Break bonds to give stable molecules or ions. Redistribution of valence electrons within the cyclic periodate gives \(\mathrm{HIO}_{3}\) and two carbomyl groups. A result of this electron redistribution is an oxidation of the organic component and a reduction of the iodine-containing component. \(\mathrm{HUO}_{3}\) (lodic acid)

First, look at the structure of a typical cyclic periodate ion and lodic acid. A cyclic periodate ion has the basic formula, \(\mathrm{IO}_{4}\), and contains iodine (I) and oxygen (O) atoms. Lodic acid (\(\mathrm{HUO}_{3}\)) is an inorganic compound, featuring hydrogen (H), iodine (I) and oxygen (O) atoms.

Now we need to break the bonds of the molecules and redistribute the valence electrons. This process will result in an oxidation of the organic component and a reduction of the iodine-containing component. The cyclic periodate ion undergoes cleavage of the I-O bonds and rearrangement, resulting in new molecules.

By following the electron redistribution after breaking the bonds, we can identify the stable molecules or ions formed. In this case, the electron redistribution results in the formation of a \(\mathrm{HIO}_{3}\) and two carbomyl groups.

We know that the electron redistribution causes oxidation of the organic component and reduction of the iodine-containing component in the cyclic periodate. In our case, the oxidation of the organic component happens when the carbomyl groups are formed. The reduction of the iodine-containing component results in the formation of iodate ion (\(\mathrm{IO}_{3}^{-}\)). Furthermore, the lodic acid (\(\mathrm{HUO}_{3}\)) is an inorganic compound that does not undergo oxidation or reduction in this process. To summarize, the electron redistribution in the cyclic periodate results in the formation of \(\mathrm{HIO}_{3}\) and two carbomyl groups, following the oxidation of the organic component and reduction of the iodine-containing component. Lodic acid (\(\mathrm{HUO}_{3}\)) is an inorganic compound that remains unchanged in this process.