Balancing disproportionation reaction of hypoiodous acid

In aqueous solution, the acid $\ce{HIO}$ disproportionates according to the following equation where m, n, p and q are simple whole numbers in their lowest ratios.


$$\ce{mHIO → nI2 + pHIO3 + qH2O}$$


This equation can be balanced using oxidation numbers.


In $\ce{HIO}$ and $\ce{HIO3}$, the oxidation state of iodine is +1 and +5 respectively. Therefore in $\ce{HIO}$, iodine lost 4 electrons to iodine atoms, therefore n is 2 and finally p is 1. However, when the electrons are received by the iodine atoms, they should form iodide ions instead of molecules. How would you tackle this question?




The disproportionation rection of $\ce{HIO}$ cannot produce $\ce{HIO3}$ and $\ce{HI}$ (or $\ce{I^-}$ ions) in acidic solution. The reason is that $\ce{I^-}$ and $\ce{IO3^-}$ ions do react with one another, if $\ce{H^+}$ ions are present, to give : $$\ce{IO3^- + 5 I^- + 6 H^+ -> 3 I2 + 3 H2O}$$ If no ion $\ce{H^+}$ is present, the previous reaction does not occur, and the solution contains simultaneously the ions $\ce{I-}$ and $\ce{IO3^-}$


 2