The formal charge of an atom in a molecule is the charge that would reside on the atom if all of the bonding electrons were shared equally. We can calculate an atom's formal charge using the equation FC = VE - [LPE - ½(BE)], where VE = the number of valence electrons on the free atom, LPE = the number of lone pair electrons on the atom in the molecule, and BE = the number of bonding (shared) electrons around the atom in the molecule. Created by Sal Khan.
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- Hi thanks for the clear explanation.
But in the 'About' below the video, its says that 'FC = VE - [LPE - ½(BE)]'.
Shouldn't it be 'FC = VE - [LPE + ½(BE)]' ??(3 votes)
- the number of any valence electron that can be found in any element is not the same in the periodic table, because it is not there at all, or is there a formula to find it, if so please include it in the video.(1 vote)
- A quick trick I use to find out how many valence electrons elements have is by counting, from left to right, how far right an element is in a period (or row). For example, say carbon, it's in the second period and it's the fourth element in the period counting from the left so it has four valence electrons. Or sulfur, which is in the third period, which is the sixth element from the left so it has six valence electrons. This process works for transition metal elements too. An element like iron is a transition metal from the fourth period and is the eighth element from the left so it has 8 valence electrons.
Hope that helps.(9 votes)
- at 3.49, how in the upper structure the oxygen has 3 bonds?(1 vote)
- Because it can form three bonds? Oxygen isn't limited to forming just two bonds. Really the only thing oxygen is limited by is having no more than 8 valence electrons.(8 votes)
- What is actually going on physically between the electrons and the nuclei in the molecules when there is a resonance structure? In the example, the double bond could be either be shared between the N and the right O, or the left O, but in both structures, all the atoms have their valence shells filled so why would the double bond switch between the right O and left O?(2 votes)
- the electron probability cloud merges side by side for this structure, it forms a π bond, but the bond is a giant bond formed between the three atoms, we call it the delocalized π bond or the coniugated π bond, you'll catch a better understanding of this concept in the unit about bond kinds and hybrid orbitals.(2 votes)
- Why can't we connect the hydrogen to the nitrogen, how do I know that it should be bonded with the oxygen?(2 votes)
- Well first, Sal mentions that it is nitrous acid so we should know it’s structure from the name. It has a set structure. All Sal is doing with his formal charge discussion making different resonance structures, or moving electrons around to get new valid Lewis structures.
From a more fundamental view, acids generally have hydrogens attached to oxygen atoms. This is because when the hydrogen leaves, it leaves behind a lone pair on the oxygen. The resulting conjugate base will have a -1 formal charge on the oxygen. If the hydrogen were attached to the nitrogen, this -1 formal charge would be on the nitrogen instead in the hypothetical conjugate base. Oxygen is more electronegative than nitrogen so having a -1 formal charge on oxygen is more stable.
Hope that helps.(2 votes)
- I'm very confused by the term "lone pair electrons". If we say two lone pairs then it's going to be four electrons, but if we say two lone pair electrons do we mean just two electrons in a lone pair?(2 votes)
- We usually just say a lone pair (singular) or lone pairs (plural). Each single lone pair is composed of two spin-paired electrons. So don't really concern ourselves with the individual electrons, rather just the lone pair as group.
Hope that helps.(1 vote)
- Why don't individual atoms have full shells? They are bonded, and both covalent and ionic bonds have constituent atoms with full shells.(2 votes)
- Individual atoms are capable of having fully filled electron shells. The simplest examples being noble gas elements like helium and neon.(1 vote)
- Are there any simple ways to remember formal charge rules?(1 vote)
- How the formal charge was discovered??(1 vote)
- [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.