I am upgrading the grade 12 chemistry credit and have to use the school supplied algorithm to build the Lewis diagram of $\ce{IO3-}$ ion. It seems to me that iodine oxidation state in this ion should be 5, therefore it would have one single and two double bonds with oxygen. When I follow the steps below, I am arriving at a different formula.
I do not need to answer whether the ion is polar or not and what is its shape.
Step 1: Count all of the valence electrons in the molecule or ion. In the case of an ion, add or subtract electrons to account for the ionic charge.
$$\ce{I = 7e^-, O = 18e^-}; \text{ionic charge} = \ce{1e^-}; \text{total} = \ce{26e^-}$$
Step 2: Arrange the peripheral atoms symmetrically around the central atom. Use a pair oil electrons to form a bond that links these atoms to the central atom.
6 electrons placed on $\ce{I}$, $\ce{26e^- - 6e^- = 20e^-}$ remain
Step 3: Add pairs of electrons to complete the octet of the peripheral atoms.
$3\cdot 6$ electrons placed on $\ce{O}$, $\ce{20e^- - 18e^- = 2e^-}$ remain
Step 4: Place any unassigned electrons on the central atom.
2 electrons placed on $\ce{I}$, $\ce{2e^- - 2e^- = 0e^-}$ remain
Step 5: If the octet of the central atom is incomplete, move a lone pair of electrons from a peripheral atom to a new position between the central and peripheral atom.
After completing step 4 of the rules above I am arriving at $\ce{I}$ having 8 electron dots. At this point all electrons have been used up, the octet of $\ce{I}$ is complete, and I can only create a structural diagram with two single and one double bonds.
In the course I have taken previously we were taught that $\ce{I}$ was an exception to the octet rule, but when I asked this teacher, I was told that I was not allowed to use the exceptions and had to apply the octets only.
Am I missing anything? Is it possible to comply with the algorithm above and come up with the correct answer?
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