Question

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

Answer 1

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

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Let's consider a general polyatomic ion, `"A"_m"B"_n^(cpm)`, containing two distinct elements `"A"` and `"B"`, of ratio `"A":"B" = m/n`, and having charge `""^(cpm)`.

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 least three---but not necessarily three---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 `color(blue)(""^(-2))`, since it is 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, `color(blue)("NO"_3^(-))` is the formula for nitrate, and its charge is `color(blue)(""^(1-))`.

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