 Hello guys, good evening Yeah, okay. Just what have we done last class just one minute will start. Do you have any exam going on the school? Yes, sir. We have two semesters It's going on Yes, sir Chemistry it is done or you have to write Next yeah, what is the portion Full organic Yeah, and solid state and coordination And DNF DNF blog is also there So DNP So last class we had discussed Okay, so last class we had discussed about the various types of unit cell. Okay FCC, BCC, HCP also we have discussed and we have also discussed the arrangement of atoms like ABC AB type arrangement We have discussed location of void also we have done So today we are going to see the ionic crystals this part actually you have to know memorize because You know the position of cations and anions you have to keep in mind Because based on this only they frame questions in the exam. So Our understanding is not very much Required over here. If you know the position suppose any cell the example I'm considering If you know the position of any plus and cl minus in any cl crystal based on that you can do all the questions Okay, so here you have to just memorize the position of Cations and anions. I want to show you some questions here. Okay Try this question first This one done No, it's not That yeah No, it's not that also Richard Yeah, Richard. You got it right. Richard's saying you got it right Richard Parik that is not correct See this question why I've given you here that the kind of question that they ask here You should know the position of these voids and that we had discussed already last class. Yeah, that's correct Pradeep Yeah See answer is c here a b. Okay Let's see. I'll do this A compound is made up of particles a and b a forms FCC packing. Okay, so when it says FCC packing, it means a is forming FCC packing or they can also say ccp both are same thing Okay FCC or ccp packing it means This a is present at corners and plus face center corner plus face center B occupies all octahedral void So b is present in octahedral void and we know the position of octahedral voids That is there at the body center plus at the edge center Okay, so you imagine the cube where the body center we have b and edge center We have b again if all the particles along the plane as shown in figure Below are removed now this plane. You see how this plane looks like This looks like you see the room in which you are sitting in Any you see any top edge you take anyone top edge and you come, you know, you consider a plane Which passes through that edge and it comes diagonally opposite to the another plane another edge Opposite side So from the top to bottom diagonally you have to come like The edge that you have means what you consider this edge and draw a plane like this Along this edge So this plane you see this plane obviously it passes through the two corner atoms here Two corner atoms here and these are the face centered atom Yes or no Of this face and this face And along with this face center it also passes through the body center. Yes or no Yes correct So if you imagine that plane from the top edge The diagonally opposite edge the plane that is coming down. Okay. This plane is Passing through the body center Right and the two face center two corner atoms we have here two plus two four corner atom here So could you tell me how many a atoms have been removed? How many a atoms have been removed? Four corner a and two face centered a have been removed Right so Out of six face center two have been removed So the contribution of a if you see here that would be Two face center have been removed. So four into one by two And four corner atoms have been removed so four into one by eight. Is it correct? So it is So two and half Three by two is it? Sorry two and half means five by two Correct if you talk about be here So what do we do into? Sorry come come again. Yeah, we do face center, right? Yeah, two face center have been removed your your voice is So it should be only two into half only two no face Two into half has been removed. I am considering the contribution Okay, okay. I'm considering you what you do you can consider two into half and then you can subtract right? I'm directly can you know considering here the contribution of a in the unit cell Okay So out of six face centered a atom two has been removed So I so we are left with only four face centered and four corners. Is it fine now? Yes, right so five by two is the contribution of a in the unit cell Similarly the contribution of b is what two edge center has been removed and one body center means three B B has been removed right so out of 12 edges Right out of 12 edges Two edge center has been removed. So 10 is left Contribution of a center is one by four That is it because body center has also been removed. So it is what it is again five by two. You see we are getting So the formula is what formula is a b? understood clear So this is the easier way. I think well, I usually do like this only I just count the contribution of atoms in the unit cell and the ratio finally Otherwise you can also count Whatever the contribution we had initially and after removal Like like what part you have been removed subtract that you'll get the answer in that also But this way you can do it in one single step. Is it clear? Right, so question is not difficult Why I have given you with this question so that you can understand that how they use the concept of voids here If you do not know that the voids are placed at the edge center, you won't be able to do this question Understood I'll show you two more question here. I don't want you to solve now. We'll do this question a bit later But just see these two questions This one we have done You see this one question don't try this or we'll discuss a bit later You see here in this question what is given In a solid having rock salt type structure now the question is if you do not know what is rock salt structure You are not able to do this question right Rock salt structure if all the atoms touching the body diagonal plane are removed So first of all, you should know the rock salt Structure then only you can do this question. So for that we are going to discuss the different types of structure we have Okay, one more question. You see here We'll do this question a bit later. Okay, just now you let it be this one this question you see It is also based upon the structure of csca And that's why I said that you just need to memorize this If you do not memorize this logically you cannot do this csca has bc arrangement You know, what is bc arrangement partner plus body center? But if you do not know at body center, what atom are present and at corners, what what atoms are present You won't be able to do this question again. Okay. So all these questions depends upon the Structure, okay, so you should know all these structures So any structures any crystals We have discussed we have started last class and I've given you the radius ratio Radius ratio for triangular void. Okay. You use by the with the help of that with the help of Um, what I said the geometry you can find out the radius ratio for different uh find of for different voids also like tetrahedral We have we have octahedral void. We have cubical void all these things you can find out But I'm giving you the data because the revision they have never asked. Okay. So you should know the relation here So write down this table. You draw all of you. I'll give you the entire thing here in this table Okay, just a second This is the radius ratio And then we have coordination number arrangement anions around cations And we have some example for this Okay, so The radius ratio is 0.155 to 0.225 Radius ratio means radius of cation divided by the radius of anion Which we can also write it as rc c stands for cation R a a stands for anion if this falls in this range Then the coordination number is three And the arrangement of anions around the cat around the cations are Triangular which we also call it as planar triangular And we get triangular void here b2o3 is an example Example is not important to this If the radius radius ratio falls in this range So one more consider one more 0.225 to 0.414 Coordination number is four Arrangement is tetrahedral Example is this zinc sulfide If it is from 0.4142 0.732 Coordination number is six Arrangement is octahedral These two are the important one tetrahedral and octahedral NaCl Example If it is 0 point From 0.732 To 1 Coordination number eight It is cubic That is body centered And it is cscl The Example we have So this radius ratio just you have to keep in mind here Copy this down The lower limit you must take care of To solve the question You can equate the lower limit To solve the question means You can equate this to radius ratio should be equals to 0.225 Should be equals to 0.414 you can take So lower limit you can equate In all these range did you finish it? Now the structure of crystals heading you right down Structure of crystals See we are going to understand We are going to take the reference of three types of compounds. Okay The first one is the simple ionic compounds, which is AB type And then we have AB2 type And next we have A2B type These three types of structure we have Apart from these three we have other structures also that we are going to see Okay, so first write down The first one Structure of Structure of simple ionic compound of AB type With NaCl type it is AB type structure AB type structure. We also call it as rock salt at rock salt or NaCl type structure rock salt Or NaCl type structure example we have for this We can have oxides or oxides and sulphides of of group two metals Of group two except except BES Right halides of silver halides of silver except AGI Have this type of structure Halides of alkali metals also we can consider like we have NaCl. It's there only NaCl we can consider Right, we can also consider some examples like CaO All these molecules has NaCl type structure Okay, now what is important here point wise as to write down these First point you write down in this That all I'm taking the reference of Na plus, okay If the you know the statement which I'm giving you for Na plus This will be true for all cations in these structures, right So write down all Na plus All Na plus ions It's surrounded by surrounded by six Cl minus ions And Each Cl minus Is also surrounded by surrounded by six Na plus ions Right six Na plus ion That's why the coordination number you see for each ion it is six over here Write down the coordination number of each ion is six over here Coordination number is six and hence this type of crystal. We also call it as six is to six arrangement important This question also they ask in the exam. What kind of arrangements is there? Six is to six arrangement. Okay, so first we'll take First we'll take the you know the coordination number of Coordination number of Cat ion here we write it first and then we'll write the coordination number of an ion Six is to six arrangement. We have next point. Cl minus ion forms Forms ccp lattice. What do you mean by ccp lattice? Face center plus corner Okay, forms ccp lattice in which In which all octahedral void In which all octahedral void Is occupied by Is occupied by Na plus ion Ccp lattice In which all octahedral void Is occupied by Na plus ion I want you to tell me I want you to tell me The number of cl minus present in the crystal one unit set Number of cl minus And number of Na plus Number of cl minus and the number of Na plus because you know the position of cl minus and Na plus Hence you can easily count the number of cl minus and n plus is it four? Yeah, so cl minus occupy Form ccp lattice means at corner so eight into one by eight Plus face center six into half So we have four cl minus present Na plus one at the body center right since octahedral void and 12th center we have one by four is a contribution So it is four Since Na plus and cl minus we have equal in number. That's why the molecular formula is Na cl and hence the structures Right should be equal. They don't need to satisfy the crystal Okay, that's one thing now if you look at the position of all these ions, so What we can write suppose all these are cl minus Right it is present here diagonally opposite This is cl minus Okay, and here also we have two cl minus. I'm just drawing one face here Oh One second This is the face Of the unit cell right all these are at center, right? So here we have Na plus ion present cations Here we have Na plus ion present cations. Here also we have Na plus ion present And here also we have Na plus ion present Okay, so if you look at this distance If you look at this distance, this is the edge length So edge length is a So if I ask you to write down a in terms of radius, what would you write? A is equals to we can write r plus the radius of cation this distance plus To r minus the radius of anion, which is this distance. Yes So the radius of cation plus the radius of anion r plus plus r minus Is equals to a by two If I ask you a in terms of the radius of anion, you can think of this distance This is a face diagonal. Yes or no This is the face diagonal so we can write for r minus is equals to Root under two a Okay, a we know in terms of r minus And hence any one thing if you know Out of this three r plus r minus a you can find out all the three here with this relation Clear no doubt Yes, tell me Correct. Now if you have to find out the packing fraction Of the crystal Right method is exactly the same You need to find out the volume occupied by the atom here. We don't have atom, but we have ions So volume occupied by the ions divided by the volume of the unit set, okay Usually they don't ask this question, uh, you know to find out the packing fraction But you should know the method how to do it Okay, it's not that tough. I'll just do it for this one for all other crystals We are not going to discuss this thing again. Okay, because it is not that important But you never know it's there in the portion. So what they're going to do in the exam? You never know So how to find out the packing fraction of the crystal of any seal here? So again the packing fraction would be equals to The volume occupied by by na plus plus volume occupied by cl minus and this whole thing is divided by The volume of the cube that is aq Now we have four na plus so volume will be four in two four by three pi r plus cube plus four in two four by three pi r minus q divided by aq R plus r minus all those ratio Relation you have already I've given you in the previous slide. Okay based on the data you can find out this thing, right? So expression would be this four in two four by three pi r plus q Plus r minus q radius of anion divided by aq You can solve this you'll get the packing fraction. Is it clear any doubt? We wrote an extra r cube Sorry Oh, I have written extra This one So it is four by three Pi Okay, structure the crystal of na cl. I am not drawing. Okay, you know the position of na plus and cl minus Okay, n crt all the structures are given you can see that easily you can understand Okay, all the examples of rock salt type structure. I have done. They all have this kind of arrangement But here what we get a specific value of r plus and r minus i think terms of a we won't get a range, right? Yeah, we'll get a specific value of that anion because here we have na plus no So na plus r plus value will be fixed r value will be fixed 60 minus r value will be fixed So here see the range is what to what extent the void should be there and the arrangement is tetrahedral or octahedral Okay, for that we have that range, but here we'll have the certain value specific value of r plus and r minus Okay, right They won't ask you this thing to calculate the packing fraction of ionic crystals Normally, they'll ask in our covalent compounds where the r value is same So you don't have this r plus and r minus present over there We have only r hence the calculation is easier over there that we did last class Okay, but if they ask the method is this one Okay, so this is one type of crystal. We call it as ab type ionic compounds example is na cl The another one write down The second type of crystal we have it is cesium chloride type cs cl type structure write down The arrangement for this ions are bcc type bcc arrangement we have in this cs plus cs plus ion present present at the body center body center and cl minus at the corners At the corners, okay coordination number for each ion write down Coordination number for each ion is eight ion is eight Hence it is called eight is to eight arrangement Eight is to eight arrangement The number of cs plus and cl minus if you count it is one only The number of cs plus ion is equals to it is there in the body center So one the number of cl minus there at the corners eight into One by eight equals to one hence the formula is csca Okay, since the Ions are there at the body center and corners. So along the body set body diagonal if you see The atoms along the body diagonals are in contact So body diagonal is root three eight Which is equals to two r plus Plus two r minus equals to root three Example for this type of crystal we have We have cs br cs i We can have t i c l t l c l t l br etc There's another type of crystal here The third type we have here Coppins down third type of crystal it is zinc blend type ZNS type We also call it as sphalerite type structure ZNS Write down it exists in Two different forms it exists in two different forms the first one is zinc blend a And the second one is wurzait structure Wurzait structure Write down first one write on zinc blend it has It has FCC or ccp arrangement FCC Ccp arrangement The sulphide ion s2 minus forms ccp arrangement means corner and the face center And z n plus two Z n plus two occupy Occupy half of the Occupy half of the Of the tetrahedral white TVs the number of Sulfide ion if you calculate It is again four because it's ccp for z n plus two Half of the tetrahedral white There are eight tetrahedral white we have two into n last class we discussed Since it occupies half. So this is also four and hence the formula is ZNS Next is each zinc ion Zinc ion it's surrounded by Surrounded by four sulphide ion four s2 minus ion and And vice versa means each sulphide ion is also surrounded by four zinc ion here So here the coordination number is four for each ion And hence it is four is to four arrangement All these are informations basically and this question they ask in the exam Example for this kind of arrangement Example is not that important a g i we have B e s we have this kind of arrangement Okay, cu i Etc One note you write down the bond between zinc and sulphur The bond between The zinc and sulphur has significant has significant Covalent character Like I say z n plus two present in the half of the tetrahedral white it's what it is present in alternate tetrahedral white Okay The second type of structure we have in this that is Bursite structure One small difference we have in this second structure of this Bursite structure Write down it has hcp arrangement It has hcp arrangement so s2 minus it again adopts hcp arrangement and z n plus two occupy again here half of tv's Half of tetrahedral voids Okay arrangement is four is to four Here also the arrangement is four is to four coordination number is c Ab type structure molecule ionic molecule has these three types of structure possible Okay, nscl type cscl type and this zinc blade bursite type structure. Okay now another one is Second type of molecules ab two type ab two type molecules We also call it as fluoride structure Okay, fluoride structure important this fluoride and anti-fluoride both are important Fluoride structure Example of this we have Very common and important example The ab two types of cf2 Molecule We'll see the no important points here with respect to this molecule only write down for ca2 plus Ca2 plus All cations of this type of structure This has ccp arrangement first of all ccp arrangement its coordination number is eight each Ca plus two ions is surrounded by eight f minus ion All these no informations they sometimes they ask question like this Which statement is correct for this molecule? Okay, so what we can say each ca plus two ion plus two ion is surrounded by By eight f minus ions hence the coordination number is eight. That is what I have written Okay, number of ca2 plus if you count Because it is fcc arrangement So number of ca2 plus is again four f minus you see f minus ion It occupies all TVs present in the unit set All TVs present in the unit set And that's why the number of f minus ion Is equals to eight This is eight. This is four and hence the formula is caf2 Right for the crystal the formula is caf2 each f minus ion It's surrounded by Four ca plus two ions so coordination number Of this is four each ca plus two ion is surrounded by four f minus ion Hence the coordination number is four Right, what kind of arrangement we'll write down first ket ion and then an ion So it is eight is to four type of arrangement One second guys Done So this is the ab2 type fluoride structure we have Okay One second. Let me just pause this out. This is One second guys Okay, this is fluoride structure ab2 type one more type of a structure we have which is exactly opposite to The one that we had discussed that is fluoride So this one is fluoride the another structure is A to B type the third type of structure, okay first one second one and the third one is A to B type A to B type this is structure. We call it as anti fluoride structure Anti fluoride structure. Okay, everything is just opposite Means you can see here the position of cations there in fluoride here becomes the position of anions And things like that. Okay, so suppose I'm taking one example here of any two O type Suppose any two O the example I'm taking so the position of O2 minus ion here, you see O2 minus ion it is arranged in ccp manner Ccp arrangement each O2 minus ion It's surrounded by surrounded by four It's surrounded by sorry eight Eight any plus ion Eight any plus ion The number of O2 minus here it is again four is four for any plus you see any plus ion present In all tetrahedral void All tetrahedral void and each any plus ion the coordination number directly I'll write down for any plus ion is four And that's why the molecule or the crystal is called four is two eight Type arrangement or simply four is two eight arrangement we have Like li two O k two O Are this type of structures? Okay