The mechanism for the oxidation of HBr by O2 to form
2 H2O and Br...

Question

The mechanism for the oxidation of HBr by O2 to form
2 H2O and Br2 is shown in Exercise 14.74. (c) Draw a
plausible Lewis structure for the intermediate HOOBr. To
what familiar compound of hydrogen and oxygen does it
appear similar?

Accepted Answer

welcome back everyone in this example, we have gaseous hydrochloric acid which is oxidized by oxygen gas to form water vapor and chlorine gas. We have Cloris acid which is formed as a intermediate for this reaction. And we need to draw a LeWIS structure for this intermediate Cloris acid. So we need to recall that before we can draw our lewis structure, we need to calculate total valence electrons. So looking at every atom in our Cloris acid, we can also write the formula as H 02 Cl. So sorry, that should not be cl two. There we can begin with our single atom of hydrogen which we should recall is in group one A corresponding to one valence electron. So let's make this clear, we have hydrogen where we account for one valence electron moving onto our oxygen atoms, we would count a total of two of our oxygen atoms which we will multiply by recognizing that oxygen is located in group six A. Of our periodic table corresponding to a total of six valence electrons. So we would multiply these two atoms times these six valence electrons, which would give us a total of 12 electrons contributed by oxygen. And then we have our adam chlorine in this structure where we recognize on our periodic table that chlorine is in group seven A. And that would correspond to seven valence electrons. So we would have seven valence electrons from chlorine. So taking the total of all of our electrons here, we would count a total of 20 valence electrons total for our lewis structure. Now we want to recognize which would be our central atom and we can see that we have these two oxygen atoms between hydrogen and chlorine. And so that means that these two oxygen atoms are going to be our central atoms surrounded by hydrogen and chlorine. On the other end. Now we want to begin by making our most stable lewis structure by recalling upon bonding preferences. So if we recall the bonding preference of oxygen which is to have two bonds and two lone pairs on itself, we're going to be able to connect each of these oxygen's to first the hydrogen atom and then to the other oxygen atom here for our first two bonds and then we can add its preference of two lone pairs so that this oxygen is happy and stable. And we can see that in these two covalin bonds we would count for a total of one valence electron, meaning we would count these 123456 valence electrons on this first oxygen, which is why it's neutral, happy and stable because it's bonded according to its bonding preference and has all of its valence electrons surrounding itself accordingly. And so we would do the same thing for our second oxygen atom here. We would make a second bond to it to attach it to this chlorine atom and give it its two lone pairs as it prefers. So that it has a formal charge of zero and is stable and happy. And so right now we can count a total of 246 because we recall each single bond contains two electrons. Then we have 789, 10, 11, 12, 13, 14 of our 20 valence electrons used. So we're going to subtract that to get our difference, leaving us with six electrons left over for our structure. And now we want to recall the valence electrons of chlorine which we state it is seven valence electrons because again, it's in Group seven A on our periodic table. Right now it has a covalin bond with oxygen where it has only one of its attached valence electrons to itself. So we can fill in the rest of our electrons as lone pairs on this chlorine atom. So we would count again that one valence electron here and then we would have 23456 and seven. So we have three sets of lone pairs and adding these three sets of lone pairs. Used up our last six electrons here. So we would just subtract that out and we have zero left meaning our structure is complete and we can confirm that because each of our atoms are bonded according to their bonding preference and valence electrons that this is our most stable and best lewis structure. So this entire structure would be our final answer to complete this example. I hope everything I reviewed was clear. If you have any questions, lead them down below and I'll see everyone in the next practice video.

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