Drawing the Lewis Structure for CH_{2}OViewing Notes:
Transcript: OK, this is Dr. B. Let's do the Lewis structure for CH2O, methanal or formaldehyde. Start with the valence electrons. Looking at the periodic table, Carbon has 4. Hydrogen, in group 1, has 1 and Oxygen, in group 6 or 16, has 6; but we have two Hydrogens, so let's multiply that by 2. Let's add them up: 4 plus 2 plus 6 equals 12 total valence electrons to work with. All right: so we know Hydrogens always go on the outside, and in this one, Carbon's the least electronegative, so that should be in the center. So let's put our Carbon right here, put our Hydrogens on the outside, and then let's put our Oxygen up here. We have 12 valence electrons to work with, so let's spread those out and form some chemical bonds. We'll put the bonds between these atoms so now we have chemical bonds, they're all joined together, and we've used 2, 4, 6, and we want to have 12. So 2, 4, 6, 8, 10, 12. We can check now and see if we have octets. Hydrogen only needs two valence electrons, so each of those Hydrogens, they're OK. Oxygen needs 8, it's fine with 8. Carbon needs 8, as well. It only has 6. What we'll do is take these two up here, and let's move them between the Oxygen and Carbon; we're going to share them. And so now we can check and see if we've fulfilled our octets. We have 2, 4, 6, 8, 10, 12, so we have the same number of valence electrons; but now Oxygen has 2, 4, 6, 8 still. Hydrogens are good with two each. And the Carbon has 2, 4, 6, 8 as well. So we've used the 12 valence electrons and we've fulfilled the octets. We can also write this in a structural formula that'll look like this right here, where the double bond here is represented by this—these two lines here. And we'll spread our valence electrons a little nicer there, so they're spread out. That's the Lewis structure for CH2O. This is Dr. B., and thanks for watching. 
