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NOTE: Alll parts of tnis question must be answered correct to Move the next question Consider the following partially completed pair resonance contributors, where t...

Question

NOTE: Alll parts of tnis question must be answered correct to Move the next question Consider the following partially completed pair resonance contributors, where the lectrons havebeen cesbonds lone Pairs but the formz chargesmissingResonance ContributorResonance Contributor BINSTRUCLQNS;Complete the resonance contributors by addlng formal charges where appropriate answei the followlng questlons based on these partlally complete structures(2) Whatformal charge on the Jtom labelled X? Use the dro

NOTE: Alll parts of tnis question must be answered correct to Move the next question Consider the following partially completed pair resonance contributors, where the lectrons have been ces bonds lone Pairs but the formz charges missing Resonance Contributor Resonance Contributor B INSTRUCLQNS;Complete the resonance contributors by addlng formal charges where appropriate answei the followlng questlons based on these partlally complete structures (2) What formal charge on the Jtom labelled X? Use the drop Gown menu to select Your answer Wha: the forma charge on the atom labelled 2? Use the drop-down menu select your ans ver Which the resonanc contiouors LOWER IN ENERGY? Use drop: down meru lec Vovc Jnser



Answers

Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom
(a) $\mathrm{O}_{3}$
(b) $\mathrm{SO}_{2}$
(c) $\mathrm{NO}_{2}-$
(d) $\mathrm{NO}_{3}-$in each of the resonance structures:

So here we just looking out of residents structure so she can see here we have the system that is allowed to d localize its electrons. And so we would see an hour push. Like such Now, this would shift our double bond in our electrons and we would generate some charges on our residents structure here. So we've moved our double bond. As you can see, our electron density is moving in the direction off the head of the arrows. We would generate a positive charge, and then our oxygen would have a negative charge. Don't forget, we still have loan past, present, And now, in order to get the original structure, which is over on the left, who would simply just push those hours back like so.

Here we are continuing to draw out some lower structures. So firstly phosphorus oxygen, oxygen, oxygen and another oxygen, where they all have three line powers here, apart from one of them has a double bond. And so we have three miners charge. And of course we can have resonance using any three of these formal negative charges. So that means that we've got three different forms of resonance. Well, overall we have three minus charge on the species. Next we've got C N minus. So we've got carbon, we can have a triple bond to the nitrogen where we need a formal negative charge on the carbon to have it with a full lone pair. Next we have also 3 to-. So we have a sofa where the double bond oxygen and then we've got to oxygen's that have minus charges that give us our two minus. And again here you can see that we will have resonance because we have negative charges on the oxygen's. We've got resonance from two different positions that will take place in separate events, and they will have a knock on effect with this double bond. Next we have cielo to minus. So what we have is chlorine, double bond oxygen, single bond oxygen too long pairs on the chlorine. The oxygen with the single bond has a formal negative charge, and here we have resonance that can bounce between the two oxygen's.

Hello. So did they will be looking at this compound right here and drawing its residence forms. So first, let's draw Lewis structure. So let's cannot the electrons it has. So nitrogen has five electrons, valence electrons and has two of them now. Accident has six and has three of them. So two times five is 10 three times six 18. So 28 valence electrons. Now let's take a look at how many it needs. So nitrogen means eight valence electrons and has and there are two of them. Oxygen needs eight and there are three of, um so that's a total of 40 electrons. Easy way to know how many bombs you need is to subtract how maney need from Haman you have and divided by two. So in this case, we need six bonds. So let's draw the this way first. And let's kind of how Maney bonds we have. We have 1234 bonds. We need six. So what can we do? Well, we can ad bond here and a bond here and then we just add electrons. So nitrogen has four bonds. So has a full valence already. This one has story, so needs a lone pair. Oxygen needs to long pairs to have a full octet. This one needs two lone pairs. A swell. And then we had three zone pairs to the last one. So here is one resident structure. Now let have out Leach. Look at its formal charges. So Oxygen started off with six electrons. Has two lone pairs, so it kind of owns thes four. Electrons. Has it to bonds. And of those four electrons, that hasn't those two bonds. It's sort of owns two of them. Let's say so. Four plus two a. Six. It started off with six. It kind of ended up with six. So has a zero formal charge. Nitrogen has three bonds, of which we can say that it air quotes, owns three of those and has two and a lone pair. So has five Started out with five ended up with five so formal charges. Zero night that this my Trajan here has four bonds, so it's, we say, that it owns for electrons. It stood it off with five electrons and it ended up with four. So has a plus one formal charge. It lost an electron while it was sharing Let's look at this oxygen, right? The top oxygen we'll see. It has two lone pairs and two bonds. The two bonds contribute to up Trans. The long payers contribute for has ended up with six electrons. It started out with six, so zero formal charge. Oxygen on the bottom has three lone pairs switch. It is kind of like, which is six electrons and one bond, which we say contributes one electron. So it started off with six electrons and it ended up with seven. So has a negative one formal charge. So we see that if we add all the formal charges together, we get zero. So this is a neutral species, so that's correct. Now let's see if we conduce a different resonance form. Well, let's see. So let's say that instead of the nitrogen being double bonded to the oxygen on the top, it's double bonded to the night oxygen on the bottom. That would be a valid residence form. So we've got my chin on a tour nitrogen, contented oxygen and double bonded to an oxygen, and we will have similar formal charges to what we had on the top. Just that the top and bottom Oxygen's are now switched and formal charge. Nitrogen is still plus one. The one on the top is negative one. Now bottom 10 However, there is still 1/3 residence form. So a way you could think about residence form is actually the movement of electrons. So they get from the top to the bottom. While we sort of did as we moved this lone pair here. And then we moved the electrons in this pond up so we can't move these electrons here because then this nitrogen would have more than eight electrons. So we can't have this nitrogen double wanted to these oxygen's. But this night, but the nitrogen that's only bonded to one oxygen it has alone pair. So let's say that we moved the Sloan pair here and then we move this to the left Trans in this bond up there. So let's draw over resident appointments form like this then and we just add lone pairs were necessary. And if we look, we'll see that every, uh, Adam has an architect and no one exceeds in an octet. So this is a valid resonance form. This is possible. This is a this follows all the rules. This is another way to write this, so this oxygen still has zero charge. We didn't really touch it, But now, instead of having lone pair, this nitrogen is bonded four times, so it's sort of owns four electrons. It started off with five. So it has positive one charge because it sort of lost one electron by sharing. No, he look here. This nitrogen is still bonded four times, so it's also a positive one. Formal charge. Now let's look at the oxygen up here. Three lone pairs, one bond. So it's sort of owns seven. Electrons started off with six negative one formal charge, and the one on the bottom is also a negative one formal charge. And if he can up the formal charges, you will get zero. So it's neutral. So as you can see these air, three ballad residence forms and all of them contribute to what the molecule actually

Strong resonance structures of each of the following irons. For a. We have the story, no sign eight iron S. E. C. N. Minus, which has three residents, forms the first one S. E. Double bond, Seiko bond in has a residence form spc triple fund and Mhm. Mhm. And some of the residents form Mhm. S. E. Trip upon C. There would be the three resonance forms. We need to assign formal charges and select the residents structure that provides the best description for the iron, assigning formal charges here. You would have zero zero minus one -100 plus one, zero and minus two. So, if we look at um okay, the first structure that has the larger charges. So it's less likely if we look at the structure in the middle versus the structure on the end, the negative charge is on the more electro negative nitrogen atom here. So this would be the most likely structure for B. Look at the residence forms of the bio for meat. Hi Inn. Mhm. Bio format would be. Yeah. Yeah. It's like here. Yeah, Something like this one. And what is there another resonance former UFC. What? Double bond? The Oh, okay. Mhm. Yeah, mm hmm. And we'll find formal charges here -100, 0 And -100 and zero. So oxygen is more electro negative than sulfur. Therefore, first form here would be the more most prospered residents structure. And proceed Looking at the di Feo Harbin eight I in which has two residents forms. Yeah. Mhm. Okay. Okay. Mhm minus. And it's next form here would be. Mhm. Yeah. Yeah. Mhm. Yeah. This will be to minus courage overall. Mhm signed formal charges -1 -100 This week -1000. And if we look at so sorry, this should be minus one there. And so the structure on the left has the formal charge on the more negative oxygen atom. Oxygen's more electro negative of the sulfur. Therefore, the first configuration would be the preferred stretch resident structure.


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