Molecules of chlorine do not decompose into atomic chlorine. Why?

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Molecules of chlorine do not decompose into atomic chlorine. Why?

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Answer 1 · 2018-05-14T07:46:29

Chlorine is more stable as $C l_{2}$ $Cl_2$ than $C l^{.}$ $Cl^.$ .

Explanation:

Assuming that this question is asking about why $C l_{2}$ $Cl_2$ doesn't become the chlorine radical, $C l^{.}$ $Cl^.$ , under normal circumstances:

Chlorine is one of the diatomic elements—elements that, essentially, only exist in nature in as a diatomic molecule.
Other diatomic elements are $B r_{2}$ $Br_2$ , $I_{2}$ $I_2$ , $N_{2}$ $N_2$ , $H_{2}$ $H_2$ , $O_{2}$ $O_2$ , and $F_{2}$ $F_2$ .

This is because the $C l_{2}$ $Cl_2$ molecule is much more stable and much less reactive than the $C l^{.}$ $Cl^.$ atom.

In a regular atom of $C l^{.}$ $Cl^.$ , there are $7$ $7$ valence electrons.

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It needs just $1$ $1$ more valence electron to achieve noble gas configuration ( $8$ $8$ electrons, in this case), so it is reactive, because it wants to gain that $1$ $1$ electron.

In the $C l_{2}$ $Cl_2$ molecule, both of the chlorine atoms have noble gas configuration:

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So, chlorine is much more stable as $C l_{2}$ $Cl_2$ , compared to $C l^{.}$ $Cl^.$ , because $C l^{.}$ $Cl^.$ wants to gain $1$ $1$ more electron to achieve noble gas configuration—which $C l_{2}$ $Cl_2$ already has.

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Molecules of chlorine do not decompose into atomic chlorine. Why?

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