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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