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I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $\ce {O-I}$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $\ce {I}$ atom has an oxidation state of -2.

However, I am slightly confused as to why various sources assign the oxidation state of $\ce {I}$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $\ce { I (2.66)}$ is indeed slightly more electronegative than $\ce {C (2.55)}$. Based on this premise, we would reasonably assign the $\ce {I}$ atom in IBX with a +3 oxidation state as the pair of electrons in $\ce {C-I}$ bond would go to $\ce {I}$ rather than to $\ce {C}$.

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Image taken from here

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    $\begingroup$ Yeah, it should be +3. $\endgroup$
    – Mithoron
    Sep 27, 2019 at 14:34
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    $\begingroup$ @Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $\ce {PhI(OAc)2}$, why is it +3 and not +1? It seems that the $\ce {C-I}$ bond is treated rather differently by these source. $\endgroup$ Sep 27, 2019 at 15:09
  • $\begingroup$ Turned out @Mithoron was right after all. I'm going over yo Space Exploration where things are more down to Earth. $\endgroup$ Sep 28, 2019 at 0:02

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This is a written reply from Professor Zhdankin*, author of Hypervalent Iodine Chemistry:

Indeed, many literature sources classify pentavalent iodine compounds as iodine in the oxidation state +5. It makes sense because the electronegativity of iodine is almost the same as carbon, or even less than carbon. According to different studies, electronegativity of iodine varies from 2.21 [A. L. Allred and E.G. Rochow, J. Inorg. Nucl. Chem., 5, 264 (1958).] to 2.66 [A. L. Allred , J. Inorg. Nucl. Chem., 17, 215 (1961)]. The original Pauling's value of iodine electronegativity is 2.5, exactly the same as carbon. This terminology reflects the similarity of iodine in IBX to iodine in $\ce {I2O5}$ or $\ce {HIO3}$ by chemical and structural properties.

Personally, I prefer not to use the "oxidation state" terminology regarding organoiodine compounds. Pentavalent iodine is a perfectly correct classification for IBX.

*He has given his permission for his reply to be shared here.

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