For each of the following compounds and ions,
1. Draw a ...

Question

For each of the following compounds and ions,

1. Draw a Lewis structure.

2. Show the kinds of orbitals that overlap to form each bond.

3. Give approximate bond angles around each atom except hydrogen.

g. Lewis structure of acetic acid showing carbon, oxygen, and hydrogen atoms with bonds and lone pairs.

Accepted Answer

Welcome back, everyone draw the Lewis structure of the given molecule and label the orbitals that overlap to form each bond, including carbon Twohy bonds and the bond angle around each non hydrogen atom. And in this case, we are given prono acid, this is a carboxylic acid. And what we want to do is just through the Louis structure. So let's draw our carbon chain. We have a total of three carbons, 12 and three. Then we're going to introduce our oxygen hydrogen bond at the end. And of course, let's add car bono. So it's oxygen double bonded to carbon. Let's add the missing hygen, there are three bonded to the first carbon. So 123, then there are two bonded to the second carbon. And that's it. Now to have a complete Louis structure, we want to add two lone pairs to each oxygen because based on the octet rule, every oxygen will have two bonds and two lump pairs in a neutral form. So we have got our low structure. Now we want to consider the orbitals. So to get our orbitals, we first of all have to understand the hybridization of each carbon and possibly oxygen. Now, the first carbon is sp three hybridized because it has four electron regions, right. The second one has for electron regions, it's also SB three hybridized. So if we label our carbon atoms, let's suppose we label them from right to left. One, two and three, the carbon number three is sp three hybridized because it has four single bonds. Carbon number two is sp three hybridized that also has four single bonds. So it has four different electron regions. But carbon number one is sp two hybridized because it has three electron regions, two single bonds and one double bond. So what we're going to do is just recall that hygiene is not going to be hybridized. It is going to use its S orbital to form bonding. So essentially what we're going to do here is just indicate that carbon number three is bonded to three hygen. So each blue bond corresponds to the, the overlap between SP three carbon and an S orbital of hydrogen. So we can show that every blue bond in the structure is going to be an overlap between SP three and orbital. So we have 123456, a total of six. Now, if we move on, we have our carbon carbon bond and we know that carbon number one is SP two hybridized and carbon number two is sp three hybridized. So here we are going to have a green bond illustrating the overlap between an SP three hybridized carbon and sp two hybridized carbon well done. Now, we can essentially move on and consider our carbon oxygen bonds. First of all, we're going to say that oxygen is going to be SB two hybridized because it contains two long pairs and one double bond. So that's three electron regions. So first of all, if we consider our double bond, we have two types of overlaps. One of them is the SP two sp two overlap. So let's choose another color illustrating that illustrating that. So let's choose the black color. And let's indicate that it represents the SP two SP two overlap because oxygen and carbon, they are both SP two hybridized. What we're also going to observe here, let's consider a red pi bond which essentially illustrates the overlap between P and P orbitals, right. So we will have un hybridized P and P orbital overlap because it essentially corresponds to our pi bond within the double bond, right. Well done. Now, if we move on, we're going to observe another SB two SB two overlap between carbon and oxygen. Now, the question is why, well, essentially, even though oxygen has two lone pairs and two bonds, it's actually SB two hybridized because the lone pairs are within the conjugate system, we can form resonance, right? And therefore, we're going to draw a black bond illustrating the same SB two sp two overlap. And then of course, oxygen hydrogen that gives us SP two S overlap. Previously, we had SP three S but oxygen has sp two hybridization due to resonance. So the final overlap would be SP two S. Let's go ahead and label it. Let's say that we have SP two S overlap and that's it, right. We have labeled all of the overlaps in this problem. And now from here, what we are essentially going to do is just consider the bond angles. What about the bond angles? While we have to recall the VSE pr theory and consider our central atoms. So starting with the first carbon or basically carbon number three, if it's sp three hybridized, it's going to form tetrahedral electron geometry. And we are expecting to observe 109.5 degrees angles around this carbon both electronically and in terms of molecular geometry because there are no old arm pairs, which is also applicable to the second carbon. It has an sp three hybridization old on pairs. So that would be 109.5 degrees moving on to the sp two hybridized carbon. We are expecting to observe around 120 degree angles because it's trigonal planar geometry for an sp two hybridization, which is also true for the carbon oxygen, hydrogen bond, for example, because oxygen is sp two hybridized that would be 120 degrees. And that's it right. We have labeled all of the required angles. So we have our answers. Let's review what we have. So first of all, we have our Louis structure, then we have our orbital overlaps with different color coding so that we can keep it neat and simple. And finally, we have our bond angles for the same structure so we can just add another answer. That's it. Thank you for watching.

For each of the following compounds and ions,
1. Draw a Lewis structure.
2. Show the kinds of orbitals that overlap to form each bond.
3. Give approximate bond angles around each atom except hydrogen.
g.

Key Concepts

Lewis Structures

Orbital Hybridization

Bond Angles

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