begingroup$ Because $\ce{F2}$ is so reactive and the $\ce{C-F}$ bond is so stable, it's not considered useful in most synthetic pathways. $\endgroup$ – ringo Jan 15 '17 at 8:54
  • $\begingroup$ It’s not fully clear what you’re actually asking. The halogenation is, indeed, most reactive with fluorine. However, the halogen radical (which is what sunlight does) is most easily generated with iodine. I suspect you mean the former but could you clarify? $\endgroup$ – Jan Jan 18 '17 at 0:43
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    $\begingroup$ Why did you decide to roll back ron’s edits which were improving your post’s grammar and title? $\endgroup$ – Jan Jan 19 '17 at 12:49
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The reactivity of halogens in sunlight

OK, this tells us that we are dealing with a free radical reaction. Sunlight breaks the halogen bond ($\ce{X-X}$) to generate halogen free radicals.

$$\ce{X2 ->C[{h\nu}]\ 2X.}$$

In the next step the halogen radical abstracts a hydrogen atom from any hydrocarbon present to generate $\ce{HX}$ and a hydrocarbon free radical (note that multiple hydrocarbon radicals may be generated depending on the structure of our hydrocarbon).

$$\ce{X. + R-H -> HX + R.}$$

Finally, the hydrocarbon radical can react with the halogen free radical to produce a halogenated product.

$$\ce{R. + X. -> R-X}$$

Let's now consider the energetics of these steps. The first step, generation of the halogen free radical, uses the sunlight to break the $\ce{X-X}$ bond, so no additional energy is required for this step. Then we break a $\ce{C-H}$ bond (so we must add energy) and make an $\ce{H-X}$ bond and a $\ce{C-X}$ bond (so energy is given off). The various bond strengths are contained in the following Table.

\begin{array}{|c|c|c|c|c|c|} \hline \ce{C-H}~ \text{Bond Strength} & \ce{H-X} & \text{Bond Strength}& \ce{C-X} & \text{Bond Strength} & \text{Overall} \\ \text{(kcal/mol)}\ & & \text{(kcal/mol)} & & \text{(kcal/mol)} & \text{(kcal/mol)} \\ \hline \ 99 & \ce{H-F} & -135 & \ce{C-F} & -116 & -152\\ \hline \ 99 & \ce{H-Cl} & -103 & \ce{C-Cl} & -81 & -85\\ \hline \ 99 & \ce{H-Br} & -88 & \ce{C-Br} & -68 & -57\\ \hline \ 99 & \ce{H-I} & -71 & \ce{C-I} & -51 & -23\\ \hline \end{array}

(bond strengths taken from Bill Reusch's table)

We can clearly see that all of these free radical reactions will occur (e.g. they are exothermic) in the order you suggested, with fluorine being, by far, the most reactive.

$$\ce{F2 >> Cl2 > Br2 > I2}$$

Saying that chlorine is the most reactive is incorrect.

Note: 99 kcal/mol was used as a typical $\ce{C-H}$ bond strength. We could refine our model by using more exact $\ce{C-H}$ bond strengths (see this Wikipedia table) and calculate the relative rates of halogenation at the various carbon atoms in a complex hydrocarbon.

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Reaction with fluorine are strongly exothermic and mostly during reaction explosion takes place and reaction generally not used for synthetic purpose.

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