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I know that chlorine can have a positive charge in the case of oxides. But what I am wondering is if chlorine dioxide can form a ring and if so how likely it is.

$\ce{O=Cl=O -> O- -Cl-O- -> (cyclic)O-Cl-O}$

This reaction would form a triangular ring with each oxygen being bonded to 2 atoms so the charge would once again be zero. But right in the middle of it you have an ion. It isn't chloryl or chlorite since it has a charge of 2-. This ion then bends to where the oxygens are close enough to form a peroxide bond giving you cyclic chorine dioxide.

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    $\begingroup$ Three-membered rings are not very favorable. I wouldn't say it's imposible though because of the size of chlorine. $\endgroup$ – Dissenter Sep 15 '15 at 17:55
  • $\begingroup$ @Caters Sorry but your reasoning doesn't make sense. $\endgroup$ – Mithoron Sep 15 '15 at 18:25
  • $\begingroup$ Why? Doesn't forming a ring from a molecule with double bonds at the ends require that you form a single bonded ion first and then have the ion bend in just the right way that the negatively charged atoms can bond with each other? I mean sure this might be very unlikely but negative ions come towards each other all the time and react so negatively charged atoms within an ion would do the same thing, especially if the bond angle means that they have to come towards each other. $\endgroup$ – Caters Sep 15 '15 at 18:37
  • $\begingroup$ @Caters No - why do you think that has to be the case. For a start, where do the two extra electrons come from to form the ion? $\endgroup$ – bon Sep 15 '15 at 19:40
  • $\begingroup$ They come from the double bonds being broken. Since oxygen is more electronegative it will take both electrons from the bond and have a negative charge. This causes chlorine to have a positive charge as a result. And why wouldn't the ion need to form? I mean a double bond on both ends of a molecule with 3 atoms means that they have to be broken in order for them to bend enough to form a ring. This likewise means that at least for molecules with 3 atoms an ion does need to form. $\endgroup$ – Caters Sep 16 '15 at 1:04

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