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

May 12, 2016

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

Let's consider a general polyatomic ion, ${\text{A"_m"B}}_{n}^{c \pm}$, containing two distinct elements $\text{A}$ and $\text{B}$, of ratio $\text{A":"B} = \frac{m}{n}$, 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 $\text{N}$. 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 (${\text{ClO}}_{3}^{-}$), perchlorate (${\text{ClO}}_{4}^{-}$), sulfate (${\text{SO}}_{4}^{2 -}$), carbonate (${\text{CO}}_{3}^{2 -}$), 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 $1 {s}^{2} \textcolor{g r e e n}{2 {s}^{2} 2 {p}^{3}}$, which means it has $\setminus m a t h b f \left(5\right)$ 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:

• $\stackrel{\textcolor{g r e e n}{5 +}}{{\text{N")stackrel(color(blue)(6-))("O}}_{3}^{-}}$
• stackrel(color(green)(5+))("N")stackrel(color(blue)(8-))("O"_4^(3-))} Higher overall charge
$\implies$ generally more unstable

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

So, $\textcolor{b l u e}{{\text{NO}}_{3}^{-}}$ is the formula for nitrate, and its charge is color(blue)(""^(1-)).

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