# How could you figure out how many oxygens a polyatomic ion has?

##### 1 Answer

This of course depends on the polyatomic ion you choose, but...

Let's consider a general **polyatomic ion**, *distinct* elements *charge*

Suppose we wanted to look at the **nitrate** ion. I won't tell you the formula, and we'll figure out what its formula is.

Consider breaking down the word into the **"nitr"** stem and **"-ate"** suffix.

- "nitr" is the stem for "nitrogen", whose
**chemical symbol**is#"N"# . So, we have one of the elements figured out. - The "-ate" suffix implies that the polyatomic ion has
*more than one oxygen*. In fact, it must have at least**three**.

This is just a conclusion based on relating back to other known "-ate" polyatomic ions, such as chlorate (

#"ClO"_3^(-)# ), perchlorate (#"ClO"_4^(-)# ), sulfate (#"SO"_4^(2-)# ), carbonate (#"CO"_3^(2-)# ), and so on. These all have at leastthree---but notnecessarilythree---oxygens.

Now, to have a clue as to how *many* oxygens makes sense, let's think about the **valency** of nitrogen.

- Its
**electron configuration**is#1s^2 color(green)(2s^2 2p^3)# , which means it has#\mathbf(5)# valence electrons. - That means it can have a
*maximum*oxidation state of#\mathbf(""^(+5))# by losing all#5# .

Finally, recall that oxygen has a *common oxidation state* of **two** columns away from the noble gases.

So, we really only have one or two possibilities for nitrate, where the *total* oxidation states are marked atop:

#stackrel(color(green)(5+))("N")stackrel(color(blue)(6-))("O"_3^(-))# #stackrel(color(green)(5+))("N")stackrel(color(blue)(8-))("O"_4^(3-))}# Higher overall charge

#=># generally more unstable

The first possibility is reasonable, given that it has the *lower* overall charge.

So, **nitrate**, and its **charge** is

(The second is very unstable, though it's real.)