Lewis Structures

Since the MCAT is a multiple-choice exam we don’t have to know exactly how to draw out Lewis Dot Structures. We will walk through the process but ultimately we need to be able to recognize correct and incorrect structures. From those structures, we should then be able to determine the formal charge, oxidation number, resonance structures, and free radicals.

Creating a Lewis Dot Structure

To create a Lewis Dot Structure we will need to have our periodic tables handy. We will see how this whole process works by using NO3

Step 1: Count Up Valence Electrons

First up we begin by counting the total number of valence electrons each atom has and add them all together. To do this simply count across the periodic table skipping the transition metals. Here we can see that nitrogen has 5 valence electrons and each oxygen will have 6. For a total of 23 valence electrons.

Step 2: Adjust The Electron Count for Charge

Up next we will need to adjust our total electron count for the overall charge of the molecule. Since NO3 has an overall negative charge we need to add one valence electron to our total for a final number of 24 valence electrons. If this were NO3+ we would subtract one valence electron from the total leaving 22 valence electrons to arrange.


In most cases, the total valence number of a molecule will be even after balancing out the electrons for charge. In rare circumstances, however, the electron count will be odd. When this occurs the molecule has an unpaired electron and we call it a free radical. For example, NO2, NO, and ClO are all free radicals.

Free radicals are exceptionally reactive and cause all kinds of havoc on biological organisms. This is why our bodies utilize antioxidants that freely give up a single electron protecting our cells and tissues from damage.

If you happen to get a question about radicals and Lewis structures you don’t need to draw them out. Instead, look out for an odd number of valence electrons since you can’t make pairs of two out of odd numbers. As a result, there will always be a lone electron lurking around the molecule somewhere.

Step 3: Arrange Your Atoms

Third, we will want to arrange all of our atoms such that we have one central atom. To determine the central atom we want to find the element that can form the most bonds. These are either elements that can expand their octets such as sulfur or elements in groups 3A, 4A, and 5A. In that case, this means that nitrogen is the central atom.

Step 4: Create Bonds

Now we will add electrons to form bonds. To begin always form single bonds first then check to see if the central atom is ” happy” with the number of bonds it has. To determine the number of bonds an element wants to form we can count backward starting at the halogens. So oxygen would like to form two bonds and nitrogen would like to form three. In this case, we will start by forming three single bonds to satisfy the central nitrogen.

Step 5: Assign Lone Pairs and Make Bonds

After we have satisfied the number of bonds our central atom wants to form we add the remaining electrons as lone pairs to the most electronegative atoms first until they have a full octet. This means that they will have a total of 4 electron lone pairs or bonds when we sum both together.

We can see here that the top oxygen has one bond and three lone pairs which sum to four meaning that the top oxygen’s octet is completely filled. In contrast, the rightmost oxygen has two lone pairs and one bond for a total of 3 things surrounding it making its octet unfilled.

In our example, we used 6 of our 24 electrons by forming bonds with other molecules so we still need to assign the remaining 18 electrons as lone pairs.

Now all of our oxygens have completely filled octets, but our nitrogen is still unfilled. We want to try and fill everyone’s octet though, however, we are all out of electrons so we can’t assign it any lone pairs. Instead, we will need to start converting lone pairs into bonds until nitrogen’s octet is full. Since nitrogen is missing one of its four lone pairs or bonds all we have to do is take one of the lone pairs from oxygen and turn it into a bond.

Now each atom in the molecule has a complete octet and we are almost finished.

Step 6: Check Formal Charge

Before we give ourselves high fives for creating the NO3 lewis structure we need to make sure it is a decent representation. To do this we will calculate the formal charge of each atom within our structure. In order to do this we use the following equation:

formal\;charge = #\; of\; valence\; electrons\; – bonds – electrons\; in \; lone \; pairs
formal\;charge = #\; of\; valence\; electrons\; – surrounding\; things

To illustrate how this works let’s calculate the formal charge of the atoms in NO3.

Now try working the formal charge calculations for the rest making sure that your answer matches the number next to the atom.

Our goal is to make sure that our overall formal charge matches the molecule’s formal charge and that the number of atoms with a formal charge other than zero is minimized. We can see that when we sum our current structure’s formal charge they all add to -1 which is exactly what we wanted.

Additionally, three out of the four atoms have a formal charge other than zero. Let’s see if we can do better. To do this we will convert a bond into a lone pair. Just remember that we can’t violate any element’s octet when we do this. For example, we can’t form 5 bonds to the central nitrogen since 4 bonds fully fills its octet.

Here we can see that we haven’t violated any atoms octet and we have increased the formal charges of several elements in the process. This means that this lewis structure is more unstable and therefore the incorrect Lewis Structure.


Couldn’t we switch around the double-bonded oxygen though? Of course, we could and each Lewis Structure would be equally valid. In this scenario, we would say that NO3 has resonance. Meaning the electrons can be arranged in multiple valid ways amongst different bonds and elements. It turns out that no one resonance structure exists out in the real world, but instead molecules exist as a hybrid of all of the different resonance structures that are possible.

We call this structure that actually exists the resonance hybrid and it can affect the structural properties of a molecule. For example, the peptide bonds that hold proteins together have resonance. This gives the bond holding amino acids together partial double bond character and as a result, they end up being fairly rigid and unable to rotate, unlike other single bonds.

Here the peptide bond is shown in green with the dotted lines demonstrating resonance between the carbon that is attached to an oxygen and nitrogen.

Oxidation Number

To make matters more confusing we can also calculate an oxidation number which is similar to, but slightly different from formal charge. We ultimately concluded that formal charge is a measure of how close an atom is to its ideal number of bonds. In contrast, an atom’s oxidation number is a measure of how many electrons an atom gets on the basis of its electronegativity.

When calculating or considering the oxidation number we will always give all of the bonding electrons to the most electronegative atom. In the case of NO3 all of the electrons in the oxygen-nitrogen bonds will go to the oxygen atoms since it is farther to the left in the periodic table and thus more electronegative.

Again we have a formula for calculating the oxidation number of different atoms that starts with the valence electrons as before.
Oxidation \;#= Valence\; Electrons – Total \; Electrons

Therefore the oxidation number of each oxygen atom would be -2 and the nitrogen would have an oxidation number of +5. Again the sum of all of the oxidation number will equal our overall charge of -1.

\[ Overall\; Charge = 3(-2)+5 \ to -1\]

Many elements have pretty consistent oxidation numbers and by memorizing them we can more quickly calculate the oxidation number of different elements within a molecule. While we could memorize the different oxidation numbers in list or table form it is far easier to memorize them from the periodic table instead.

First off, the only elements with consistent oxidation numbers are those found on the far right and far left of the periodic table excluding the noble gases. So the alkali metals, alkaline earth metals, oxygen group, and halogens are the only ones that we should have memorized.

Additionally, the electronegativity increases as we go from the left of the periodic table to the right so the elements on the left side of the periodic table are going to have positive oxidation numbers since they don’t have a strong grip on their electrons. In contrast, the elements on the right side of the periodic table will have negative oxidation numbers because they are really good at stealing electrons.

Spotting Correct Structures

As I discussed previously we don’t absolutely need to know how to draw Lewis structures. Instead, it is usually enough if we can spot correct and incorrect structures. To do this we will look to see if the following pieces of information are true.

  • The number of electrons in the Lewis structure matches the number of predicted valence electrons
  • No atom in the structure has a violated duet or octet unless it is an atom that is able to expand its octet
  • The formal charges are at their minimum (check this last since you don’t always have to use this step)

Getting comfortable with Lewis Structures is a matter of practice and repetition work through this brief problem set to make sure you understand how to approach the problems.