Use formal charge to select which resonance structure makes the l...

Question

Use formal charge to select which resonance structure makes the largest contribution to the resonance hybrid. (LO 7.16) (a) Structure I (b) Structure II (c) Structure III (d) All structures are equivalent and make the same contri-bution to the resonance hybrid. (I)
(II)
(III)

(II)

(III)

Accepted Answer

welcome back everyone identify the resonant structure that contributes the most to the hybrid. So we're going to begin by recalling that the factors that are associated with major resonance hybrids are going to be the following recall that we would have minimum separation of charge in our structure. So separation between the positive and negative charges, if any, we would have full octet, it's for our atoms. We would have any negative charges on the most electro negative atom. And we know that that makes sense because recall that when we have a more electro negative atom, such as flooring, which we recall is the most electro negative atom on the periodic table, it's going to be capable of handling that negative charge. We would then associate a positive charge on the least electro negative atom in a major contributing resident structure. And then lastly, we can recall that we would want to have minimum formal charge in our structure. So with these factors outlined, let's go ahead and analyze each of our structures so we can see that looking at this first structure here in choice a We have a carbon atom which has a total of only three bonds around itself which results in its formal charge of plus one here, we know that carbon is not bonded by its bonding preference, which is to have four bonds. And that is why we have this plus one formal charge. And so carbon doesn't have a full octet here because it only has six electrons surrounding it with the three bonds that it has in the structure. So that's already a factor that we don't like. We can already say that we have no full octet on carbon. Now looking at our nitrogen atom recall that nitrogen has a bonding preference of having three bonds and one lone parent itself. And that is exactly what we see here on this nitrogen, which is why we have a formal charge of zero for this nitrogen, so it's neutral. So that's good. We can note that down octet on nitrogen And looking at carbon here, this third carbon, we see that it has a total of four bonds. So it does have a full octet. It does have a formal charge of zero but we're going to say that this is going to be therefore a minimum residents contributor because we don't have that full ahead on carbon. So let's go ahead and continue on to the next option, which is structure be. So looking at structure, be looking at this first carbon atom, we see that it's neutral. It has its four bonds. So it has a full octet. So we have carbon with a full octet again. And then now looking at our nitrogen atom, we see that it has a plus one formal charge, which we can understand because it has a total of four bonds around itself and no lone pairs. And so in this case we would count total valence electrons around this nitrogen atom. So it has a full octet, but it has a formal charge of plus one because it directly has +1234 valence electrons attached to itself where it would prefer to have five valence electrons attached to itself directly. Which is why we have that plus one formal charge. So recall that nitrogen is located in group five a. On the periodic table. And so it's actually pretty electro negative. So it's not going to really like having this positive charge on itself. So this is going to be a factor that makes us look at the structure as a minimum resonance contributor. But now let's look at this carbon here we see that the carbon is happy with four bonds just like this one. And then looking at this oxygen, we see that we have an oxygen which has just one bond and a total of three lone pairs around itself. Recall that oxygen has a bonding preference of having two bonds and two lone pairs on itself because we have three lone pairs around this oxygen atom, we have a formal charge of minus one because it has directly attached 1234567 valence electrons. Which is one more than I would prefer to have. Now, oxygen we recognize as in group six a. On the periodic table and it's a pretty electro negative element. So let's note that down oxygen is in Group six A C. Group six element. So it's pretty electro negative based on our trend since it's closer to flooring and it's happy or we can say it's capable of handling the negative charge. Now, out of all the factors we noted down, I would say the worst would be the second factor here with nitrogen having a positive charge. But we're not going to say that this is a minimum residents contributor just yet. Let's continue to analyze the rest of our options before we rule out choice B or not. So moving on to structure, see we can see that in this structure carbon is happy with four bonds in both of these areas. But now we have a nitrogen that has a total of 12345 bonds around itself. And so it actually has an expanded octet because we can count a total of 10 electrons around this nitrogen atom. But we want to recall that nitrogen is a period two element And we only understand that Pier three or below elements on the periodic table can have expanded octet. And so because my Children is a period two element, it cannot have an expanded octet. Now looking at we said carbon has a full octet. And then we have the oxygen atom where we see we have oxygen bonded to a responding preference with two bonds and two lone pairs on itself. So oxygen is also neutral and has a full octet. But because we see that nitrogen is drawn in with an expanded octet, we would say that structures C we can obviously rule out as an incorrect choice. It's going to be therefore based on this first factor that we wrote out, not a valid structure. So let's move on to our last structure which is in choice D. We can see that our carbon atom here has a total of three bonds and two lone pairs around itself. It directly has 12345 valence electrons attached to itself, where, which is one more than I would prefer. So we have that minus one formal charge on this carbon. So even though it does have a full octet, we have carbon with full octet but a negative charge and recognized that carbon is a group for a element on the periodic table. And it's not happy with the negative formal charge because carbon is a lot less electro negative than nitrogen and flooring. So now let's look at this nitrogen atom, we see that this nitrogen has a total of four bonds around itself and it does not have a lone pair. So it has directly attached 1234 valence electrons, which is one less than it would prefer to have. So it has that formal charge of plus one. Now the nitrogen does have a full octet but it has this positive charge or positive formal charge. And even though we know that nitrogen is a group five a element, it's going to be not as happy with this positive formal charge because we understand that nitrogen is much more electro negative than carbon. So this is going to be a factor that also contributes to the structure being a minimum resonance contributor. And lastly, we can clearly see that that oxygen has a full octet here in this structure for choice D. So overall we're going to based on these two factors here, say that this choice T is a minimum resonance contributor. And then going back to Choice B. We left that unchecked or unmarked. But we're going to confirm that although we have nitrogen with a positive charge here, we at least have oxygen which we know is more electro negative than nitrogen because it's further to the right and closer to flooring on the product table, we at least have this oxygen with a negative charge here. And so based on this factor that we've outlined for oxygen, we can say that Choice B is a structure that is a major resonance contributor. And overall we can rank the structures and say that from order of most to least contributing. Let's say residents contributing. We would list first Structure B, which is more contributing to residents than the next best structure which is structured D. And then the least contributing. Destruct to residents would be structure A. Because as we saw structure A did not have a carbon atom with a full octet which is worse than having a carbon atom with a negative charge as we saw in structure D. So overall to complete this example, our final answer is going to be that structure B is the best choice as a major resonance contributor for the resonance hybrid. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Use formal charge to select which resonance structure makes the largest contribution to the resonance hybrid. (LO 7.16) (a) Structure I (b) Structure II (c) Structure III (d) All structures are equivalent and make the same contri-bution to the resonance hybrid. (I) (II) (III)

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Use formal charge to select which resonance structure makes the largest contribution to the resonance hybrid. (LO 7.16) (a) Structure I (b) Structure II (c) Structure III (d) All structures are equivalent and make the same contri-bution to the resonance hybrid. (I)
(II)
(III)