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I noticed that the reaction of ozone with $\ce{I_2}$ gives iodic acid and not periodic acid. Upon searching, I found this question:

Why does iodine get oxidized to iodic acid and not periodic acid by nitric acid?

The answer says, nitric acid isn't strong enough to oxidise it, but my question is, even ozone is also not strong enough? And, is there no other reagent that could oxidise iodine to periodic acid except the electrolytic oxidation?

The reaction is: $$\ce{I_2 + 5 O_3 + H_2O ->2 HIO_3 + 5 O_2}$$

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Ozone is not mentioned in the synthetic techniques reported by Wikipedia, citing Greenwood and Earnshaw[1].

Summary:

  • Orthoperiodate ion, $\ce{IO6^{5-}}$, is obtained by alkaline oxidation of iodate either electrolytically on a lead dioxide anode or chemically with chlorine. The product may then be acidi-fied to $\ce{H5IO6}$. Alkaline conditions favor formation of the oxygen-rich orthoperiodate ion; note the inclusion of hydroxide ions in $\ce{IO3^- + \color{blue}{4 OH^-}->H2IO6^{3-} + H2O +2e^-}$ (the periodate ion is listed partially protonated, as it appears at pH 14 in the Pourbaix diagram from [2]).

  • Metaperiodic acid, $\ce{HIO4}$, may be obtained by dehydration of the ortho acid. This can be accomplished by heating to 100°C (but below 150°C) or by treatment with dilute nitric acid.

Reference

1. Greenwood, N. N.; Earnshaw, A (1997). Chemistry of the elements (2nd ed.). Butterworth-Heinemann. p. 872. doi:10.1016/C2009-0-30414-6. ISBN 978-0-7506-3365-9.

2. Voutchkova, Denitza & Ernstsen, Vibeke & Kristiansen, Søren & Hansen, Birgitte. (2017). "Iodine in major Danish aquifers." Environmental Earth Sciences 76. Doi: 10.1007/s12665-017-6775-6.

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There was a previous question in Chem.SE which discussed the reaction between iodine and ozone. The product was found to be tetraiodine nonoxide, $\ce{I4O9}$. This reaction was performed in gas phase and published in a paper1. The reaction was done in a flow system, with $\ce{N2/O2}$ mixture as carrier gas at a pressure of 100 kPa and temperature range of 293-370 K. It was noted that $\pu{3.9 \pm 0.2}$ molecules of ozone reacted with one molecule of iodine. The reaction was of first order with rate constant, $\pu{ln k = ( 14.7 \pm 0.6) -( 2050 \pm 230)T^{-1} dm^3.mol^-1.s^-1}$.

Reference

  1. Reaction of iodine with ozone in the gas phase, A. C. Vikis and R. MacFarlane, The Journal of Physical Chemistry 1985 89 (5), 812-815, DOI: 10.1021/j100251a019
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    $\begingroup$ I have been taught that the product is $\ce{I4O9}$ when carried out in absence of water, and it forms $\ce{HIO3}$ in presence of water. Is that correct? I feel the OP was asking about the reaction in presence of water. $\endgroup$
    – TRC
    Jun 6, 2021 at 5:44
  • $\begingroup$ @TRC Yes, you are correct. $\endgroup$ Jun 6, 2021 at 7:13

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