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

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

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Answer 1 · 2016-05-12T02:15:20

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

Let's consider a general polyatomic ion, $A_{m} B_{n}^{c \pm}$ $"A"_m"B"_n^(cpm)$ , containing two distinct elements $A$ $"A"$ and $B$ $"B"$ , of ratio $A : B = \frac{m}{n}$ $"A":"B" = m/n$ , and having charge $^{c \pm}$ $""^(cpm)$ .

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 $N$ $"N"$ . 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 ( ${ClO}_{3}^{-}$ $"ClO"_3^(-)$ ), perchlorate ( ${ClO}_{4}^{-}$ $"ClO"_4^(-)$ ), sulfate ( ${SO}_{4}^{2 -}$ $"SO"_4^(2-)$ ), carbonate ( ${CO}_{3}^{2 -}$ $"CO"_3^(2-)$ ), 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 $1 s^{2} 2 s^{2} 2 p^{3}$ $1s^2 color(green)(2s^2 2p^3)$ , which means it has $5$ $\mathbf(5)$ valence electrons.
That means it can have a maximum oxidation state of $^{+ 5}$ $\mathbf(""^(+5))$ by losing all $5$ $5$ .

Finally, recall that oxygen has a common oxidation state of $^{- 2}$ $color(blue)(""^(-2))$ , 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:

$5 + N 6 - O_{3}^{-}$ $stackrel(color(green)(5+))("N")stackrel(color(blue)(6-))("O"_3^(-))$

$5 + N 8 - O_{4}^{3 -}}$ $stackrel(color(green)(5+))("N")stackrel(color(blue)(8-))("O"_4^(3-))}$ Higher overall charge
$\Rightarrow$ $=>$ generally more unstable

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

So, ${NO}_{3}^{-}$ $color(blue)("NO"_3^(-))$ is the formula for nitrate, and its charge is $^{1 -}$ $color(blue)(""^(1-))$ .

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

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

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