and finally let's focus on stereo isomers and there are going to be two types for of geometric and Optical so for geometric isomers to occur that's going to happen when atoms or groups can either be arranged on the same side or opposite side of the compound relative to the central metal ion there's two things that would help us to distinguish whether or not it's s or trans first is going to be coordination number so we're dealing with mainly coordination numbers of four which we see here and it's always going to be that square planer geometry or coordination numbers of six which is going to give us that octaedro geometry also we must be dealing with um Lian groups okay that are identical so we have to have at least two mainly it's just going to be two Lian groups that are identical so we're looking at this Square planer geometry for example here we have two Li groups here that are the same and these two Li groups here and then for octahedral we have this plane here and then we have two Li groups are the same and on this plane here we have these two Lian groups are the same so we have to make sure that we're dealing with coordination numbers of four or six and we have those two lians that are identical now let's go ahead and talk about these structures so in square planer we're talking about around a plane right and so the plane is going to be flat as we see here and so when we have something that's going to be Cy geometry these two identical lians are going to be on the same side that's where the sits comes from and then when we have those identical lians across from one another that same plane for square planer that is going to be train and so this is how we actually get these Cy or trans isomers based on where these identical Li groups are arranged now with octahedral the plane here is going to be vertical you see and then we have our plane across here again when we have identical Lions on the same side it's going to be Fifth and when those identical lians are cross from one another in the same plane it's going to be trained okay so you can use this schematic and understanding for studying compounds and making sure you make these clear distinction and as many classifications as you can of course this is all about you stopping and taking a flash car moment write something small on the front and back about geometric stereo Isom get an example in detail down so let's go ahead and do an example for ourselves which of the following compounds can exhibit Cy trans isomerism so first let's go ahead and gather our information about CIS and Trend we know that we have to have things that have coordination number of four or six and we know we have to have two identical liks so with the approach for this question we just want to go through each compound and make sure that it has four or six um coordination number and two identical liing so let's go ahead and start that okay so for this one we have nickel and unfortunately we only have six of the same Lian so we definitely can get rid of that not four or six next we have our iron and we do have two different lians but there's only one of this o and O attachment so we need two identical ligans here yes we do have the coordination number of six but it w won't work also we have our chromium attached to the co and NH3 and they're both going to be the identical amount and we only need two identical lians okay and then here we have our copper attached to co and there's five of them yes we have two lians but we're going to need more than just this one single one and so we cannot use that one either our process of elimination of course a is the correct answer let's see why well first we have our coordination number of four and then two so that means we have a coordination number of six and we also have bromine here and this bromine is going to be constituting our two identical lians so that's the reason why a would be our choice again sometimes it's about recognition and sometimes just about looking at compounds and saying can this actually exhibit FIS trans isomerism that's really how you want to approach these question next we have Optical is or an anterior and this is going to be coordination compounds that are Mirror Image Arrangements of liing in a chairo environment and it must be nons superimposable structures that are considered optically active so what is optically active Okay an optically active substance are those that can rotate the plane of polarized light mainly to the left or right so that means you have a substance that is optimally active it can be a solution and you add a ray of light through it it can actually polarize that light or bend the light to the left or right okay all based on the arrangements that you see in these stereo is so it's going to be many drugs and molecules that are using Pharmaceuticals um that actually use these chyro and optic isomers that can really affect their biological activity so of course we can use that for research science of the W now one example will be like ibuprofen they have Optical isomers one that's going to Aid in pain relief and also one that's for side effects as well so imagine having that same medicine that's going in your body breaking down and doing two different things all because of the iers that are in it so let's break let's break this all down okay what we're going to usually start with with an octahedral complex okay so octahedral means that's going to have six different liing or six different things attaching to that Central node so as you can see here we have the complex of CR en N2 v+ and so what you would need to do is go ahead and figure out and calculate how much of each of these are going to be contributing to the coordination number for our chromium that's actually going to be a br2 here in plus so you have our chromium and then we're going to have our ethylene diamine liin which we know is B Dente so there's two of them so two * two is going to give us four and then two bromine so that gives us six things surrounding that chromium and so when we actually have our overall octahedral complex we'll have our um CR here and then we'll have that plane here and then of course the vertical axis here as well now with our Ethylene diamine or is by Dente so the donor atoms are going to be these two nitrogens here and that means that they're also going to make a kind of loop with one another which again when it comes to bente or things that are poly Dent they can create those ring structures so you will see a line going from this nitrogen to the other one and they always have to stay that way so that means whenever you rewrite something you have to make sure that you are connecting this nitrogen to that nitrogen no matter how much it moves around so our goal is to take this o theed complex and ask ourselves one question does it have an enan or Optical ISO so how do we do that well we need three things one we're going to need the mirror image two we're going to need that mirror image needs to be nons superimposable and then three it must be HRI okay folks so there the three things we need to figure out if this complex has a stero is so what do we do well let's go ahead and break down each prodct first let's get our Mirror Image now technically since there's going to be three things for me to do I need to make sure that I'm going to have three different structures the first structure is going to be my mirror image and then I'm going to have to create take that same mirror image and rotate it by 180 degrees to figure out if it's nons superimposable and then once I have that overall compound I'm going to compare it to my original compound to figure out if things are chye okay folks but first let's get our mirror Im so I'm just going to draw a line to dictate that this is going to be mirror let's pretend that this molecule is looking in the mirror and so let's say it's right here and it's looking through the mirror let me go ahead and rewrite that better apologies so let's just say we have its eye right here and it's kind of looking in the mirror this way so that means you can kind of Orient this as left and Orient this as right if it's looking in the mirror you're going to have the same structure let's go ahead and write it you're going to have your CR these vertical bonds are going to stay the same and then you're going to have one Bond going in the back of the page this way another Bond going to the back of the page and then one Bond coming out and one Bond coming out so as you can see I kind of just wrote the schematic of the octahedral shape of this overall molecule now when it comes to figuring out what would be in the mirror image anything that's in the front is still going to be in the front so this bromine is here and when you look in the mirror image you kind of face the mirror portion and that bromine is also going to be here this bromine is also going to be here so you see they're in the front anything that's in the back is going to stand in the back so this nitrogen here is going to stand in the back this nitrogen here is going to stand back and again because these nitrogen are connected to one another they also must be connected to one another in the mirror image so we now have our Mirror Image cool that's always going to be our first step go ahead and get that mirror image next we must verify if this mirror image is nons superimposable so how do we do that we take basically this axis here here with the chromium here and we're going to rotate our liance by 180° another way to write that is going to simply just put our vertical axis and then we're going to rotate it 180 degrees and that means every single Li in here is going to be rotated by 180 degrees so what do we need to do we need to first set up our new compound here chromium our vertical um attachments are going to stay the same and then we're going to have have two attrations going to the back of the page two coming out the front of the page and now we need to do our rotations so when you're looking for the non-s superposable image this is probably the hardest part because we need to rotate each and every Lian and every time we make a rotation we need to make sure we are verifying that rotation on the brand new molecule so because we have some of the same liance here we can label this as Lian a and this one as Lian B and all we're going to do is basically again move each of these lians in 180 Direction so for this bromine here is now going to move mov over here okay so it's going from here and it's doing a 180 and moved over here that means we now have bromine a over here let me get a better Mark okay now we have bromine a over there and I'm going to be erasing it each time just so that we are on the same page and there's no extra confusion confusion here because yes we are moving Lian at by L so again we're going to have bromine this bromine moving 180 which is going to be now in this orientation so we have a bromine here next we're going to take this bromine and moving into this orientation again 180 so again it's going from this orientation now to over here for both our bromines are now in two different orientations here next we're going to have our nitrogen here go into a 180 so let me get a different color let me go ahead and erase the previous attachments so now we're going to have this nitrogen here move 180 so it's going to now be in this attachment now because this nitrogen is attached to this top nitrogen here we still have to make sure rewrite that attachment there so even though it may look weird now this nitrogen is in the front and it must be forced to attach to the nitrogen that is in the upper plane here in the vertical part just to verify that we're keeping everything the same all right let's move forward we now have this nitrogen that we're going to move to 180 so this nitrogen is now going to move over here 180 and now we have this nitrogen here let's go ahead and get a different color let erase those boom boom now we're going to have our final nitrogen that we already moved here okay which was this one here and we have moved it over here to this one but remember the connection still has to be connected to the nitrogen down here in the bottom so I'm just going to make sure that I make a connection to this nitrogen okay we have effectively rotated our mirror image and now this is the compound we get now we need to verify if it's nons superimposable so that means I can take this compound and layer it to our original compound here and if they equal to one another then they're seen as superimposable so there will not be Optical Isom but if the connections are different then they are Optical isomers because they're nons superimposable so as you can see here the bromines are the same so bromine the bromines are the same for both of them but the connection here this nitrogen is connected now to nitrogen in the front so this yellow connection to the nitrogens have changed and this pink connection to those nitrogens have changed as well so because of that this structure is not superimposable for this structure so they are definitely optically active we now have our mirror image and we know they're non superimposable now the final thing we have to ask is is it Kye that's how we're going to verify it is it Kye and so to answer this question of is it Kyro we have to understand what it means for these compounds to be in chyro environments so a chyro environment is basically an environment where matters if you have a chiro molecule or if a symmetry exists so think of like a baseball gloves okay there're specific gloves for the right and the left hand and those can be considered as chiro environments because they're only specific hands that can fit into each glove so that means once you have that nons superimposable mirror images we have to make sure that they're also asymmetrical so if I'm looking at this image here now it's to write like a line going through it you would see that either side are not equal so when both sides are not the same they're going to be seen as asymmetrical that asymmetry is going to help us realize that we're dealing with something that is chyro you also will see the other um Optical isomer or an antim here is also asymmetrical now if you do see things that are symmetrical we know that cannot be an optical is if you see things that are superimposable and can layer on to one another and equal look the same we know it's not an optical ising okay folks um so you just want to keep that in mind as we're going along and so because of this a because they're asymmetrical because they're nons superimposable when they're in chyo environments like um the substance and the plane of light going through this the optically active substance of course they're going to display different property okay folks and so you can change the different High environments it doesn't have to just be about a plane of polarized light but that's going to be our Focus for our class so again if you just want to quickly spot if something is chyal just look at the Symmetry because you can actually write a plane going down this mod molecule on either side looks different you know they're not symmetrical they have to be a asymmetrical okay so symmetry is going to be our signal for if the molecule is going to be chyro because sometimes we're not really focused on the specific environment that's in all right so we basically have proven that we have Optical isomer because we have a mirror image we have a non-s superimposable once it's rotated and all the lians are changed and moved around and we know it's chyro or asymmetric let's go on to our last example okay let's take what we learn to another example like which is a pair of optical isers here so it says consider the following octahedral complex structure each involving ethylene diamine in two different uni Dente liance so uni is going to be the same as our monod Dente liance as well okay folks and this question asks which if any of the following is a pair of optical isomers so for optical isomer we know that we have to have that mirror image that is going to be nons superimposable and we know things have to be pyro or asymmetry so what we can do is go ahead and look at our answer choices we have a through D and then we have none of these choices are correct we have to make sure that we are we can prove every single answer when you see this kind of none of these choices are correct sometimes people just be like oh I found an answer let's move on and answer but you always want to ask yourself wait a minute let me prove that I have this answer correct because what if it is letter e for example okay folks so just by looking at these complexes we can start asking certain questions like for example which ones are chyal that means which ones are asymmetric that can help us kind of get rid of certain things so we can see that these are symmetrical they're kind of the same on either side and this one's also symmetrical so no matter what they cannot be Optical Isom so that means anything that has three and four we can get rid of now for letter D it says that we're comparing structure three and for we want to know if these are Optical isomers as you can see they're the same thing so there definitely wouldn't be Optical isomers here because they can be layered so they're superimposable mirror images and they are not pyro next let's look at one and four let me go ahead and erase this and to go ahead and compare number one and four they are not the same isomers because they both are not pyro so you need to make sure they're Cyro again this one's not chyro this one's not chyro so I'm just going to get rid of anything that has one or three and four in it because we know that they're not pyro because they're not um asymmetrical now finally we're down to letters a and letter E and A lot of times people will say well I'm just going to answer it and the answer is letter A but we have to make sure we can prove that the answer is letter A so let's go ahead and look at these structures first are there chyro yes so that's a good thing um how else can we go ahead and look at them are they mirror images well one thing you can do is take your first structure and just write the mirror image of it so I'm going to take my metal here it's attached to nitrogen here in a y it's attached to nitrogen here it's attached to X here attached to nitrogen here and it is attached to another nitrogen here these are going to be um our by Dente ligans now I'm going to do the mirror image so I'm going to have my metal as well our nitrogen and our y groups are going to stay the same all right and as you can see we're just going to create the mirror image so if I'm looking add it this way I'm going to have my nitrogen and my x value or my X Lian in the front let me go ahead and write that I'm going to have nitrogen here and X here okay folks and so this nitrogen is attached to this nitrogen we going ahead and make that attachment we also have these two nitrogens back here that are going to be the same so as you can see structure one and it's mirror image is actually structure number two so structure number two is the image of it so that means we have things that are chyro chyro and it's a mirror image now the final thing we need to do is rotate it 180° to verify that it's going to be non-s superimposable so let's go ahead and do that okay so we're going to rewrite our metal here as you can see we're going to move everything about 180 so this n is now going to be over here okay that means it's still attached to this nitrogen up here so we already Consolidated that one this x is now going to be back here okay so we already got that Ling this nitrogen is now going to be over here where this x is okay and then this nitrogen is now going to be over here with this x is so I'm wri X okay so now the big question is are they superimposable or are they nons superimposable if I take this structure and I take structure number one I see that there are a difference in those connections right even the ethylene dying the um ethylene dying biodent lians here they have different attachments and different arrangements and so that means that yes they are nons superimposable if you were layering on top of one another they would not equal so because they're non-s superimposable they're chyro and we have a mirror image our answer would be letter A that's how we kind of combine all this together okay folks so the coordination challenge for our transition metals is over and they all came in on top with Fierce coordination compounds created so we got to learn about coordination compounds and what structure they can create and then the next part we're going to focus on describing bonding in these complex ions with veence Bond Theory and hybridization and then use crystal field Theory to explain the properties of complex ions like color and magnitude see you there for our next challenge folks