Draw a Lewis structure for each of the following:
d. NH

Question

Draw a Lewis structure for each of the following:

d. NH₂O⁻

Accepted Answer

Welcome back everybody. Our next problem says provide the lowest structure of the given NH 40 plus. So we have our molecule here. It has a nitrogen, four hydrogens and oxygen and an overall positive one charge. So what are the steps we go through to try and come up with the best Lewis structure? Well, first of all, we're going to look at the total number of valence electrons in the entire molecule. So total valence electrons, I'll put e negative with nitrogen, nitrogen has five valence electrons. So let's just put a little five up here. Hydrogen has one and oxygen has six. So for the total valence electrons, we have one nitrogen. So five, we have four hydrogens. So four multiplied by one plus six valence electrons from the oxygen. So we have five plus four plus 615 from our atoms, but we have this positive one charge. So we have to add one, excuse me, positive in charge, but that means we're missing an electron. So let's note that because that's important, missing one electron, it's easy to mix those things up as I just did. So we don't add one. It's not a total charge, it's a number of electrons. So subtract one since missing one. So we have 14 valence electrons. So we know that overall, when we finish our structure, it needs to have 14 valence electrons. So the next step is we draw a sigma bond framework. So posit how these might be connected by single bonds and then we modify it based on how many valence electrons we have. And the stability of certain arrangements, we know that hydrogen will only form one bond. We have four hydrogens, one nitrogen and one oxygen. So let's put the nitrogen and oxygen together with a single bond. So no, since hydrogens can't carry more than one bond, we wouldn't end up with any hydrogens being internal. So we know the nitrogen and oxygen must be bound to each other. So we could have a structure where say there are three hydrogens on the nitrogen and one hydrogen on the oxygen. So it's dry. Ha we could also try a structure where nitrogen bond to oxygen where each of these has two hydrogens. Now, what about a structure where nitrogen has one oxygen and nitrogen has one hydrogen and oxygen has three. Well, hopefully, we remember that oxygen is usually most comfortable with two bonds. We know there are compounds where nitrogen takes three hydrogens as well as 1/4 bond. But the idea of oxygen taking four bonds is just very unusual unless we have some reason where we'd need that as a potential structure, we won't worry about that one. So let's evaluate these two. So first, let's look at the total valence electrons in each of these structures. Does it match what we need? So, from structure one, with the three hydrogens on nitrogen, let's count the number of valence electrons. So first, we need to know how many are here based on the sigma bonds. And that will let us know if there's any lone pairs that need to be placed. Keeping in mind that the whole molecule should have a positive one charge. So I have first count the total number of bonds. So I have 12345 bonds. So five bonds multiplied by two electrons equals 10 electrons in bonds are 10 bonding electrons and we need a total of 14. So that means we need more electrons that are left over. So we'll need two lone pairs. Well, nitrogen isn't going to take on any lone pairs. It's already got four bonds, but here's our oxygen with two bonds and it's quite happy to take two lone pairs to give us our four leftover electrons. So there we have 14 valence electrons do we have a positive one charge? Let's calculate the formal charges here. All these terminal hydrogens, four of them have just one bond and one valence electron. So they have a zero formal charge. Let's look at nitrogen. It's formal charge fc equals it's a number of valence electrons So five minus, it's a number of non bonding electrons, which is zero, no lone pairs here minus this number of bonds, which is four. So we get a positive one charge on the nitrogen. So we'll just put a plus by our nitrogen. In the oxygen. In this case, it's formal charge equals, we know oxygen has six valence electrons minus, there's two lone pairs that makes four non bonding electrons and it has two bonds. So minus two. So we have six minus four minus two equals zero. So oxygen carries no formal charge here. So one, that's one potential of structure with a positive one charge on the nitrogen and two lone pairs on the oxygen. Now, let's say our other lewis structure works out. This would be the one with nitrogen and oxygen bonded in the middle. And each of those carried two hydrogens. So first of all, the total valence electrons, we have 12345 bonds, five bonds multiplied by two electrons in each bond equals 10 electrons just like in our other structure. That means we have four electrons left over which will equal two lone pairs. Looking at our atoms here, nitrogen and oxygen each have three bonds each. So let's put a pair of lone pairs on each of those. So lone pair of nitrogen, a lone pair on oxygen. So now we have 10 bonds or 10 electrons and bonds plus four electrons giving us 14 valence electrons for the molecule, which is correct. So now let's just figure out the formal charges the hydron all are terminal with one bond. So they have no charge nitrogen. Its formal charge equals five valence electrons. It has one known pair. So minus two non bonding, there's three bonds, so minus three bonds, so five minus two minus three is zero. So no formal charge on this nitrogen. Then for oxygen, it's formal charge, there are six valence electrons minus there's a single lone pair. So minus two non bonding, three bonds. So minus three, we have six minus two minus three leaving us a formal charge of positive one. So in this structure, the positive charge is on the nitrogen. So now we've got two structures that have the right number of valence electrons, which one is going to be our correct lowest structure. Well, the difference is the first one has a positive charge on nitrogen. The second one has a positive charge on oxygen. Well, the first one's going to be our structure because it will be more stable. And that's because nitrogen is less electron than oxygen. So it's going to be more comfortable with a positive charge. So we'll say oxygen is more electron and therefore less stable with a positive charge. So with that in mind, our lowest structure of our given ion NH 40 plus has nitrogen with three hydrogens, a positive charge in a bond to an oxygen that oxygen has two lump pairs and it's bound to a hydrogen. So again, that means that the lone pair, excuse me, the positive charge is on nitrogen. In this case, a more stable arrangement than the alternative one we drew. And hopefully this video helps and we'll see you in the next one.

Draw a Lewis structure for each of the following:
d. NH₂O⁻

Key Concepts

Lewis Structures

Formal Charge

Resonance Structures

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