# The teacher insists that iodine oxidation number in iodate ion is 7

What started as this question http://chemistry.stackexchange.com/questions/67756/is-it-possible-to-use-the-school-supplied-algorithm-to-build-the-lewis-diagram-o continued with an email from the teacher, where they confirmed that the octet rule has to be followed and also mentioned that the oxidation number of iodine in iodate is 7, not 5 as I believe.

My thinking is based on the fact that oxygen is always 2, therefore iodine has 3*2-1=5. Two double bonds and one single bond with oxygen are present in iodate ion. Iodine has oxidation number 7 in periodate ion as 4*2-1=7. Where is my error?

EDIT:

More back and forth with the teacher, and they pointed me at an example of how they think $\ce{[ClO3]-}$ should look in a Lewis diagram: According to the teacher, both chlorate they have in the course and iodate in the assignment would have to follow the exact same steps when building Lewis diagram.

I have an issue with this methodology, as two $O$ are somehow missing their 2nd bonds. This is absurd and does not jive with any of the textbooks I studied. How should I explain this to the teacher, who does not seem to understand basic arithmetic?

• Another example of "alternate facts"... // PS - This isn't worth getting in a argument with your teacher - especially a public one. – MaxW Feb 9 '17 at 5:34
• Your extension was a completely separate question that does not touch the original; I have therefore rolled it back. If you wish to discuss how the full Lewis structure of chlorate (or iodate) you should ask it separately — but note that there are almost certainly questions (and answers) on how to draw it on this site already, so please don’t forget to search. – Jan Feb 10 '17 at 1:09
• Lewis diagrams usually use different symbols for the electrons from different atoms so that you can keep track of which atom the electrons came from. Like solid dots and circles. – MaxW Feb 10 '17 at 23:04
• Do you want to be right or do you want to be happy? Getting into a brewhaha with the teacher isn't necessarily the best way to proceed. – MaxW Feb 10 '17 at 23:06
• Smart comment, but what I want is a guarantee that the teacher cannot arbitrarily deduct points. If I commit to the wrong process used in the course, there is no such guarantee. Makes sense? – aiag Feb 11 '17 at 2:16

## 2 Answers

Plus 5 according to: https://en.wikipedia.org/wiki/Iodate

Teachers do sometimes make mistakes.

• I would be very hesitant to quote Wikipedia, as it is a resource, which can be edited by anyone with any agenda at any moment. But I do agree that there is no way in hell I could have OS=7 in iodate. Only in periodate. – ajeh Feb 8 '17 at 17:58
• @ajeh. Quote me then! I added the reference because some users seem to demand them, especially on the Biology S.E. Here's another reference but you'll have to parse it out of a reaction: books.google.com/… – bpedit Feb 8 '17 at 18:16
• @bpedit The link says that the page is unavailable for viewing or I've reached the limit for this book. But I did find a confirmation of OS=5 for iodates in several textbooks, including f.ex. Modern Inorganic Chemistry by Chambers and Holliday, p. 260 – aiag Feb 9 '17 at 0:30
• @bpedit Can you please check the link you posted as I noted above, it does not open? Thx! – aiag Feb 10 '17 at 1:03
• I don't think there's a particular need for a source for this. It's quite straightforward and if anything, an explanation of how the oxidation state is arrived at will be much more useful. Although - to be fair - OP's original question already contains the correct reasoning. – orthocresol Feb 11 '17 at 1:07

There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:

1. How many valence electrons do we have? (add up)

For both chlorate and iodate the calculation is identical:

$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$

2. How many valence electrons are required for full octets/hydrogen dublets?

This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.

$$4\times 8 = 32\tag{2}$$

3. How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$

$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$

4. How many electrons are remaining? These will form lone pairs. $(1) - (3)$

$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$

Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{X-O}$ bonds ($\ce{X} = \ce{Cl, I}$) are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure: (Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image.)

Note that this diagram is that your teacher used, too, except for explicitly noting the formal charges of both iodine and oxygen. If you want to simplify, you can draw square brackets around the molecule, write the total charge next to the closing square bracket and ignore the distribution of formal charges — that is valid (and what your teacher did). I am sorry to say that they are correct in that regard — and since they admitted their error concerning the oxidation state of iodine in iodate, they are fully back on track.

• @Greg This question has moved so much since its existence … – Jan Jun 13 '17 at 13:31