# Is it possible for the iodine atom in periodate ester to be octavalent?

Recently, I was reading Clayden's Organic Chemistry (2nd ed.) and I chanced upon this diagram on p. 443 (as shown below) showing an octavalent iodine in the periodate ester. I found this rather puzzling as I thought that the maximum possible valency of iodine was seven. I am well aware that bonding schemes such as the 3-centre 4-electron bonding scheme allow central atoms to make more than the number we would expect the atom to make conventionally. However, this is the first time I am seeing octavalent iodine and thus, I would just like to confirm if it is in fact, possible for iodine to be octavalent.

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

I would suggest that a reactive intermediate/transition state was formed, and not an actual compound. I'd say both $\ce{C-O}$ bonds are partially formed in the intermediate
The diagram in the book does not show where is the $\ce{Na+}$ during the reaction (see above picture), probably for the convenience (but therefore, somewhat misleading). Remember, a periodate ester is an ester of periodic acid. The most common periodate esters are cyclic periodate esters, form as the intermediate in oxidation of vicinal diols such as sugars and glycols. For example: The oxidative cleavage with periodic acid has been successfully applied in the constitutional analysis of sugars (see: Chemgapedia article).