 Okay, now another type of a structure we have is spinal structure fourth type spinal structure right on the spinal structures are the examples of any crystal spinal structures are the examples of any crystal with more than two types of ions spinal structures are the examples of ionic crystals with more than two types of ions these are the compound of these are the compounds of two different cations these are the compounds of two different cations in bracket you write down a two plus means we'll have a divalent cation and we'll have a trivalent cation a two plus and b three plus right and the compounds of two different cations a two plus and b three plus and oxide ion right o two minus ion the general formula for this type of a structure we have we have a b two and o four type a b two o four type right now further this spinal structure also have two different types here so in this one the first type you write down in in spinal structure we call it as normal spinal structure and the second one is inverse spinal structure normal spinal structure now in this the important point again the same position of anions and cationsium so in inverse spinal what happens o two minus has FCC packing or CCP packing next point one by eight one by eight of tetrahedral voids are occupied by are occupied by divalent metal ion metal ion that is a two plus we are assuming here divalent metal ion half of the half of the ovis octahedral voids are occupied by are occupied by trivalent ion that is b three plus so could you tell me how many a plus we have how many a plus how many b plus we have for a two plus and b three plus number of a two plus is what one one by eight into one sorry one by eight into eight that is equals to one one once again one by eight so number of the title what is one number of b three plus equals to number of octahedral void is four right and half of the octahedral void has been occupied so it is two could you tell me for o two minus number of o two minus is nothing but FCC packing so it is four so look at the ratio here what is the formula the formula is a b two oh that is a general formula okay a b two oh four this is the normal spinal structure now you have this information that how many octahedral void is vacant and how many octahedral void is occupied similarly for tetrahedral void you have the information right so they can ask you any question on this let me just give you a few examples here for example if I ask you the ratio of the ratio of TV divided by OV occupied occupied in spinal structure and when I write this spinal spinal structure it means this normal spinal structure could tell me the answer here how many tetrahedral voids are occupied one right see number of a two plus is one by eight into it one by eight of the tetrahedral void is occupied so one tetrahedral void is occupied and the number of octahedral void occupied is two so it is one by two okay they can also ask you TV by OV unoccupied in this TV by OV unoccupied that would be seven by two right seven is unoccupied and out of four two is occupied so two is unoccupied seven by two so all these questions are not tough just need to have this information that is it the second type of structure we have in this spinal structure we call it as inverse spinal structure B inverse spinal structure right down here also you see O2 minus we have no change O2 minus forms FCC packing FCC packing A2 plus the divalent cation this is also same one by eight of of TVs tetrahedral void but we have a difference in the position of trivalent ion okay it says that it is present in one by eight of TVs plus one by four of OV means this trivalent ion is present in both tetrahedral void and octahedral void if you copy this down one sec done so for number of O2 minus again you see here is four number of A2 plus is one by eight into eight that is one number of B3 plus one by eight into eight plus one by four into four and that is two okay and hence for this also the molecular formula is A B2 O4 but the arrangement of ions trivalent ions here is different tell me OV by TV occupied OV by TV unoccupied yes what is the answer two in both yes so number of OV occupied here is we have only one octahedral void occupied right one by two no is it yes one by two for yeah it's one by two one OV occupied and two TV one TV is taken up by the trivalent ion and other one is divalent ion right yeah so one by two here and similarly here also one and we have two take care of this thing in the question what they have asked have they asked for TV by OV or OV by TV okay take care of that they can bring any questions from this okay they can even mix inverse spinal and non and and normal spinal structure also okay so take care of that another type of structure after this spinal right down see the example of this is inverse spinal is Fe 304 Fe 304 remember we have done this in in the 11th grade we have two types of oxidation state of ion here right what is that right so one is divalent cation Fe 2 plus Fe O dot Fe 203 this is arranged in inverse spinal inverse spinal structure okay next slide down corundum structure two three four type no corundum the general formula of this type for this type of a structure is A2 O3 type A2 O3 type very common one you know it is Al 2 O3 example you see we have Al 2 O3 we have another structure of example is Fe 2 O3 the oxides of iron you see the previous one is Fe 304 another oxides of iron has inverse spinal structure but this one has corundum structure okay CR 2 O3 also has corundum structure okay right down in this the oxide iron that you have O2 minus it forms at CP arrangement at CP arrangement A3 plus present in two-third two-third of octahedral voids two-third of octahedral voids and then the formula is A2 O3 you can find out the number of A3 plus and O2 minus iron you will have the same formula you see here two-third of octahedral voids at CP arrangement what is the rank of this is it six right yes sir so oxygen is six and two-third of this ovis we know the number of ovis is nothing but the rank of the crystal unit cell so total number of ovis also six total number of tv's are six into 12th so here we have six ovis so out of six two-third is occupied so two-third into six that is four right so we have A4 O6 that is A2 O3 type structure clear sir actually in the wood design structure there why did we take it as four even there it should be six right this is the coordination number four I have taken okay I have written the rank of the this thing unit cell it is a coordination number yes sir that's right this is the corundum structure where this is also important well last one I write down that is structure of FeO another oxides of iron the structure of FeO after this we have one more that is the diamond structure of diamond structure of FeO and then diamond we'll see right down again here O2 minus O2 minus forms CCP packing in all these you see the anions occupy the lattice points right and cations occupies the void okay that's a general thing you can now understand here O2 minus forms CCP packing and Fe2 plus occupies all octahedral voids octahedral void it's look like looks like any cl type structure only okay but right down this oxide is found to be non stoichiometric but this oxide oxide is found to be non stoichiometric with the formula with the formula Fe 0.95 and 4 this molecule we call it as the steid WUS TIT steid okay non stoichiometric compound this is the this is the formula we have right on in this type some of the Fe2 plus iron divalent iron some of the Fe2 plus iron in octahedral void in octahedral void are replaced by Fe3 plus iron are replaced by Fe deficient sorry Fe3 plus iron 3 Fe2 plus iron and every 3 Fe2 plus iron would be replaced by every 3 Fe2 plus I'm sorry okay every 3 Fe2 plus iron would be replaced by Fe3 plus iron 2 Fe3 plus iron in order to maintain electrical neutrality in order to mean electrical neutrality just this point you have to memorize so what happens in this you see to maintain the electrical neutrality of the crystal 3 Fe2 plus iron is replaced by 2 Fe3 plus iron so that the charge will be balanced but in this way what happens 3 iron atom is replaced by 2 iron atom right that's why the system or the crystal becomes Fe or iron deficient crystal okay if you see the number of atoms then the crystal becomes iron deficient crystal and the last structure we have in this write down the structure of diamond once they have asked question on this or the packing efficiency of diamond okay the last one write down the structure of diamond write down diamonds has FCC structure it has FCC structure but along with this there are four more atoms present so FCC structure with four more atoms present in in alternate TVs alternate tetrahedral void tetrahedral okay so the number of carbon atom in diamond if you see the number of atoms or the rank if you count here for FCC it is 1 by FCC it is 4 so half into 6 plus 8 into 1 by 8 but along with this four atoms present in the tetrahedral void so the number of atoms or the rank is 8 the carbon atom write down next the carbon atom in tetrahedral void in tetrahedral void in tetrahedral void is in contact with its four surrounding carbon atoms its four surrounding carbon atoms hence the coordination number is four carbon atoms present in tetrahedral void is in contact with its four other nearest carbon atom coordination number is four next point the carbon atom in FCC lattice the carbon atom in FCC lattice do not touch each other if you look at the distance so the tetrahedral void and the corner atom the distance is root 3a by 4 so the relation of a is 2r is equals to root 3a the body diagonal and one fourth of it is root 3a by 4 so the tetrahedral void the corner atoms and the tetrahedral voids are in contact and this is the face centered atom we have three face centered atom right in the cube if you see have discussed last class also so this is the relation of a and r for this thing for diamond we have okay packing fraction you can find out easily the packing fraction is z effective is 8 so we'll write down 8 into 4 by 3 pi r cube divided by a cube a and r relation you already have you substituted and the answer that you get is 34 percent so once they have asked this question that what is the packing fraction of diamond okay it's 34 percent you can memorize this also otherwise position of atoms if you know you can calculate this easily so why would you say it doesn't touch each other the FCC lattice see in FCC lattice normally what happens the atoms are in contact right the face center and means along the body along the face diagonal if you see that's a 4 r equal to but in case of diamond the case is different the carbon atom is present at a tetrahedral void and here the space is not that much so it you know because of the presence of this atom the corner atoms and the face center atoms lose its contact okay so these two corner atoms and the tetrahedral voids are in contact these two are in contact okay here also in contact and here also it is important this one and this one in contact but this atom is not in touch with this see all these the thing we have it depends upon the radius of the atoms it's very difficult to explain this theoretically okay it depends upon the radius of the atom that how depending upon the radius we can have the distance over here right and hence we can say geometrically that whether these are in touch or not okay so in case of carbon atom what happens the placement of carbon atoms in the tetrahedral void because of this only these two atoms lose its contact that's why the carbon atom the diamond you know the structure of diamond is a bit different with the normal FCC lattice that we have and we have this relation of a and r because of this how do we get to know this structure for that we have different crystallographic you know system by which we can find out one of the method we have that we use is Bragg's law right that we'll discuss in the last we used to determine the interplanetal distance between the two layer in the structure so there are various technique by which we can find out the structure of the crystal so with that only we define the atoms or the structure of any crystal like we are discussing all this okay so logic if you ask me so why it is not in touch I don't have much to say frankly speaking the only reason is if the space here is not that enough then this will like we have in cscl the body centered atom push the outer atoms right and hence the atoms lose its contact so in diamond it does not have the normal structure like we have FCC these are not in contact the face center and the corner but all the face center and corner atoms with the atoms which is present in tetrahedral void they are in touch with yes sir okay so all these things you see whatever I have discussed so far I have just given you the information it's not like I'm not teaching anything to you but this is the information you should know the position of an an that is it and how do we get this positions by all those crystallographic technique that we have I wish we can know look at the structure of various crystals and okay so based on those studies these you know structures are given so these are the structures we have discussed all you know important one apart from this we have few more structures but that haven't they ask in the exam it's not given that's neighbors also so I'm not you know taking up those things now you look at this question which I was discussing initially this one you try yes so what is the answer you are getting all of you are getting a yeah a is correct the question is solid having rock salt structure if all the atoms I think one body diagonal plane are removed except the body center then the formula left in the unit set okay so all the atoms touching one body diagonal plane are removed so which plane it is talking about in another question the same thing right from the top edge and the diagonal your opposite bottom edge right so along this edge if you remove so we are removing a four cornered atoms four cornered atoms we are removing two edge centered atom we are removing that is it right except the body center okay so four cornered atoms but two face also yes face also two face center side adjacent volapase side volapase right two face center also we are removing okay so a is it's not given a is at the face center forming the face center it's not given so which atom you assume at the face center FCC arrangement b b rock salt okay fine so the contribution of b is what we have only four cornered atoms four into one by eight and we have four face center atoms four into one by two so two plus one we have five by two it is and a in the octahedral void so a is is we have two octahedral void so we have 12 edge centers 10 into one by four plus one so five by two plus one so it is seven by is it right so we are getting a seven b five which is which is a a seven b five the ratio is that only k so this is the one we have second one you see this is structure done 346 you are getting cscl has busy arrangements along the void ion it is in touch so we have two r plus plus two r minus is equals to root three a inter ionic distance which ian you are considering cs and cl cs plus and cl minus is it or two cl minus i think yes because two cl minus the distance is what either we can have root two a by two or a only because position of cl minus is what it is at the sorry it is at the corner only busy see we have no between two cl minus the edge length is a only correct that is the value of the edge length of the unit cell so obviously the question is about cs plus and cl minus so we need to find out root three a by two isn't it yes yes a value is given you can find out simply root three by two these are not difficult not difficult once you know the structure if you do not know the structure of cscl you won't be able to do it okay one more question i'll show you here for this one the answer is let's try this i'm coming one second yeah done a two we two oh three yeah yeah that's right roger now yeah now you've got it right roger padek i think you are missing something you check your calculation the answer is a b o three pradyoth see the the crystal should be neutral right a and b will have same a b o three so we can have will have various possibilities for part b yes a plus b should be six and then you can write down the possibilities that we have that is it yeah so how many different possibilities yeah that's right pradyoth how many different possibilities we have so um so o two minus is arranged in ccp so number of o two minus i'll write down directly is four one six of the tetrahedral voids are occupied by cations okay so cation a so number of a we don't know the charge on a right one six of the tetrahedral voids so it would be four into one by six a tetrahedral void we have so we have eight so we have eight into one by six that is four by three and for b it is four into one by three that is again four by three and hence the formula is a four by three b four by three o four right so it is a b o three type crystal we have here a b o three type crystal we have now you see we have oxide o two minus total negative charge is six here right total negative charge is c six so whatever the charge on a suppose a will have plus a we assume and b will have plus b and o we all know it is minus two three times so obviously a plus b should be six then only the molecule should be neutral the crystal will be neutral so we can have various possibilities we can have um a value as plus one b as plus five that is one possibility we can have a as plus two b as plus four another possibilities a as plus three b as plus three another possibilities a as plus four b as plus two and the last one is a we can take as plus five and b plus one so there are five possibilities we have we can have any one of the two uh one of the combinations given in the question right here the answer we can write any one of the three now in this till now they haven't asked but they can ask you numerical value question numbers there are a number of possible you know uh possible you know the charges the possible charges on a and b so that the crystal is this we have so suppose this question is an integer type then the answer for the second part would be okay so they can frame questions from this integer type one okay now the last two things uh we have to discuss in this chapter the first one is uh the imperfection in solids right uh and then the second one is bragg's law okay imperfection we have different different types completely theory okay just you need to know two things mainly they ask in imperfection of solids that is this schottky defect in frankle defect okay so we'll discuss that only before that we'll start bragg's law okay so next heading right down bragg's law so we have five two two