Which Lewis structure below has been drawn correctly?
In Lewis structures, we take the atomic symbols of the elements and lay them out with their valence electrons to indicate the nature of the bonding or the lone pairs or radicals. If we were starting from scratch, we count up the total number of valence electrons for our group. We then lay out the symbols and put our nuclei in the right places, generally, with the one that conform the most bonds in the middle. And then join them together using bonding pairs. And then try to use the remaining electrons to fill in the outer shells, using lone pairs, making multiple bonds, or inserting radicals.
We can start by counting the number of valence electrons in each structure. And then we’ll see whether it’s what we would expect based on the elements inside. The first structure involves two oxygens and a carbon. It looks very much like carbon dioxide. Each dot indicates one electron. So for this structure, we have a total of eight electrons. An equivalent Lewis structure has double bonds, double lines, in place of the four dots, the four bonding electrons. Structure B has a carbon in the middle and four satellite hydrogens. It looks very much like a methane molecule. If you count up the electrons carefully, you can see there’s a total of 26.
In the next structure, we have one nitrogen and three fluorines. It’s a structure that looks very like nitrogen trifluoride. In this structure, we have 24 electrons. The next Lewis structure looks like an argon atom. And there are seven valence electrons. And the last structure looks like sulfur dioxide, a sulfur with two oxygens either side. In this structure, we have a total of 18 electrons.
Now, let’s go back to the first structure. Carbon is in group 14, otherwise known as group four. As such, we expected to have four valence electrons to contribute. An oxygen is in group 16. So they should contribute six electrons each. So structure A is actually eight electrons short of what we’d expect. So structure one cannot be a valid Lewis structure. This is how the Lewis structure of carbon dioxide should look like, with double bonds between the carbon and each oxygen and two lone pairs on each oxygen. All three atoms have full valence shells, having eight electrons each.
Now, into structure B, we know that carbon should contribute four electrons. And the hydrogens, being in group one, each contribute one electron. So in this case, we actually have too many electrons. We should only have eight. The correct Lewis structure for methane looks like this, where we have a single bond between the carbon and each hydrogen. The carbon has a full outer shell with eight electrons and the hydrogens have full outer shells with two electrons. So structure B is not correct either.
Nitrogen is in group 15. So we know it has five valence electrons. While fluorine is in group 17. So each fluorine should contribute seven valence electrons. So in this structure, we’re missing two electrons. And this is what we’d have to do to correct the structure. Use a lone pair from one of the fluorines to form a double bond to the nitrogen, allowing the nitrogen to have a full octet. And we’d have to add the two plus charge. Alternatively, if we were looking for the Lewis structure of neutral nitrogen trifluoride, this is how it would look. Single bonds between the nitrogen and each fluorine add all atoms with a full octet.
In the next structure, we have a single argon, which should contribute eight valence electrons, given that argon is in group 18. Instead, we only have seven electrons. So we have an argon ion. This is how an argon atom would look as a Lewis structure with a full octet. This means that none of the first four structures were drawn correctly. But it’s not enough to just tick the last one. Let’s make sure.
Sulfur and oxygen are both in group 16. So they contribute six valence electrons each. That gives us the 18 electrons we’d expect if this configuration were neutral. The next thing to check is whether the electrons are in the right place. The two oxygens on either side have full valence shells, having eight electrons each. However, the sulfur has 10 valence electrons. This is actually okay. Sulfur is in period three. So it has accessible d orbitals that it can use for bonding or for lone pairs. So, while it may look a little strange, this is actually the correct Lewis structure for sulfur dioxide. Meaning that of the five Lewis structures given, the only one drawn correctly is that of sulfur dioxide.