# How would you draw all the resonance structures for nitrate, NO3-?

Oct 29, 2015 #### Explanation:

First is you need to know the number of valence electrons. You can do this by (1) drawing the electron configuration per element or (2) consulting your periodic table.

If you chose to draw the electron configuration per element, you will have something like this:

$N$ (atomic number = 7) : $1 {s}^{2}$ $\textcolor{red}{2 {s}^{2}}$ $\textcolor{red}{2 {p}^{3}}$ (5 outermost electrons)

$O$ (atomic number = 8) : $1 {s}^{2}$ $\textcolor{red}{2 {s}^{2}}$ $\textcolor{red}{2 {p}^{4}}$ (6 outermost electrons)

If you chose to consult your periodic table, just notice that $N$ belong to Group 5A while $O$ belongs to Group 6A. The group number denotes the number of outermost electrons. Now that you know the number of valence electrons per element, you need to compute the total valence electrons for the $N {O}_{3}^{\text{-1}}$ ion.

5 + (3 x 6) = 23 electrons

But since the whole molecule has a -1 charge, you need to add this too. So the total number of valence electrons is 24.

The next thing to do is draw. Normally, the first element in the chemical formula is the central atom. In this case, the $N$ atom is being surrounded by three $O$ atoms. Notice that the lone pair of electrons from $O$ are also included in the diagram. If you count the total number of electrons on the above drawing, it is already 24 electrons (single line counts as 2 electrons). But this drawing has violated the octet rule since the central atom only has 6 electrons instead of the correct number 8. So what to do?  All the resonance structures are correct since it all follows the octet rule and all have a total number of 24 electrons. Just take note that the only bond moving is the pi ($\pi$) bond or in layman's term, the double bond and one of the electron pairs from $O$ atom.