# Chemistry - Molecular structure of iodine nonoxide

Regarding its ionic form, I don' think it is as simple as $$\ce{I^3+(IO3^-)3}$$. It is much more complex than that. From Ref.1:

This compound is likewise to be considered as an $$\ce{I(III,V)}$$ oxide and reacts with alkali hydroxide to give $$\ce{I-}$$ and $$\ce{IO3-}$$.

$$\ce{3I4O9 + 12OH- -> I- + 11IO3- + 6H2O}$$

Structurally, $$\ce{I4O9}$$ is possibly an iodate $$\ce{I3O6+IO3-}$$ (more precisely $$\ce{(I3O6+)_n.nIO3-}$$) in which the isopolycation $$\ce{I3O6+}$$ has a polymeric structure and is formulated as $$\ce{I_^{III}(I^{V}O3)2^+}$$ consisting of twice as many pyramidal $$\ce{I^{V}O3}$$ groups as square-planar $$\ce{I^{III}O4}$$ groups thus $$\ce{I4O9}$$ would correspond to $$\ce{I(IO3)2^+IO3-}$$ which would become $$\ce{I(IO3)3}$$.

Reference

1. Inorganic Chemistry By Egon Wiberg, A. F. Holleman, Nils Wiberg, Academic Press, 2001