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Ch.8 - Basic Concepts of Chemical Bonding

Chapter 8, Problem 54b

Consider the formate ion, HCO2-, which is the anion formed when formic acid loses an H+ ion. The H and the two O atoms are bonded to the central C atom. (b) Are resonance structures needed to describe the structure?

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Hello. Everyone in this video. We're going to talk a little bit about our lewiS structures and resonant structures. So first of all, we need to go ahead to find what a resident structure is. Silver resonance structure is a set of two or more lower structures that collectively describe the electronic bonding of a single poly atomic species, including fractional bonds and fractional charges. Alright, so first, what we do in this problem is going ahead to calculate the total number of bands electrons. Alright, so we have our nitrate ion that's going to be n minus. So from our nitrogen atom we have one nitrogen atom. Each atom will have five valence electrons. And for oxygen we have three atoms. Each atom will have six valence electrons. And for the nitrogen it has five valence electrons because it's in group 58 and for oxygen is in group 68. So three times 6 is and now we add up these sums of five plus 18 is going to equal to 23 electrons. Alright, now, second step will be to draw the LewiS structure. Alright, so in the problem, it's just that we have a central nitrogen atom and it's bonded to three oxygen's. So we'll go ahead and put those on. All right, so we need a total charge for the molecule to equal to negative one because that's what the ion is as well. So this will need to also, collectively That is from charges to a sum of -1. How can do that is if we have one pi bon here to at least one of the oxygen's And the other ones will just have three lone pairs. This will give the nitrogen or the oxygen connected single bonds in the nitrogen with a negative one from more charge. And this will also give the central item which is the nitrogen A plus one charge. So we do on the side here. You see a -1. Last one. Negative one. So you go to zero Total of -1, which is what we want. All right now let's go ahead and draw out our Lewis structures but our resonance structure. Alright, so the first one, the structure can actually just be our lewis structure. I actually go ahead and kind of rearrange this. So we have our central atom and then connected to our oxygen's one having a double bond. Yeah, the third will be single bonded. Alright. And then we'll put brackets now and a negative charge, indicating that we have a iron and drawing the arrow pushing or are electrons for the next resonant structure. We can grab this lone pair to make a pi bond here and then grab me the pipe on to create a long pair here. And this will create the structure again with the brackets. Still we have the nitrogen at the central atom. This oxygen does not change now we have a double bind on the left oxygen and then the top oxygen will have a single bond And this still will give us a -1 from will charge for the molecule for the ion. And again, drawing the arrow, pushing we have this long pair come to make a double bond and he doubled board breaking to create a long pair in that case. Then we'll have the third and final resonant structure, still having the nitrogen as the center adam. Now this left oxygen will have a single bond and the top will have a single bond as well, making this also a negative charge. Alright, so we can see here from the resident structures that all the electrons in the nitrogen auction double bond. So all these, it's going to move to any of the other Austrian atoms and have a long pair be negatively charged and form so we see that um whenever we have a nitrogen auction double bond, we can always take this or break this pi bond and make a long pair. And I will always have one of the oxygen arms that had a single bond to nitrogen or this or this, this, this, this or this. The don't care, will always come down to make a double bond between the nitrogen and oxygen atom. So in this case we whenever we're moving all these electrons, we're also doing this while maintaining a overall charge of negative one because that's what we're trying to represent here. Right? So to answer this question is that yes, the residents structures are needed. Alright. And this is going to be our final answer for this problem.