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$\ce{Cl2}$ and $\ce{Br2}$ can disproportionate in a cold, aqueous, basic environment to form $\ce{ClO-}$ and $\ce{BrO-}$ respectively, yet $\ce{I2}$ directly forms $\ce{IO3^-}$. Why isn't $\ce{IO-}$ isolatable in solution?

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    $\begingroup$ At any temperature, IO- itself are rapidly redox reaction I- and IO3-. $\endgroup$ – user6006786 Mar 17 '16 at 10:24
  • $\begingroup$ @user6006786 Sure, but why? $\ce{ClO-}$ doesn't showcase this kind of behaviour at lower temperatures, yet $\ce {IO-}$ does. $\endgroup$ – L3ul Mar 17 '16 at 11:25
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    $\begingroup$ I think this is because the cl-o bond is stronger than i-o, but I could not find the data $\endgroup$ – user6006786 Mar 17 '16 at 14:36
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When iodine is added to water, the following reaction results:

$$\ce{I2(l) + H2O(l) -> OI^{-}(aq) + 2H+(aq) + I^{-}(aq)}$$

$\ce{I2}$ molecules and water molecules react to form hypoiodite ($\ce{OI-}$). The reaction can move both ways of the equilibrium, depending on the pH of the solution. Iodine may also occur as $\ce{I3-}$(aq), $\ce{HIO}$(aq), $\ce{HIO3}$(aq) . The compounds that are form behave differently when they come in contact with water. When iodine ends up in surface waters, it may escape as iodine gas.

$\ce{IO-}$ rapidly disproportionates to form $\ce{IO3-}$ and $\ce{I2}$ $$\ce{IO- -> IO3- + I2}$$.

See here for more details.

For the disproportionation reaction, consider the Latimer diagram for iodine:-

enter image description here

A species has a tendency to disproportionate into its two neighbor species if the potential on the right of species in a Latimer diagram is more positive than that on the left. In this case, right hand potential($\ce{+0.42}$) is greater than the left hand potential ($\ce{+0.15}$). So , it disproportionates.

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    $\begingroup$ Yet, my question still remains: why $\ce{IO-}$ rapidly disproportionates, whereas $\ce{ClO-}$ can be isolated? $\endgroup$ – L3ul May 7 '16 at 5:25
  • $\begingroup$ Well, in your edit, I see that $\ce{IO-}$ is a radical oxide. That doesn't sound right since all electrons are found in pairs. $\ce{IO}$ does have an unpaired electron, but not $\ce{IO-}$. $\endgroup$ – L3ul May 7 '16 at 8:42
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    $\begingroup$ Your answer does prove that $\ce{IO-}$ is unstable according to the standard reduction potentials of $\ce{IO^{3-}}$ and $\ce{IO-}$ - but that's not what I'm looking for. I'm trying to find the reason behind the numbers, not the numbers themselves. $\endgroup$ – L3ul May 7 '16 at 8:50
  • $\begingroup$ @L3ul the reason behind the standard reduction potentials is the lower electronegativity of iodine compared to chlorine. See my answer below. $\endgroup$ – Thawn Jun 6 '16 at 11:20
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Chlorine has a higher electronegativity than iodine. Therefore, the Cl-O bonds are less stable than the I-O bonds (generally, oxygen reacts better with elements of lower electronegativity). Thus $\ce{ClO3-}$ is comparatively less stable than $\ce{IO3}$ which means that in case of chlorine the equilibrium is closer to $\ce{ClO-}$ while - in comparison - for iodine the equilibrium is closer to $\ce{IO3-}$.

The reason for the lower electronegativity of iodine compared to chlorine is that iodine has two more electron shells than chlorine, meaning that the outer electrons of iodine are further away from the nucleus.

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From the comments, one apparent reason is that the $I-O$ bonds, being weaker than $Cl-O$ bonds, are more labile.

There may also be a steric effect. The +5 oxidation state ions are pyramidal, and we have to "squeeze" multiple oxygen atoms around "one side" of a (positively charged, thus somewhat shrunken) halogen atom.

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