Draw the Lewis structure of azidomethane (CH₃N_...

Question

Draw the Lewis structure of azidomethane (CH₃N₃) [Show two important resonance structures.]

Accepted Answer

Welcome back, everyone determine the structure and the resonance structures of the compound laz side or C two H five N three. Our first step is to make sure that we keep track of our V electrons. So let's calculate the total number of ants electrons. We have two carbon atoms. Each carbon atom contributes four V electrons because carbon belongs to the group four A. So we are multiplying two by four, we're adding five hydrogen atoms. Each hydrogen contributes one valance electron because hygiene is in group one A and then three nitrogen atoms, each nitrogen contributes five electrons because nitrogen is in group five A and we end up with a total of 28 advance electrons. The second step is to understand that our name sells a lot fl Azi. So that means it has an E group which essentially is ach three ch two group, right? That would be our ass group, meaning all of our carbon atoms and hydrogen atoms are already consumed. And this means that the next atom will definitely be nitrogen since there's nothing else to bond that carbon to. So this is what we're going to do. We're going to essentially add our ethyl group. So that would be our carbon carbon bond. Now, we're going to add the three hydrogen atoms bonded to the first carbon, then the two hydrogen atoms bonded to the second one. And now the only possibility for us to bond these three nitrogen atoms is to bond each nitrogen to each previous gen. Right from here, we want to understand how many V electrons we have in the given structure. So each bond has two vance electrons, we have 123456789, nine, multiplied by two. This gives us 18 electrons. So we have 18 electrons assigned but we have 28 total. So if we subtract 28 minus 18, this gives us 10 electrons remaining. We always start with the terminal atoms. We have our terminal nitrogen which has two electrons only. So to satisfy the oat rule, we're going to add six electrons onto that nitrogen. This leaves us with four. So we can essentially add or electrons on the other nitrogen and think if it is reasonable. So now the terminal nitrogen has eight electrons, the other Ngen to the left has 12345678 electrons, but the other ngen only has four electrons. So the octet rule is not satisfied even though we have assigned all of our electrons. So because our nitrogen lacks a total of four electrons to have an Octa, we can first will use both of the lone pairs on the adjacent nitrogen to introduce a triple bond. So let's go ahead and do that. We're going to remove both lone pairs and add two pie bombs so that we end up with a triple bond, right? And now this means each Ngen has a, so this is a reasonable structure. And what we want to do is just to calculate the formal charges for nitrogens A B and C. The formal charge formula, it tells us that we are taking the number of electrons subtracting the number of bonds and the number of lone pair electrons. So the formal charge for nitrogen M I'm sorry, the formal charge for nitrogen A would be five, we have four bonds and no lone pair electrons that be positive one. The formal charge for nitrogen B would be five minus four, right? We have four bonds and no loan pair electrons. So that's another positive one. The formal charge for C would be five minus one bond minus six loan pair electrons that gives us negative too. So we can essentially assign positive one are basically just a positive charge, another positive charge. And then we have a negative two. Of course, we have to understand that this resonance structure is not very stable or not stable at all because one of the nitrogen atoms has a negative two charge. So this is significantly large, right? Meaning we want to think about possible resonance structures. Now to perform resonance, we have to use one of the lone pair elect one of the lone pair soft electrons. So thinking about the terminal nitrogen, we can simply use that lone pair of electrons to form a pi bond. And that means we are breaking the adjacent pi bond adding a lone pair onto nitrogen. So this would give us our second Rison structure. We can simply draw our ethel group. Now, what we have is nitrogen with a long pair of electrons, there would be a double bond because we're breaking our pi bond and there's some that are double bond, right? Then the terminal nitrogen has two lone pairs. And now let's see what we have based on the formula. For the first nitrogen, we have two bonds, we have three bonds, I'm sorry, and two lone pair electrons. So five minus three minus two, this gives us no formal charge for the nitrogen atom, five minus four minus zero. This gives us positive one. And for the other one, based on the neutrality, we must have negative one. Let's see if that's correct five minus two minus four, that's negative one. Well done. Now, what else can we do? We can actually form a triple bond. Why not? So we can use another loan pair of electrons and let's go ahead and do that. So we're going to take another loan pair of electrons from the terminal nitrogen and we're going to form a triple bond now, and if we're forming a triple bond, that means we have to break a pie bond between the other two nitrogen atoms. So we will get our third and final structure. So that would be our L group. Then what we noticed is that we are forming an additional lone pair on the first nitrogen, there would only be a single bond, then we have our triple bond. And finally, that nitrogen has only one lone pair remaining, assigning the formal charges starting from the left side, five minus two minus four. This gives us a negative one formal charge. Now the central has a positive one charge because it has four bonds and no loan pairs. So five minus four, this gives us positive one. And that means the other ngen should have a zero formal charge five minus two minus three. That's correct. So we have our structures, we can eliminate the arrows and any unnecessary labels so that we can clearly illustrate resonance. All that we have to do is just add resonance arrows to illustrate that these are resonance contributors. And in addition to that, we are just going to add brackets indicating resonance. So that would be our final answer. We end up with three resonance contributors for flaz. I, one of them is significantly unstable. Generally, we would not even show it because it has a negative two formal charge, right? We definitely want to minimize our charges. So the first one is a real unstable and sometimes you will not even see it in reins because it's not significant. But we were asked to draw all of the reins forms. So we also included that one. Thank you for watching.

Draw the Lewis structure of azidomethane (CH₃N₃) [Show two important resonance structures.]

Key Concepts

Lewis Structures

Resonance Structures

Hybridization

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Draw the Lewis structure of azidomethane (CH3N3) [Show two important resonance structures.]

Key Concepts

Lewis Structures

Resonance Structures

Hybridization

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Draw the Lewis structure of azidomethane (CH₃N₃) [Show two important resonance structures.]