# Formal charges on polyatomic ions

Why is it that sometimes a polyatomic ion's Lewis structure is adjusted to remove the formal charge on the central ion, but sometimes it is not?

For example, in the $\ce{I3-}$ ion (triiodide), the Lewis structure is the following:

This gives a single formal negative charge to the central iodine. In my textbook, it says to remove formal charges from the central atom whenever possible, so couldn't we form a double bond with one iodine and therefore move the formal charge off of the central atom like this?

Comparatively, in $\ce{IO4-}$ (periodate), the Lewis structure is:

There, the single formal negative charge is on one of the oxygens instead of on the central ion, making it a resonance structure.

So basically what I'm asking is why is there a formal charge on the triiodide ion's central atom but not on other ions' central atoms? Like why is it okay that the triiodide ion has a formal charge on the central atom when it could be moved off to create a resonance structure, such as in $\ce{IO4-}$?

• Ouch, all those structures hurt my eyes D=
– Jan
Dec 16, 2017 at 16:08
• Yeah, sorry :( I had to create the second two using Google drawings because I couldn't find an appropriate picture. Getting all of the electrons to line up well is hard! Haha sorry about that Dec 16, 2017 at 16:11
• They hurt because they are so wrong, not because of the way you drew them (I have seen much worse attempts at drawing structures; yours’ are actually legible, rather well-spaced, etc.)
– Jan
Dec 16, 2017 at 16:13

$$\ce{I-I\bond{...}\overset{-}{I} <-> \overset{-}{I}\bond{...}I-I}\tag{1}$$
In periodate, there is no question that oxygen is far more electronegative than iodine and that iodine’s oxidation state of $\mathrm{+VII}$ does not go well with a formal or partial negative charge. Thus, it is only logical for the outer oxygen atoms to bear the negative charge. A better way to draw it would be in adherence to the octet rule, i.e. four $\ce{O-}$ substituents and a central iodine with a $3+$ formal charge.
There is nothing inherently wrong with a formal charge on the central atom, though. Take for example tetrahydridoborate $\ce{BH4-}$, the addition product of borane and hydride. Here, boron should only have three valence electrons but if bonded to four hydrogens it must have four. Thus, it carries a formal negative charge. (In reality, the hydrogens carry the negative charge, though, since they are again more electronegative than boron.)