# Molecular structure of iodine nonoxide

A question in an exam was as follows:

Iodine reacts with ozone gas to form a dark yellow solid $$\ce{X}.$$ Let the number of lone pair of electrons in un-ionised form of $$\ce{X}$$ be $$m,$$ number of lone pair of electrons in the anionic moiety of $$\ce{X}$$ be $$n$$ and the positive charge on the cationic moiety of $$\ce{X}$$ be $$p$$ units. Then what is the value of $$\displaystyle\frac{m - p}{n}?$$

This reaction of iodine with ozone is:

$$\ce{I2 + O3 -> I4O9 <=>I^3+(IO3^-)3}$$

Therefore the anionic moiety is $$\ce{IO3-}$$ and cationic moiety is $$\ce{I^3+}$$. However the first part of the question states un-ionised form. My assumption is that this is $$\ce{I4O9}$$ and they ask for the number of lone pairs on the molecular form.

The answer to this question takes into consideration that $$\ce{I4O9}$$ is an equimolar mixture of $$\ce{I2O4}$$ and $$\ce{I2O5}$$ which gives the answer to be $$2.5$$

However, from the abstract of J. Raman Spectrosc. 1985, 16 (6), 424–426:

The Raman spectrum of $$\ce{I4O9},$$ formed by the gas‐phase reaction of $$\ce{I2}$$ with $$\ce{O3}$$, has been measured. Freshly prepared samples of $$\ce{I4O9}$$ gave broad band spectra characteristic of an amorphous solid. Vibration bands at $$780,$$ $$740,$$ $$620$$ and $$\pu{450 cm−1}$$ were observed. It was established conclusively that $$\ce{I4O9}$$ is a distinct molecular species and not a mixture of $$\ce{I2O5}$$ and $$\ce{I2O4}.$$

If $$\ce{I4O9}$$ is a distinct molecular species, what is the molecular structure of $$\ce{I4O9}?$$

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

Tetraiodine nonoxide ($$\ce{I4O9}$$) has been prepared by the gas-phase reaction of $$\ce{I2}$$ with $$\ce{O3}$$ but has not been extensively studied.
However, looking at the reaction $$\ce{I4O9 <=> I^3+(IO3^-)3}$$ allows us to venture a good guess at the structure, which I have drawn below:
The central atom's orbitals are $$\mathrm{sp^3d}$$ hybridized and the molecule is expected to adopt a T-shaped structure: