Draw the important resonance forms for the following molecules...

Question

Draw the important resonance forms for the following molecules and ions.

(d) NO₃^–

(e) NO₂^–

(f)

Accepted Answer

Hey everyone. Let's solve this problem. And says, do these ions have other resonance forms? Draw the structures of possible resonance forms in each case. So let's remember that residents usually occurs when we have pi electrons and nearby charges could be positive or negative and in this case we have three ions and resonance is common in ions having pi electrons and charges. So let's see what we're working with here. Our first two compounds, It only has their condensed formula. So let's draw those out in their lewis structures. So first we have to calculate the number of valence electrons adding an extra electron for that negative charge. Right? So cl seven electrons, Three oxygen would be 18 plus one for that negative charge. Giving us 26. What happened there? 26 valence electrons. So that would look something like this chlorine, three oxygen's 246 electrons. So I still have 20 other electron. So let's add those as long pairs. 246, 8, 10, 12, 14, 16, 18, 20. Alright, Is this the final form of this compound? Is this very stable? Well, let's check the formal charges. I see that all these oxygen's have seven electrons touching them. So they're all going to have negative charges and chlorine has 12345 electrons touching it. It likes to have seven. So seven minus five is plus two on that chlorine. So if chlorine wants to more electrons and there are some oxygen's here that have some extra electrons. We can form two double bonds from two of these oxygen's and the residents occurs because there are more than just two oxygen's in which those electrons can be donated. So we can create a few different combinations or residence forms of this structure, let's say the first two oxygen's donate their electrons. And then that last oxygen has a negative charge. Or we could have the two outside oxygen's have double bonds. And that center oxygen has a negative charge. Or last one we could have there were two right side, oxygen's have the double bonds and the left side oxygen has a negative charge. Giving us these three residents structures four C L 03 minus. Okay, Let's do a chess 04 part B. And it's H S 04 minus. So we have one plus six plus 24 for oxygen plus another one for that minus charge, Giving us 32 valence electrons. Okay, so sulfur is gonna be my central atom here. I'm going to have three oxygen's. Oops, that's not oxygen. Okay, and we can still add 22 more electrons. Right? 2468 10. So we have 22. Let's start with these oxygen atoms. 2468, 10, 12, 14, 16, 18, 20 22. And let's see what our formal charges are. So all these oxygen's that don't have A H attached are going to have a negative one formal charge. Right? They have seven electrons touching it. How about sulfur sulfur likes to have six Adams touching it. It only has 412341 from each sigma bomb. So it's going to have a plus two formal charge? Kind of looks like our first compound? Right part A. So similarly, we can form two double bonds from two of these oxygen's to give that sulfur the two electrons that it desires. Okay, so let's see what that looks like. Let's see, I'll draw it this way. So how about we take these two oxygen atoms from the top and bottom? So, oh, double bond, S single bond O double bond. Oh, O H. Okay. And we can repeat the same thing with again, different combinations of those oxygen atoms. So let's say the top and the right hand oxygen double bond and the oxygen without a double bond gets a negative charge. And we can do it again right the bottom and the right hand oxygen with the top oxygen getting a negative charge. And are there any other ones? Well, we can redraw the first structure, but that will be it. Let me redraw it. So it's more clear. So we have the top. Oh, I'm sorry, not the original structure, we can have sulfur with just having a plus one formal charge, which would be only one double bond oxygen And the other two single bond. And having that negative charge. Do you think that's going to be the more stable resonance structure? No, but it's still a possible structure. So we'll include it. Okay, and then we have our original structure over there. Okay, So are you catching onto the pattern here when you have different oxygen's, you can sort of swap them around. And lastly, let's look at our compound, we have CH two double bond into C. H bond to carbon yo group And Ch two, which has a lone pair giving it a negative charge. So we see a negative charge neighboring that pi bond. So those electrons are going to go towards our partial positive carbon in these electrons can go onto that electro negative oxygen, giving us The oxygen with three lone pairs and a negative charge In a double bond down here with CH two. Okay, that's it for this problem. I hope this video is helpful. And thanks for watching.

Draw the important resonance forms for the following molecules and ions.
(d) NO₃^–
(e) NO₂^–
(f)

Key Concepts

Resonance

Formal Charge

Electron Delocalization

Watch next