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:
$$\ce{I2 + O3 -&gt; I4O9 &lt;=&gt;I^3+(IO3^-)3}$$
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 $\ce{IO3-}$.
$$\ce{3I4O9 + 12OH- -&gt; I- + 11IO3- + 6H2O}$$
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

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