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.

d. CH₃–CH=CH₂

e. HC≡C–CHO

f. H₂N–CH₂–CN

Accepted Answer

Hey everyone. Let's solve this problem. It says for these compounds and ions right. The appropriate lewiS structures show the types of orbital's that overlap to form each of the chemical bonds and determine the bond angle around each atom. So first it says to do to write the LewiS structures. So for lewis structures we know that we count the number of valence electrons. Then we draw a sigma bond framework and we can also draw the specified double bond since we know that they will be in those spots on the compound and lastly will complete the octet by adding loan pairs with any remaining remaining electrons as long pairs. Okay, so we're also looking at the orbital, so something to remember is that a single bond is made up of S orbital's a double bond is made of one S and one p overlap. Right? So one from each atom, so technically two S and two P from each atom, but it's in one overlap and one s overlap and two p orbital overlap. Okay, so we don't have p orbital's overlapping unless we have a double or triple bond. Okay, so let's draw our compounds our LewiS structures. So first we have H two C double bond ch two. So adding up the number of valence electrons, we get eight for the two carbons and four from the two or sorry, four from the four hydrogen. So that gives us 12 valence electrons. So let's draw it out. We have see let me draw the angles C h c h double bond, c h h 2468, 10, 12. So we used all of our valence electrons. So let's write out the types of orbital's that overlap here. Well, I have two types of bonds happening. The ch bonds on the outskirts and then this carbon carbon double bond in the middle. So those C. H bonds, the carbon is S. P two hybridized because it has 123 charge centers. And then the hydrogen we know is just an S orbital. So the green bonds will be S. P to S bonds. Okay, so s. p two s and that same on this side S. P two S And then the blue bond here in the middle. We have sp two sp 2 that covers the s orbital's. And since it's a double bond, we have a P p orbital. Okay, let me write that down here actually. So we have more room and then we have to determine the bond angles. Right? So we have technical difficulties. SP two s. P two and pp Orbital overlap and the bond angles on both sides are going to be 120 Degrees because of that sp two hybridization. So all of these will have 120°. Okay, cool, let's move on to be our second compound, we have ch three carbon double bond to C C H O. So we have three plus four plus four is eight plus four plus four is another eight plus one plus six, which gives us 26 valence electrons. So let's draw it out. Ch three, c double bond. C. See I know the ending C. H. O. Is an al to hide. So I can already draw this as C double bond O. H. Okay, so that's the benefit of knowing functional groups. You already know that it's going to be C double bond O. H. So that gives us 2468, 10, 12, 14, 16, 18 20. Let's add someone pairs. 22 24. Oh, sorry, that's a triple bond here, 26. There we go. Okay, so what type of orbital overlap do we have going on here? Well, these ch bonds are all the same and this carbon is sp three hybridized and the hydrogen czar just s so those would be S. P three S orbital's overlapping. For those three bonds. Now we have our sp three carbon. And this carbon in the triple bond. So this carbon has two charge centers. So that will be S. P. So we will have Sp three sp hybridized orbital's overlapping. And then again another. Now we have to S. P groups right, the carbons that are triple bonded to each other. So this is going to be S. P. S. P. And how many P orbital's to write? two PP overlapping Orbital's? Okay, right there in the middle and then we move along, we have again the sp carbon with this carbon that has 123 charge center. So that's S. P. Two. And This bond, the carbon hydrogen bond, s. p two s. And also count the carbon oxygen bond, which is S. P. Two and this oxygen has 123 charge centers. Also including the lone pair. So that's sp two sp two and it's a double bond. So that is also p orbital overlapping. Cool, so let's write the bond angles. So these bond angles R .5. This triple bond makes this whole part 180°. Trying to find room to draw here and then this carbon that's S. P two hybridized. We'll have bond angles of 20°. Okay, I know it's a lot of writing there. So I hope you are following along. Okay. And lastly we could draw this one a little bigger. We have C H three and H C H two C. N. Which is four plus three plus five plus one plus four plus two plus four plus five, Giving us 28 valence electrons. And if we draw that out we have C. H three and H C H two C. N. We know as a night trial group. So I'm going to draw that triple bond and then we can add two more lone pairs. So those will go on our nitrogen. Okay, now let's write out the hybridization. So we have again, a methyl group is going to be three S. P three S. So S. P three S. Overlap. We have three of those 123. And then we have The Sp three Carbon With our 123, 4 charged groups on our nitrogen. That's also sp three. Then we have The Sp three Nitrogen With our 1234 also sp three carbon. And we can't forget about the hydrogen bond. So we have The nitrogen to hydrogen would be s. p. three s. And the two carbons here, or carbon hydrogen bonds would also be sp three S. Right 1234 charge centers. Makes it S. P. Three, the hydrogen czar s orbital's. And we have our carbon and nitrogen, the carbon has two charge centers. So that's S. P. And again bonded to that S. P. Three. So sp three sp and then our carbon is S. P. R. Nitrogen is also sp and we have a triple bond. So this will be two P. P orbital overlaps. Now let's look at the bond angles. So our triple bond is 180 degrees right? It's linear. Then R. S. P three is going to be one oh 9. to tra he drel And our nitrogen has four charge centers as well. So this will be 10. And this carbon will also be 109.5 because there both sp three hybridized. Okay, that's it for this problem. I hope this video is helpful. And thanks 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.
d. CH₃–CH=CH₂
e. HC≡C–CHO
f. H₂N–CH₂–CN

Key Concepts

Lewis Structures

Orbital Overlap

Bond Angles

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