(b) With what allotrope of oxygen is it isoelectronic?

Question

Accepted Answer

welcome back everyone in this example, we're asked whether the a side ion and three minus and the tri iodide ion I three minus our electronic. So we're going to recall that we need to draw out the lewiS structures in order to determine whether these two ri so electronic And remember that electronic means that we should have species with the same number of electrons. So beginning with our side we're going to calculate total valence electrons before we draw a structure by recognizing that nitrogen on the periodic table is located in Group five A. We have three nitrogen atoms And because we recognize that it's in group five a. We recall that this corresponds to its number of valence electrons being five valence electrons that we multiply by. And so this would give us a total of 15 valence electrons. But because we have that minus one and I in charge we should recall that we gain one electron. And so when we take the total of these two things, we're going to get a total of actually 16 valence electrons total for a lewis structure. So based on our formula, we have three nitrogen atoms, recall that nitrogen has a bonding preference of having three bonds and one lone pair. So we can use that for one of our nitrogen atoms here. They can have three bonds to the central atom and one lone parent itself That uses a total of 24678 of our 16 valence electrons. So we would say That we have eight valence electrons left. And so we can use those last eight electrons as a single bond between the central atom and this third nitrogen atom where we can use the rest of our valence electrons as lone pairs on that third nitrogen atoms. So we would have one lone pair, a second lone pair and a third lone pair here. Now this is against nitrogen bonding preference and the central atom is also against nitrogen bonding preference because it has four bonds. So we would see that nitrogen in the center will have a formal charge of plus one because we see that it has a total, sorry, a total of 1234 valence electrons in Covalin bonds. And it prefers again to have five valence electrons to be stable. So that is why we have that plus one formal charge. Whereas our nitrogen here on the right has a total of 1234567 valence electrons directly attached to itself, which is two extra than it would prefer to have being five to be stable. And so that would contribute a minus two formal charge on this third nitrogen atom. Now, because we have these two extra lone pairs on this third nitrogen atom And this -2 charge. This negative charge will contribute these two extra lone pairs to residents to form a second arrangement of our structure. And so we can say that we have a second resident structure where we would have the formation of our three nitrogen atoms which are all connected by double bonds where the outer nitrogen atoms will have two long pairs And formal charges of -1 since they will have six directly attached valence electrons, which is one more than they would prefer. And the central nitrogen atom will again have a formal charge of plus one because it still has those four directly attached valence electrons shared in covalin bonds between the other nitrogen atoms, which again is one less total of valence electrons than it would prefer to have being five to be stable. So this would complete our resonant structures for our lewis structures of our azide ion. And now we want to compare this to our tri iodide ion. So that's I three minus and we'll use a different color here. So it's more clear. So we use purple for tri iodide And again we want to calculate total valence electrons. So we recognize that we have three I died atoms which we will multiply by the valence electrons for Iodide, which we see because I died is located in group 78 on the periodic table that corresponds to seven valence electrons. Again we have that and I in charge which contributes an extra one valence electrons to our total and calculating everything together, we're going to get a total of 22 valence electrons. Now we can already see that we have a total different number of valence electrons. 22 is not equal to our calculated total of 16 valence electrons for a side. So we can already agree that this is not going to be ice Elektronik. But let's continue and draw the lewis structure to see if that changes anything. So to jar loose structure, we have three aydin atoms surrounding each other. We're going to make our base connections by creating single covalin bonds between each of these in atoms And recall again that as we stated, iodine should have seven valence electrons. So right now each has one valence electron in a co valent bond in our Lewis structure where we can fill in the six extra electrons around our outer aydin atoms. So we would have two or three lone pairs around the outer iodine atoms where we would count a total of 1234567 valence electrons for a stable formal charge of zero. For this first outer iodine atom and then following the same pattern for the second outer iodine atom, we would have three sets of lone pairs which would again give it a formal charge of zero because it has seven directly attached electrons in the structure. Now this would use a total of 24, 6, 8, 10, 12, 14, 16 of our 22 valence electrons. So we're going to subtract 16 which would leave us with six valence electrons left where we can use these last six valence electrons as lone pairs on that central iding. So we would have three lone pairs on the central iding. Which would definitely violate our octet rule. But recall that again, hiding is a period five elements on the periodic table which is below period three and anything below period three is okay to violate the octet rule. So calculating formal charge, we would see that we have a total of valence electrons, which is one more than it would prefer to have to be stable as seven. So that would create a minus one formal charge on this central idea. An atom And so this would be our completed lewis structure for tri iodide. And as we stated, try I died an ion and our azide, an ion have different number of electrons and so therefore they are not I so electronic. And so this would be our final answer to complete this example. I hope that 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

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