In this example, we draw the Lewis structures for nitrate, assigning formal charges to all atoms.
Our first step is to sum the valence electrons.
Nitrate has one nitrogen atom with five valence electrons and three oxygen atoms with six valence electrons each.
The charge of nitrate is one minus so there's one additional electron.
Adding these together, we have 24 valence electrons.
Next we'll draw the structure with only single bonds.
Nitrogen is the central atom and our three oxygen atoms are peripheral atoms.
Creating these three bonds has used up six electrons, so we have 18 valence electrons left.
Our next step is to satisfy the octet rule by adding one pairs.
We start by satisfying the octet rule for our peripheral atoms, which are the three oxygen atoms.
Each oxygen atom wants six more electrons.
Adding these lone pairs, we've used up all of the remaining 18 electrons.
Notice that nitrogen, our central atom, is not yet satisfied.
Since we've run it out of electrons and not all atoms are satisfied, we need to create multiple bonds.
Nitrogen wants two more electrons, so we can use one lone pair of electrons to create a double bond.
Now all atoms are satisfied.
However, instead of creating the double bond with the oxygen on the right,
we could have created the double bond with the oxygen on the left.
Or we could have created the double bond with the oxygen on the bottom.
These are three resonance structures for nitrate, because they differ only by the distribution of electrons.
The actual nitrate ion does not have one double bond and two single bonds.
Rather, it has three identical bonds somewhere between a single and a double bond.
The actual polyatomic ion is an average of these three resonance structures.
Our last step is to evaluate our resonance structures using formal charge.
Since all three of these resonance structures have the same bonds, we'll focus on assigning formal charges to the atoms in one structure.
To calculate the formal charge on nitrogen,
we take the number of valence electrons in a neutral nitrogen atom
subtract the number of lone pair electrons on nitrogen and the number of bonds with nitrogen.
A neutral nitrogen atom has five valence electrons, there are no lone pair electrons on nitrogen,
and there are four bonds with nitrogen.
So our formal charge on nitrogen is plus one.
Next we'll calculate the formal charge for the oxygen on the right with a double bond.
A neutral oxygen atom has six valence electrons, there are four lone pair electrons on this oxygen, and there are two bonds.
So we have six subtract four and two.
The formal charge on this oxygen is zero.
Last, we'll calculate the formal charge on the other two oxygen atoms, which each have just one single bond.
A neutral oxygen atom has six valence electrons,
there are six lone pair electrons on each of these oxygen atoms, and there's one bond.
So we have six subtract one and six.
The formal charge on each of these oxygen atoms is -1.
Notice that if we add together the formal charge on all of our atoms, we'll get -1, which is the charge on nitrate.
Although we would like more atoms in the polyatomic ion to have zero formal charges,
this is the best Lewis structure with oxygen and nitrogen following the octet rule.