4
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Possible duplicate of Iodate ion (IO3-). Is it polar or non-polar? $\endgroup$
    – Mithoron
    Commented Feb 5, 2017 at 2:01
  • $\begingroup$ @Mithoron How's that relevant? $\endgroup$
    – aiag
    Commented Feb 5, 2017 at 2:02
  • $\begingroup$ Read precisely question and answers (at least Jan's answer) and see. $\endgroup$
    – Mithoron
    Commented Feb 5, 2017 at 2:11
  • $\begingroup$ @Mithoron My question is quite different. I do not care about the shape and polarity, only of how I must follow the school's algorithm. Did you actually read my question? I am going through the course where sulphite ion has 3 single bonds with O, so I really have to put up with weird stuff. $\endgroup$
    – aiag
    Commented Feb 5, 2017 at 2:20
  • $\begingroup$ The algorithm is naturally as flawed as the Lewis-theory for these kinds of molecules and one structure alone will never sufficiently describe the bonding situation. However, in this particular case, the description of all single bonds is the closest to the actual bonding. The so-called octet expansion through the involvement of d-orbitals has been disproven many times. Most of the times the octet rule prevails. Similarly 'sulphite ion has 3 single bonds with O' is correct and not at all weird. Double bonds in those compounds are weird. $\endgroup$
    – Martin - マーチン ♦
    Commented Feb 5, 2017 at 9:57

1 Answer 1

Reset to default
-1
$\begingroup$

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.

enter image description here

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.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ The structure is in no way problematic; instead it is the most correct. $\endgroup$
    – Jan
    Commented Feb 13, 2017 at 0:22
  • $\begingroup$ @Jan The idea of resonance structures in this sort of thing is entirely incorrect then? $\endgroup$
    – Jason Patterson
    Commented Feb 13, 2017 at 5:23
  • $\begingroup$ Resonance structures would imply octet expansion which in turn would imply iodine d-orbital participation. We are talking about the 5d orbitals for an element of the fifth period, i.e. the other relevant orbitals are 5s and 5p. (4d are fully populated and function as core orbitals.) But these are energetically far away; even 6s has less energy than 5d (yet nobody ever suggests any $\mathrm{s^2p^3}$ hybridisation). Look around a few answers by Martin to see the d orbital participation theory disproven. $\endgroup$
    – Jan
    Commented Feb 13, 2017 at 21:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.