Write a hybridization and bonding scheme for each molecule that contains more than one interior atom. Indicate the hybridization about each interior atom. Sketch the structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7. c. C2H6 (skeletal structure H3CCH3)

Video transcript

Hey everyone in this example, we need to draw the molecule ethane showing the orbital overlap and label of the hybridization of the bonds. We need to give the hybridization of the orbital's in the carbon carbon single bond within this molecule. So our first step is to draw the outline of our molecule. We should recognize that we have two carbons which will be in the center and they're surrounded each by two hydrogen. And then our next step is to recall the orbital that our hydrogen are within. So we should recall that the configuration, the electron configuration for hydrogen is in the one s. one orbital. And we should recall that. This is going to hold a maximum of two electrons. So this orbital is fully filled in for the notation for hydrogen. And so therefore we would say that hydrogen is in the S orbital and it's unhygienic ized because we can't hybridize s orbital's. I'm sorry. That should be spelled accordingly. So we can go ahead and recall that we represent these s orbital's as circles around our hydrogen here. And the next step is to figure out our hybridization for our carbon atoms. And we should recall that. We can use a X. E notation for our central atom to determine its hybridization now because we have two carbons in the center. We can choose either one of these carbons to go with the hybridization notation. So let's go ahead and just choose the carbon on the left. So we'll say that we're going to do H two C and then C. H. Two. So to represent the hybridization for carbon on the left as our central atom. We're going to say that A represents carbon as our central atom and then X. Represents the electron regions or electron groups bonded to this central atom. So we're gonna have to count how many of our electron regions we have bonded to our central atom here. So this is our central atom we chose. And we can say that because we know that hydrogen should only should only be able to form single bonds to the atoms here. We would count one electron region here which is this first hydrogen atom two as the second electron region. Because we know it's bonded to this carbon to as well. And then we would count this entire CH two group as the third electron region bonded to our carbon, this carbon here as our central atom. And so we would go ahead and write A X. Four, which we would recall therefore tells us that we have S. P. three. We're sorry, sp two hybridization for our carbon atom. And this would technically apply to both of these carbon atoms since this molecule is symmetrical. So we need to go ahead and draw in these p orbital's and these are going to be hybridized because as you can see they're going to be somewhat overlapping with our s orbital's from each hydrogen in this molecule. And these are also represented or rather representing our single bonds that are connecting the hydrogen to the carbon atoms in the molecule, we next want to recall that these single bonds here are also what we can represent as sigma bonds represented by this symbol here. And we can go ahead and label the hybridization by recognizing that these are hybridized. And so they're going to have the combination of the hybridization of our carbon atom as well as of our hydrogen atom. And that means that the hybridization of this bond here is going to be S. P. Two, which comes from both the hydrogen and the carbon. And so that bond there with the arrow that we just drew has this hybridization for the sigma bond or the single bond again, sigma and single bonds are interchangeable terms. They mean the same thing now, as we should recognize from earlier. We also have technically the bonding preference for carbon, being that it wants to form four bonds. And right now, we've only drawn in two of the bonds here. So we should recognize that to get to the bonding preference for carbon, we want to have a double bond in the center of this molecule and we're going to show the orbital's how these orbital's create this double bond. So because we did state that the hybridization of our carbon atoms is sp two, we're going to go ahead and draw in our p orbital's for the central carbon atoms here. So we'll draw in one here and we'll draw another here and as you can see, they're also overlapping each other. And when this happens, when we have the overlap of two sp two orbital's, this is what forms our sigma bond, A. K. A. Our single bond connecting these two carbons. But as we stated earlier, we also have another bond because we have a double bond here. And so how do we form a double bond? We want to recall that a double bond Contains σ bond and one pi bond. And so we should recognize how pi bonds are represented by orbital's are created by orbital's. Well, what we can say is that because we have this sigma bond in the center, we also should understand that above and below the sigma bond, we're going to have to an hybridized p orbital's and these two purple unharmed sized p orbital's are going to be overlapping with each other to form our pi bond, which a. K. A. Is our or is within our double bond. Because we recall here on the left as we set a double bond contains one sigma and one pi bond and our one sigma bond was formed from our overlapping p orbital is here, as well as the un hybridized p orbital's that are overlapping on each side of the carbon here to form our pi bond. And so we can show the hybridization of this carbon carbon bond here by showing that the carbon on the left has an unhygienic sized p orbital and then the carbon on the right also has an unhygienic sized p orbital as we've drawn in purple here. And so to make sure that we fully answer this question, we want to give the hybridization of the orbital's in the carbon carbon bond so we can go ahead and expand upon our answer by also adding in the label here for the hybridization of the sigma bond that is contained within our double bond. And we can say that this sigma bond here, let's make that leader has the following hybridization where for our first carbon, we know that it's sp two hybridized and for the second carbon that is bonded to We also know that it's sp two hybridized because as we stated earlier, carbon corresponds to X four notation which we recall gives us sp two hybridization for each of our carbons in this molecule. And so as our final answer for this question, that is going to be this entire diagram that we've drawn with the three labels that we've given for the hybridization of each of our types of bonds in the molecule here. So we have the hybridization of the pi bond and the sigma bond for our carbon carbon bond here, which again, we stated as a double bond in the center of the molecule. And we have the hybridization for each of the carbon hydrogen bonds in this molecule. So they all have this hybridization here. Each of these four carbon hydrogen bonds, so this will complete this example. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.