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Formal charge | Molecular and ionic compound structure and properties | AP Chemistry | Khan Academy


5m read
·Nov 10, 2024

  • [Instructor] In this video, we're going to introduce ourselves to the idea of formal charge, and as we will see, it is a tool that we can use as chemists to analyze molecules. It is not the charge on the molecule as a whole; it's actually a number that we can calculate for each of the individual atoms in a molecule.

As we'll see in future videos, it'll help us think about which resonance structures, which configurations of a molecule will contribute most to a resonance hybrid. So before going too deep into that, let's just give ourselves a definition for formal charge, and then as practice, we're going to calculate the formal charge on the various atoms in each of these resonance structures for nitrous acid. These are both legitimate Lewis diagrams.

They're both legitimate resonance structures for nitrous acid, but we'll think about which one contributes more to the resonance hybrid based on formal charge. So the definition of formal charge, and we're going to do this for each atom in our molecule, for each atom, we're going to calculate the number of valence electrons in a free, neutral atom. From that, we are going to subtract the number of valence electrons allocated to a bonded atom.

And so your next question is, what does it mean to be allocated? Well, I will break up this definition a little bit. So if we want to think about the valence electrons that are allocated to a bonded atom, these are going to be the number of lone pair electrons plus one half of the number of shared electrons. So let's try and make sense of this by applying this definition of formal charge to the constituents of nitrous acid.

So let's start with this hydrogen over here. What's the number of valence electrons in a free, neutral atom of hydrogen? Well, we've seen this multiple times; you could look at this on the periodic table of elements. Free neutral hydrogen has one valence electron. Now how many valence electrons are allocated to the bonded atom?

Well, one way to think about it is to draw a circle around that atom in the molecule, and you want to capture all of the lone pairs, and you want to capture, you can think of it as half the bond. You could say for each bond, it's going to be one electron 'cause it's half of the shared electrons. Each bond is two shared electrons, but you're going to say half of those, and then you have no lone pairs over here. So the number of valence electrons allocated to the bonded atom, in the case of hydrogen here, is one, and so we are dealing with a formal charge of zero for this hydrogen.

Now what about this oxygen here? Well, we do the same exercise; I like to draw a little bit of a circle around it. The number of valence electrons in a free, neutral oxygen we've seen multiple times—that is six. Then from that, we're going to subtract the number of valence electrons allocated to the bonded atom. So the bonded atom has two lone pair electrons, and then it gets half of the shared electrons.

Half of the shared electrons would be one from this bond, one from that bond, and one from that bond. So you add them all together: two, three, four, five. So six minus five is equal to positive one, and so the formal charge on this oxygen atom, in this configuration of nitrous acid, is positive one. Now what about the nitrogen?

Well, we'll do a similar exercise there. A free neutral nitrogen has five valence electrons; we've seen that multiple times. You can look at that from the periodic table of elements. From that, we're going to subtract the number of valence electrons allocated to the bonded to nitrogen. Well, we see one, two, three, and then two more lone pair electrons, so that is five.

You have zero formal charge there. Then let's look at this last oxygen. This last oxygen—a free neutral oxygen has six valence electrons. From that, we're going to subtract the number of valence electrons allocated to the bonded atom, so two, four, six lone pair electrons, plus half of this bond. So that's seven allocated valence electrons: six minus seven equals negative one.

So this oxygen has a formal charge of negative one, and I really want to remind you, we're not talking about the charge of the entire molecule. Formal charge is really a mathematical tool we use to analyze this configuration. One way you can kind of conceptualize it is, in this configuration, this oxygen on average has one more electron hanging around it, one more valence electron hanging around it than a free neutral oxygen would.

This oxygen has one less valence electron hanging around it than a neutral free oxygen would. Now let's look at this configuration down here. Well, this hydrogen is identical to this hydrogen; it has no lone pair electrons and just has one covalent bond to an oxygen. So we would do the same analysis to get that its formal charge is zero.

But now let's think about this oxygen right over here. A free neutral oxygen has six valence electrons. The number of valence electrons allocated to this one is two, four, five, and six. So six minus six is zero—no formal charge. Then we go to this nitrogen. Free nitrogen has five valence electrons; this nitrogen has two, three, four, five valence electrons allocated to it, so minus five, it has zero formal charge.

And then last but not least, this oxygen right over here. A free neutral oxygen has six valence electrons; this one has two, four, five, six valence electrons allocated to the bonded atom, and so minus six is equal to zero.

What we see is this first configuration, or you could say this first resonance structure for nitrous acid had some formal charge; it had a plus one on this oxygen and minus one on this oxygen, while the one down here had no formal charge. Everything had a formal charge of zero.

As we'll see in future videos, the closer the individual atom formal charges are to zero, the more likely that that structure, that resonance structure, will contribute more to the resonance hybrid. But we'll talk about that more in future videos. The whole point of this one is just to get comfortable calculating formal charge for the individual atoms in a molecule.

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