 The second one, second type of structure of ZNS write down, it has HCP arrangement, hexagonal flows packing, okay, hexagonal and high flow. It has HCP arrangement, hexagonal flows packing, S-2 ion, S-2 ion forms are arranged in HCP manner, HCP manner and ZN plus 2 ion occupy half of the tetrahedral point, occupy half of the tetrahedral point, alternate tetrahedral point we can also call it, okay. Next write down the structure of ionic compounds, fourth one no, fourth one after this, the structure of ionic compounds of AB2 type, AB2 type. Sir, everything for the horizontal structure is the same like the, for the, yeah, yeah, 4-to-4 arrangement we are here also, for this also it is 4-to-4 arrangement, only the difference is it has HCP arrangement or here a CCP arrangement, okay, that is the only difference, AB2 type structure you write down, this is structure we also call it as fluoride structure important, F-L-U-O-R-I-D, fluoride structure, example is C-A-F-2 calcium fluoride, C-A-F-2, okay, example write down C-A-F-2, next line, C-A plus 2 ions, C-A plus 2 ions are arranged in CCP manner, are arranged in CCP manner and F-ion occupy all tetrahedral point, F-ion occupy all tetrahedral point, no, in generally, in general what happens, the size of an ion is bigger than R minus is more than R plus, okay, so like I said generally cation occupies tetrahedral point, in general, okay, but we have some you know, exceptions you can say like C-A-F-2 or we can have another structure of oxides of ion, there also we have some anions present in the void, okay, but generally voids are smaller in size, no, suppose if the atoms are arranged this way, okay, so voids obviously it is smaller inside, so whatever smaller ions are there that is present in the void, in general, that's why we say in general the size of cation is smaller and cations present in the void, but for some of the structures like arrangement is in such a way that for some structures anions are also present in the void, the first one is this one AB2 type and that's why it is you know important, two types of structure we have AB2 type you also call it as structure of C-A-F-2 where F minus is present in tetrahedral point, okay, another one is E2B type which is anti-chloride structure, so now we have to know, anti-chloride structure, this one is chloride, this one is anti-chloride structure and the example of this is NE2, right, on this structure, right, so it is possible that in the void there are anions present, okay, for some of the structure, but in general we say cations are smaller in size and it occupies point, okay, with some exception, this one is the first exception we have where anions are present in the void, okay, okay, what did you write, C8 plus 2 occupies form CCP arrangement and F minus occupies all tetrahedral void, how many F minus is there then, 8, how many C2 plus, C2 plus will be 4 because form CCP arrangement, so what is the arrangement now? 4 is to 8, see what is the, how many F minus is surrounded by 8 CA2 plus, no, it is different, see CAF2 we have, correct, CAF2 and number of F minus, number of F minus is to 8, it is not the coordination number, number of CA2 plus is what, it is not the coordination number, right, so write on next time, CA2 plus ions, CA2 plus ions is surrounded by 8 F minus ions, CA2 plus ions is surrounded by 8 F minus ions and each F minus ions is surrounded by 4 CA2 plus ions, correct, you can understand this, this is a general molecule, right, so CA plus 2 ions is surrounded by 8 F minus ions and each F minus ions is surrounded by 4 CA plus 2 ions, so 4 CA plus 2 means what, the product charges 4 into 2, 8 F minus is what, product charges 8 into minus 1 that is 8, so you do not get confused with this, do not take right 8 CA2 plus and 4 F minus, which is not possible, because 8 CA2 plus means what, charge is total what, plus 16, 4 F minus is minus 4, so this would not be neutral, okay, you do not compare this way, you do not have to get confused with this, because 8 F minus and 4 CA2 plus neutral, so this is the arrangement we have, we cannot change this 2, right, 8 CA2 plus and 4 F, that is not possible, okay, so when you get the question like this and the option if you get confused 4 of 8 or 8 or 4, you can think like the charge, it should be neutral or whatever, okay, so hence the arrangement is what, 8 is to 4 arrangement, because each F minus sorry, each CA2 plus ions is surrounded by 8 F minus ions, so arrangement is 8 is to 4 arrangement, okay, next structure you write down, examples of this you write down first, CRF2, BAF2, BACL2, etc. CRF2, BAF2, BACL2, etc. Next one is structure of A2B type, A2B type, anti-fluoride structure, this is structure, fluoride they have asked many times in the exam, so this one is fluoride structure, AB2 type, the other one is A2B type, it is anti-fluoride structure, example is NA2O, NA2O, now in AB2 type F minus is present in the void, so here NE plus is present in the void, reverse, here where N ions will be, here where K ions will be, just pull down, okay, so write down all NE plus ions are present in tetrahedral void, are present in tetrahedral void, are present in tetrahedral void and each NE plus ion is surrounded by 4 O2 minus ion, the charge is 2 minus, each NE plus ion is surrounded by 4 O2 minus ion, next line O2 minus, it is not O2 minus, it is O2 minus, charge is 2 minus, O2 minus forms CCP arrangement, now again CCP arrangement means what? FCC, corner plus phase center, corner plus phase center, each O2 minus ions is surrounded by 8 NE plus ion, 8 NE plus ion, the previous one is 8 is to 4 arrangement, so this one is 4 is to 8 arrangement, it is 4 is to 8 arrangement, fluoride anti-fluoride is very important, examples LI2O, K2O, etc., A2B type, next write down structure of diamond, write down it has FCC structure, it has FCC structure and 4 more atoms and 4 more atoms are present in the alternate tetrahedral void, the alternate tetrahedral void, okay, can you tell me what is the Z effective for this rank of this crystal, for diamond, see Z effective, FCC, 4 8 into 1 by 8 plus 6 into half, and 4 more atoms are present in alternate tetrahedral void, okay, so the 4 atoms of alternate tetrahedral void, what is the contribution, 1 tetrahedral void to 1, this is for tetrahedral voids, it should be Z effective for this, you should know this, FCC and 4 more atoms are present in alternate TVs, okay, tetrahedral voids, now what is the relation of KNR, so here is an atom present, okay, and these two are in contact, right, so this and this atom is in contact, so what is this distance, what will be the distance of the center, corner and center distance, root 3A by 2, because bore diameter is root 3A, so this is root 3A by 2, what is half of this, root 3A by 4, this distance is root 3A by 4, so here root 3A by 4 is equals to 2R, diamond, they have not asked question till date, but R or E carry relation, okay, Z effective, you should know what is the atoms present, okay, so FCC plus 4 at alternate tetrahedral void, it is more than FCC, obviously FCC plus 4 more atoms present, okay, so here we have packing efficiency, packing fraction, number of atom is 8, so 8 into 4 by 3 pi RQ divided by AQ, okay, A or R by relation to R, so here we have 8 into 4 by 3 pi RQ, for eco we can write 8R by root 3, so it is 3R by root 3, packing efficiency of time is 34% approximately, what was in FCC, FCC was 34, yes, you can say because in diamond the carbon atom are not in contact, okay, or use extra here the void we present, okay, so in this write down, in diamond the carbon atom, the carbon atom in FCC lattice are not in touch, are not in touch, but the carbon atoms which are present in the tetrahedral void, in the tetrahedral void touches the surrounding 4 atoms, surrounding 4 atoms, okay, so that is why its efficiency is not more than FCC, FCC make up face centered around crescent, all these atoms are not in touch, it is obviously if it is not in touch, it means what the atom is not present into the lattice as it was in FCC crystal, that is why the packing efficiency is less, okay, so you must remember this, this relation if you know, this if you can find out, okay, they haven't asked any question in diamond, but it is important, okay, next one I write down, spinal structure, next one, spinal, S-P-I-N-E-L, spinal structures, write down, these are the structure of ionic crystals, these are the structure of ionic crystals, ionic crystals which has more than 2 types of ions, which has more than 2 types of ions, next line, hence spinal are the compounds with 2 different cations, with 2 different cations, in bracket you write down, divalent and trivalent, how does it exist, F-E-O dot F-E-2-O-3, so F-E-2-O-3 may the oxidation instead of F-E is what, plus 3, and F-E-O may oxidation instead of F-E-O, plus 2, so it is the compound F-E-3-O-4, in this we have a trivalent ion, F-E-3-plus, a trivalent ion F-E-2-plus and an oxide, this is a spinal compound, okay, it contains 2 different cations, in bracket write down, divalent and trivalent and an oxide as an anion, an oxide as an anion, okay, the general formula is a-b-2-O-4-tide, it is a-b-2-O-4-tide, okay, a-divalent, b-trivalent and oxygen-2, see it should be neutral, right, so confusion, neutral only because oxygen with the minus 2 we always have, minus 2 into 4 is what, minus 8, our a-plus-3 over, so it is not neutral, okay, a-plus-3 over or a-plus-2 over, so plus 4 plus plus 3 is plus 7, it is not neutral, which is not possible, so if it is plus 3 and plus 2, so plus 3 into 2 is plus 6, plus 6 plus 2 is plus 8, plus 8 minus 8 is 0, so in this structure we can write a is always trivalent and b is trivalent, general structure is a-b-2-O-4-tide, okay.