Draw a Lewis structure for each of the following:
b. HN...

Question

Draw a Lewis structure for each of the following:

b. HNO₂

Accepted Answer

Welcome back, everyone write an appropriate Lewis structure for the species C H three N 02 or nitromethane. Our first step whenever we want to draw a le structure for a compound is to calculate the number of valence electrons present in our structure. In this case, we have one carbon which contributes four V electrons, we hyen atoms, each of them contributes one electron and then nitrogen which contributes five ions electrons, two oxygen atoms which contribute six von selections. Each we know these numbers from the periodic table, right? Because the number of ions electrons is found from the group. A number well done. So if we do the math, we end up with the total number of ions electrons, that would be 24. Our next step is to usually assign a central atom. But in this case, we have an organic molecule, we will have two central atoms, carbon and nitrogen. Due to the fact that there are two oxygen atoms which are more electron than nitrogen and hydrogen atoms, which can only be bonded to carbon. In this case, to satisfy the Octa rule. In addition to that, because it's an organic structure, it sells us the bonding in, in it. So we would be a metal group and then we have N 02, which is a nitro group. So carbon will be directly bonded to nitrogen due to the fact that carbon must have four bonds, we already have three hydrogens bonded set. So definitely we will have ac N bond in the structure. Our next step is to draw single bonds between our central atoms and terminal atoms. So what we're going to do here is simply add the remaining atoms. Let's draw our C M bond. Then we're going to draw three hydrogen atoms bonded to carbon. We know that because it's given under the name C three right and nitrogen will have two oxygen atoms. So we're going to draw two single bonded oxygen atoms for now, our fourth step is to satisfy the octet rule for the terminal atoms. In our words, we are not worried about the central atoms yet. We just want to make sure that we add our electrons to the terminal atoms, hydrogens follow the rule. So we don't need to worry about them. We simply want to add we alone pairs to each oxygen because each oxygen already has two electrons from the bond. Now, our next step is to calculate the remaining number of valence electrons that are not assigned. So first of all, if we calculate the number of electrons present, we have 123456 bonds. So that'd be six multiplied by 12. First of all, let's write down that the number of electrons unassigned and we are going to take our original number 24 minus, as we said, we have six bonds. So that'll be six multiplied by two. And then we're going to subtract our loan pair electrons. So that's another six multiplied by two, which gives us no electrons unassigned. Meaning none of our central atoms will get any lone pair. But now we want to make sure that we also satisfy the Octa rule or our central atoms, carbon has it has four bonds, nitrogen has three bonds only. And that means we're going to take one of the lone pairs from oxygen and form a pie bond out of it to make sure that nitrogen gets an a because any of the two auction atoms can be used, we will of course get resonance structures well done. So we got an Octa on nitrogen. And of course, we can show that we can do that two ways. One of the ways is to use the lone pair from one oxygen. So let's modify our loan pairs in a better angle. And then another structure can be obtained by using that loan pair from the other oxygen. So we are simply going to remove that long pair and draw a double bond. And of course, our next step is to calculate the formal charges. Now, carbon has four bonds, meaning it will have no formal charge each hyen has one bond. Of course, no formal charge. Now, oxygen and ngen this is where we have to pause and think so. Step number seven is to recall the formal charge formula, the number of ions electrons minus the number of bonds minus the number of long pair electrons starting with nitrogen five because it belongs to group five A for bonds, no long pairs. That gives us a formal positive charge, formal charge on the single bonded oxygen. You may already guess that'd be six minus one bond minus six lump pair electrons that's negative one. Apparently, the structure has no net charge. So the other auction will have a formal pos a formal charge of zero. Apparently that's true because there are two bonds and two loan pairs. So there's no necessity to worry about an additional calculation, but you of course, can perform it if you want. But now we are ready to add our formal charges and get our final answer. So if we want to draw a Lewis structure correctly, we are first of all going to add our resonance arrow once again. And then we are going to add a formal positive charge on nitrogen in each case because it has four bonds and no lone pairs. Of course. And now each single bonded oxygen will have a formal negative charge. Well done. We have successfully obtained our Lewis structures for nitromethane. Thank you for watching and I hope to see you in the next video.

Draw a Lewis structure for each of the following:
b. HNO₂

Key Concepts

Lewis Structures

Valence Electrons

Formal Charge

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