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Ch.12 - Solids and Modern Materials

Chapter 12, Problem 128a

Although polyethylene can twist and turn in random ways, the most stable form is a linear one with the carbon backbone oriented as shown in the following figure:

The solid wedges in the figure indicate bonds from carbon that come out of the plane of the page; the dashed wedges indicate bonds that lie behind the plane of the page. (a) What is the hybridization of orbitals at each carbon atom? What angles do you expect between the bonds?

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welcome back everyone in this example, we have an image of a section of a polyvinyl alcohol shown below. We need to identify the hybridization of the carbon atoms and the expected bond angles. So we're going to focus on each of our carbon atoms in the center and we want to see the bonding regions or bonding groups. So looking and let's actually label these carbons. This is carbon 1234 and five. We can recognize that Carbons One, carbon three and carbon five are all bonded the same way and we can see that they're bonded to other carbon atoms around themselves where our central atom sorry, is surrounded by other carbon atoms. So that's two bonding regions and then we have a bond to a hydrogen and a bond to oxygen or hydroxide here. And so we would say for this group we have again, two bonds to carbon and a bond to oxygen and a bond to hydrogen. So we would have four bonding regions. Let's actually write this in blue. And then now we want to look at carbon two and carbon four where we recognize that they're also bonded in the similar way and we have bonds where we have the central atoms which is carbon here for carbons two and four bonded to two other carbons, where they're bonded to two hydrogen atoms as well. So they wrote H two and condensed form. But we can see that they're bonded to two carbons and two hydrogen. So we would have the following group where we have C. C. H. H. Again, for a count of four bonding regions, We want to recall that when we have four bonding regions around our central atoms, we would therefore say that our hybridization Is going to be s. p three. We also should take note of the fact that we have no lone pair electrons around these central carbon atoms. So we have zero lone pairs and so that would correspond to a molecular geometry which should be touched or he'd rel and we want to recall that the bond angle for tetra hydro molecular geometry is a value of 109.5 degrees. So for our final answer, we're going to confirm that the hybridization of these carbon atoms, carbons 135 and as well as carbons two and four all have the hybridization of sp three And their bond angles are going to be 105. So what's highlighted in yellow are our final answers. I hope everything I explained was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video