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  • - [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,

  • and 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 free,

  • in free neutral,

  • neutral atom,

  • atom.

  • From that, we are going to subtract the number

  • of valence electrons allocated,

  • allocated to bonded,

  • bonded atom.

  • And so you're next question is,

  • what does is 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,

  • number of lone pair electrons

  • plus one half of the number of shared electrons.

  • So lets 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.

  • So 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,

  • 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 gonna say half of those,

  • and then you have no lone pairs over here,

  • so the number of valence electrons allocated to 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.

  • And so the number of valence electrons

  • in a free, neutral oxygen we've seen multiple times,

  • that is six, and 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,

  • so 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,

  • and then 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,

  • and so you have zero formal charge there.

  • And then let's look at this last oxygen.

  • So 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,

  • but 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 it just has one covalent bond to an oxygen,

  • so we would do the same analysis

  • to get that its formal charge is a 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,

  • and 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.

  • And so 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,

  • and 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 pint of this one

  • is just to get comfortable calculating formal charge

  • for the individual atoms in a molecule.

- [Instructor] In this video, we're going

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