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Which one of the following has an optical isomer? [2010](a) $left[mathrm{Zn}(mathrm{en})left(mathrm{NH}_{3}ight)_{2}ight]^{2+}$(b) $left[mathrm{Co}(mathrm{en})_{3}i...

Question

Which one of the following has an optical isomer? [2010](a) $left[mathrm{Zn}(mathrm{en})left(mathrm{NH}_{3}ight)_{2}ight]^{2+}$(b) $left[mathrm{Co}(mathrm{en})_{3}ight]^{3+}$(c) $left[mathrm{Co}left(mathrm{H}_{2} mathrm{O}ight)_{4}(mathrm{en})ight]^{3+}$(d) $left[mathrm{Zn}(mathrm{en})_{2}ight]^{2+}$$(mathrm{en}=$ ethylenediamine $)$

Which one of the following has an optical isomer? [2010] (a) $left[mathrm{Zn}(mathrm{en})left(mathrm{NH}_{3} ight)_{2} ight]^{2+}$ (b) $left[mathrm{Co}(mathrm{en})_{3} ight]^{3+}$ (c) $left[mathrm{Co}left(mathrm{H}_{2} mathrm{O} ight)_{4}(mathrm{en}) ight]^{3+}$ (d) $left[mathrm{Zn}(mathrm{en})_{2} ight]^{2+}$ $(mathrm{en}=$ ethylenediamine $)$



Answers

Balance each of the following:
$$\begin{array}{l}{\text { a. } \mathrm{Zn}+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Pb}+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}} \\ {\text { b. } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O}} \\ {\text { c. } \mathrm{Al}+\mathrm{CuSO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{Cu}}\end{array}$$

This question is asking us if three different metal complexes are Cairo and we start out by looking at this tetra he'd roll, uh, think based farm complex. So we're going to draw this out as think where we have one thing coming out at us and one thing going away from us, we have, uh, quit our voyage to oh, age to Corinne. Cory, the way that we know something is Cairo is whether or not its mirror image is super imposible or not. Uh, and so we'll draw that mirror image Come quarrying queen. Oh, it's too wage to you. And from this mirror image, if you were to take a modeling kit which I highly recommend using and looking at the plane that goes through but the O. H two and the metal then rotate about that access you could get back to this this structure from this mere image indicating to us that this molecule is not Cairo for a part too. We have, uh, this octave. He drill trans complex where if we draw yes, we start with go that are you down? Then we have, uh, about us. There we go away from Les away from us going out at us now. This is the Trans I simmer. So that's gonna put these 50 which I'm gonna draw using just line going across this because they are on the opposite sides of each other again. The Korean on opposite sides of each other as the trance structure. And if we draw the mirror image of that again, are you two note the most again we have. Yeah, 50 Corinne's. Yeah. And these two are identical. And so this a molecule is not Cairo. Uh, part C is the same as part B, but it's the system by summer. And so we're gonna start by drawing that, Are you? Do you know we have the bitch be running from here to here? That the running from here to here? Chlorine. Corinne, if we draw this mirror image, are you? Yeah. Look where we have the chlorine here courting here. And then we have That's baby. Unless and again. I highly recommend using a modeling hit for this because it will be easier to visualize where these two molecules are no longer super imposible Because of the directionality of this by don't eat lichen here and here. And so this molecule is Tyrell

So here we have the problem. 23.46. This is determining if each of the following medal complex is Cairo and therefore, if it has an optical I summer. So first we want understand Cairo Karaleti is the property of the compound in which the eye summers are not super impossible. So non stupor impossible. All right. And the optical ice summers are the compounds in which the arrangement of the substitute in the three D dimension is different. And they had a saying chemical formula. However, there mirror image are not super impossible to each other. So the 1st 1 we have part A. We have a square planner, P e N cm, too. All right, so we're draw the molecule into a three d model. It will look like this. This is the end, See? And and see it. This is a planner, and it does not have any Carol. So this is a no for party part B. We have the monk, you octu. He'd roll Nicole heat and in a 34 two plus so draw this complex into a three d model. It will be anedge three legal on a three on three on a three. And this is the ethylene thiamine right here. All right, so for this monkey, this is the only possible structure. So for this one, the answer is also no. It has no Carl and the optimal ice over part. See, we have a hunk to hear. Draw, sis. Well, a dea ea n to see l br. All right, So for this molecule, draw it into a three D model. We have center Adam Palladium and and this is ethylene Dia me Another up early. I mean, c l br. All right. So for this molecule, it does have a non optimal by summer. And the optimal ice armor for this molecule will be three. And and and then he and yes. So this is this molecule is Cairo, and it has optical isil er, which is this one?

And this question were asked to explain whether a complex exhibits geometrically Someone call for your definitions. A geometric Osama riz um, has Liggins bonding to the metal with different special or entrance. So for simplicity, let's see, we were looking at a square plainer complex with the general formula of and a two be too. Okay, then we would have a medal in the center connected to a CZ and beats again. Let's keep it simple and say that this is a square plane or complex. So makes bonds like this. And I have two options. I can either make my A's what we call trance to each other there across from the metal, which means that the bees are also trance to each other. And then my second option, I can have the A's and what we call a cyst position, where their next to each other and because the A's are nice to each other, the bees must fill in the remaining spots like so we call this sis based on the connection between the ace, the bees air also says to each other in this other compound we have the bees is trance and the phases transfer as well. Um, so these air to geometric ice tumors, Um, because they have the same elegance with the Legans have different special rate for agents. Now, one question that comes up often is whether following is also a geometric. I simmer, let's say of the metal in the center, and I have Hey, here on the quote unquote top. He's on the corner of growth sides. Now, the question is, are these two geometric eyes and ears of each other on the answer is no. Although we can draw them differently. There, actually, the same representations, different representations of the same compound because you could take this compound complex. You're on the left and rotate it 90 degrees. You draw. Really? If we rotated 90 degrees, we're gonna get back, sir. Visual complex. So this is not a new I simmer. It's just a rotated version of this versus so this particular formula Emmy to be too, and a score planner. Geometry has thes two ice climbers, and that's the basic idea that we're gonna use to look through that each of these following problems important, eh? We're given cobalt die aqua to ox awaits. Now this Bo X represents, Um, and Oxo Late Ligand, shown here on the right. The important thing to know about the Aqsa Late Litigant. It's that it's relatively short. It binds at the's oxygen sites. And because this molecule is very small, the distance between these this small. So this is what we call a short Legans and this'll. Ligon will always be, ah, bond at six sites because it's short so way can only ever pick up their assess. This looking. So that's drawn with this compound complex. More look like we've got the cobalt in the center and we have eight on the insights. So these are all in the plane, and then I have this site coming out of the plane. This I'm going into the plate, and so my options are to look at the water. The water can either be trance or cysts. Let's start out drawing it trance. Then that means that the Aqsa Late litigants have to be in these he pictorials. It's and they had to be connected. In this way, I would not be able to connect these two sites with an ox, a late Liggan because those would be trance to a joke, so that's not an option. So this is war. Nice summer in my alternate ice, summer would be tohave the waters sis to each other, and the way that that might look can cobalt in the middle. Here's my four sites in the plane, this one coming out of a plane that was going into the plane. I would draw the water one of the waters here and here. So now notice that these are cysts to each other, whereas before this, I somewhere they were trance. And when I fill in the remaining Aqsa Lates, I would put them in these of these sites, and this would be my complex. So there's two Different is a MERS, one that's trance has the water like in strands, one that has the water, like insists, and you might try to draw them differently the way that we did with the example with the different rotations. But what you'll find is these were the only two options do this imagery, even if you were to draw, um, for example, the connection between here and here and here and here, even though those Airtran's those were still represent the same my summer So really, this is for hold me option. So the answer is, does this complex have geometric? I saw more ism. The answer is yes. It has two geometric I summers. Next up, we've got, um, cobalt with three s. Linda. I mean hooligans like the ox a late. This is a relatively short ligand. It binds at thes nitrogen tze and because it's short again, it always always appears as sis. The Ligon binds at two sites that versus to each other. And no matter how you do this, you will find that there is the single I simmer and it looks like this again. You might connect the Liggins a little bit differently, but as long as they're all sis to each other, you will find that there is exactly one way to represent this leg. Now I've been drawing the Octa hydro complexes with, um with these straight lines showing you the four sites that are in the plane and then one site that's coming out of the plane on one side that's going back another way that this might be showing. And I think this is the way it's shown in your textbook is with these two sites in the plane and then showing the equatorial ones with two sides coming out of the plane towards you and thes coming into the plane away from you. That's another way to draw it. I was gonna represent this this way. Then I would have thes sites here, um, this forward one connected to the bottom one and finally each other sites here. So these air equivalent, um, I like using the representation of Alaska's is a little bit clear that your book might use the representation, right? They represent the same thing moving on to part. See, we have this complex. So we've got three different Legans again. We've got this Aqsa late here that is always going to be says as we discussed before. And so we have a few options. We have the combination where the waters for the waters resists the combination where the ammonia sze groups persists. And then the combination the combination where they're both trans. Let me show you what that looks like. So first, let's have the water Ligon species to each other. And I look something like this again. You can rotate tomato molecule in any which way. Oh, But this is one of the options, and what will happen is the ammonia is we can be trance to each other. So actually, perhaps a better way to think about this is to consider the combinations. Were these air trance? Forget me. So let's consider the option where the water molecules or trance to each other. Then that I said it would look something like this. There's no way to put the ammonia Sze also trance to each other because this ligand, the absolutely and has to be sits so limits have water. I look like next up, we can consider the option where the A morning of groups or trance to each other. But then this ice rumor looked like this with members trance. And then finally we have the options with both groups are found says to each other thes air the only three ice rumors that possible again. If you think you can come up with another one, I challenge you to rotate it until it looks like one of these. So to answer the question um, yes, this complex sacks geometrically summer for the previous one. The answer was no. There is no geometric. Hi, summary is, um, now moving on to a party party, Uh, has four legions first. Excuse me. Three Liggins. One of them is in ethylene diamond, and it's a D ate complex. So that's how you know that it is square plainer. So we're gonna drive with these four sites again. Remember that, um, the ethylene die mean must bind to cece sites. You might be tempted to include this Iceman. It may be possible for long religions, but not for ethylene dyke. So again, this complex would be possible if I was working with a longer leg end. But that's a lead. I mean, is not one of them. So this is not an option for this particular complex. Which is why there is a single hi summer. And there is no geometric. I saw. Marie is, um, finally, we have this complex and party nickel with two carbon eels and two chlorides. Um, these were the only two possible litigants again, you might be tempted to draw rotated version of one of these like this, but what? You'll find this. If you rotate this leg in 90 degrees in it in one of these directions, then it's just gonna be equal to this later. So, yes, there is your guy Summers here, and there's two ligaments that are possible.

Here for the solution For this is Dad. The Onley spiral complexes exerts as up and Sumers here for the F e C 204 hole size. All three negative exist as a and Anne Sumers heard the structure for this. Yeah, for the part B. Uh huh. This is the structure for the part B and with the c E o S E N and three whole 43 Positive has a symmetry plan and a grill, so it has no and anti on Mars. This is for the structure on Now, for the part C C O N s three to hold twice as three Old wise in three Positive has an Carola structure and exist as an and NGO Mars. This is Kyra Structure. Yeah. Now for the party plan off symmetry for the CRS duo Foreseeable, to have a plan off symmetry that is this and Onda So it is Ah, a Kriel enhance no and yeah and then to Mars present. So all the options are explained with the structure in which only Carol complexes exist as an and an NGO Mars. For this part a is an an informal ERISA Carol structure and Barbie is not an Isa Mars. It is symmetry plan on part C then And I am so much because her style structure and the party at the somatic plant So it is not and and and summers So this was the explanation of the question. Go through this. Thank you.


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