Is it possible to use the school supplied algorithm to build the Lewis diagram of the iodate ion? 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?
When I follow your school's algorithm for iodate I get a central iodine atom with three single bonds to oxygen.  I've shown my work below.  This is problematic, as you well know, but I can potentially see where your teacher is coming from.  



When I teach formal charge and resonance structures and the like, I often have students use their intro chemistry rules to determine the structure of a polyatomic ion such as this.  If they follow the rules they get a pretty structure that looks proper - I then introduce the idea of formal charge by pointing out the wild imbalance of charges in the ion and how energetically unfavorable that would be, then we work out a way to rearrange the electrons to minimize that problem.

I also have a lot of students who immediately go to google to find Lewis structures rather than drawing them themselves, so when a student tells me that they think the structure should be something different, it takes an actual argument from them before I believe that they know why that should be.  I don't know that that is the case here; it also seems entirely possible that your instructor is just wrong.