Why is the hypoiodite anion unstable in solution, whereas hypochlorite is relatively stable?

$\ce{Cl2}$ 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?




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](http://www.lenntech.com/periodic/water/iodine/iodine-and-water.htm#ixzz47wLtzV81) for more details. 


For the disproportionation reaction, consider the [Latimer diagram](https://en.wikipedia.org/wiki/Latimer_diagram) for iodine:-


[![enter image description here](https://i.stack.imgur.com/qry3G.png)](https://i.stack.imgur.com/qry3G.png)


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.


 1

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.


 1

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.


 0