 Let us write down ionic crystals see till now whatever we have done there is a size of atoms are same ok what we have done for simple cubic unit cell this is the edge length right edge length is this and we have taken A is equals to 2 R. So, we are taking a common radius here for FCC, VCC, simple cubic unit cell for all those the radius of atoms are same because it is a overhead crystal all atoms are of same size, but in ionic crystals we will have ions that is cations and anions and their size will also be different ok we will have two different radius we will get Rc is the radius of cation or we also write it as R plus R a the radius of anion or R minus we have two different radius here radius of cation and radius of anion ok. Generally cations are write down generally cations are smaller in size cations are smaller in size and presence in void presence in void ok. Now, for the different arrangement of atoms we will have different radius ratio of this means for a different arrangement like octahedral radius ratio Rc and R a will have a fixed range means this range remember Rc by R a come and you will have then the void will be octahedral void. Similarly, we have a fixed range of Rc and R a Rc the ratio of Rc and R a the fixed range for tetrahedral void triangle void also correct. So, that range you should memorize ok sometimes in the question they will ask you what is the radius of cation is the radius of anion is this for the formation of octahedral void. So, once you know the ratio of Rc and R a you can find out that value ok. So, that we can derive also geometrically but the derivation is not important. So, I am not doing the derivation ok I am giving you the range of radius of cation and anion radius ratio ok. So, write down for different arrangement of ions for different arrangement of ions we have a fixed radius ratio a fixed radius ratio of cation and anion right down the column you draw and write down here we will have a radius ratio one side radius ratio is a ratio of Rc by R a radius of cation by radius of anion coordination number coordination number of ions what kind of arrangement we have. So, if the radius ratio is 0.155 to 0.225 if this is the ratio of Rc and R a then the coordination number will be 3 and the arrangement will be triangular planar triangular ok. Next one is 0.225 to 0.414 2 to 5 to 0.414 coordination number is 4 arrangement is tetrahedral arrangement is tetrahedral and we get tetrahedral void here right where it is 0.414 0.732 it is 6 octahedral 0.732 2 1 it is 8 and it is BCC body centered BCC structure. So, triangular planar triangular is this atoms if you arrange like this is a void. So, what is the void? It is triangular void 2D void is this mainly the important thing is this 2 ok this ratio you must means if you keep on increasing the size and this range for this range the arrangement will be this. If the range change in this all this cation and anion will change in such a manner. So, that the arrangement becomes tetrahedral ok that is again 3 dimensional it is difficult to visualize but you have to memorize these 2 issues most important is this for the example got it ok. Next you write down different ionic crystal we have different structures ok. So, types of structure write down types of structure of ionic solid type of structure of ionic solid. So, we have used Pythagoras theorem. Yes yes yes we can use this is a void ok and this is what this distance is it is cation it is an ion and this is cation. So, this distance is r c plus r a ok this distance is from here to here this distance is what say this triangular will draw this is what let me this is what it is r a ratio of a time r a this is what cost 30 30 we can write this. So, cost 30 is what this divided by this base by half means this is what r c r c plus r a cost 30 is root 3 by 2 ok. So, it is root 3 by 2 or we can write r c cross multiply this 1 plus r a by r c is equals to what 2 by root 3 ok. So, r a by r c is what 1 point cost 30 is this by this taken r c by r c plus r a here cost 30 is what root 3 by 2 r c by r c plus r a and then we can write root 3 r c plus root 3 r a is equals to 2 r c ok r a divided by root 3 r c by r a minus 2 r c by r a is equals to what minus root 3. So, solve this r c by r a is equals to what we get minus of root 3 divided by 2 minus root 3 negative when you solve this you will get this one 0.155 ok. So, with geometry you can do all this derivation you can do ok, but that derivation is not for solving questions you just know this range ok tetrahedral you have to you know again imagine the geometry and draw the structure and then you can do the derivation with the help of geometry ok. Why is it like that is 0.4 and 4 that is root 2 minus 1 exactly this is 0.73 that is root 3 minus 1 exactly. So, it comes when you solve this is for tetrahedral when you do this for tetrahedral you get this value expression will be same root 2 minus 1 root 3 minus 1 derivation is not in point ok if you want again give you that also, but that derivation is not in point ok. So, like this triangle void we use a we take the help of geometry and we can find out the. . . . . . . . . . . . . . . . . . ;) like this is the first structure we write down the first structure we have rock salt rock salt or NaCl type structure rock salt or NaCl type structure now in this like I said position of cations and anion you should know right so what is the position of Na plus what is the position of Cl minus you should know right right on the next line all Na plus ion all Na plus ion it's surrounded by six Cl minus ions six Cl minus ion and all cl minus ions chloride ion it's surrounded by six Na plus ion okay so just one more thing here the maximum number of ions by which one particular ion is surrounded that we call it as the coordination number of that ion okay so coordination number of Na plus is what? Na plus is six Cl minus is six that's why this NaCl type structure or rock salt type structure we call it as six is to six arrangement e question which is like that what kind of arrangement is this okay so it is six is to six arrangement what is the coordination number of Na plus six Cl minus six right so one type of question is coordination number of cations and anion another type of question is what kind of arrangement it is six is to six arrangement third type you write down the position of this cations and anion right now all Cl minus all Cl minus are arranged in CCP manner all Cl minus are arranged in CCP manner in which in which all octahedral voids are occupied by all octahedral voids are occupied by Na plus ion okay occupied by Na plus ion so when you look at the face of this crystal Na type or rock salt type structure now one face looks like this just one face and drawing all the cornered atoms are occupied by see all the cornered atoms are occupied by Cl minus okay and face center also we have Cl minus right and we know at the L center what we have octahedral void and all octahedral void is occupied by Na plus present at this point okay this is the arrangement we have this is minus minus minus minus plus plus plus k times okay okay body center will be Na plus over and center will be Na plus okay now if I find out if I ask you what is the relation of A and R here can tell you what is this distance this edge net is it to R minus plus to R plus right kind of so a is equals to what we can write a is equals to 2 into R minus plus R plus okay so R minus plus R plus means radius of the sum of the radius of an cation is the most important thing is the location of Na plus and Cl minus you'll get many questions in which they'll say there's a AB crystal in which all B atom occupies which has rock salt structure and all B atom occupies octahedral void A forms CCP lattice means A is equivalent to Cl minus B is equivalent to Na plus so we can compare that and find out the whatever thing is required okay now packing efficiency if you have to find out your packing efficiency for that what we require this just we'll do right on the expression we don't have to find out the packing efficiency okay they won't ask you all students packing efficiency is what it is a volume occupied by the ions divided by volume of units in okay how many Na plus are present into this number of Na plus can you tell me Na plus at octahedral void so one at the body center plus L center 12 into 1 by 43 see the volume of ions so volume of Na plus and 4 by 3 pi R plus Q volume of Cl minus 4 by 3 pi I minus Q but how many Na plus and Cl minus are present that we have to find out correct we have 4 Na plus present so any plus case like we have the volume occupied by Na plus is what 4 into 4 by 3 pi R plus Q similarly number of Cl minus is what FCC A into F center and corner so it will be again 4 8 into 1 by 8 plus 6 into 1 by 2 4 into 4 by 3 pi R minus Q you have to substitute 4 into 4 by 3 pi R plus Q plus R minus Q this divided by volume of unit cell is 4 by sorry this AQ or you can 2 into R plus into R minus when you substitute this A to Q into R plus plus R minus okay but this you don't have to do now they won't ask but why I have done this because it is same thing here we have two different ions so we'll find out the volume occupied by k ions plus volume occupied by n ions so volume of ions will be that divided by volume of unit cell is AQ AQ plus R minus so this part they won't ask you okay most problems we can leave this part okay but it is not that difficult also expression looks like of it's no complex but it is very similar okay only we have two different radius so we have to write down the volume of each ions separately okay three things are important the most important is position of cations in n ions rock solid structure may come back and over come back and over so that I will be at the in the octahedral void and I will be at the forms ccp lattice ccp means face center and corners okay arrangement will be 6 to 6 any plus is 100 by 6 CL minus and CL minus 100 by 6 so one two three position of ions important right now all those crystal which has a similar structure right on examples for that any CL we have discussed already LICL AGCL AGCL AGBR okay RBI KBR etc all these crystals forms FCC lattice okay now next structure you write down cesium chloride type structure CS CL type structure cesium chloride type structure right down in this time the arrangement is VCC type body centered cubic cesium the arrangement is VCC type next line CL minus present at the corners CL minus present at the corners and CS plus CS plus is at the body center so get I will be at the body center CS plus how many CS plus will be there so body center may have one over right number of CS plus will be one number of CL minus will be eight eight K so CS CL structure is one is two one you know so number of CS plus and CL minus will be same right or CL minus as a video for the corner pair present is it eight into one by it ticket CL minus be one over so number of CL minus and CS plus is one up I got it so you packing a piece of this you need a lot of work volume occupied by the ions support each other one one into four by three fire plus cube plus our minus cube or key or all the relation you have to pick up a bc pattern atoms along the boy diner in contact right so this boy diner what it is R minus 2R plus and R minus okay so it is 4 into no it's not for root 3a is equals to 2R plus plus 2R minus here instead of a we have root 3a okay body dialogue a long I don't touch correct so boy dialogue a little ticket or he will be R minus and R plus fit R plus and R minus so 2R plus plus 2R minus was root 3 but again like I said this packing efficiency they won't ask this question okay so arrangement is what arrangement is okay the correlation number of CS plus is what CS plus present CS plus here Pena is a coordination number yeah okay get new items are on today in CS plus 8 a 4 top layer 4 bottom okay so the coordination number of CS plus is 8 and CL minus is also 8 coordination number of CL minus is also 8 and hence it is 8 is to 8 arrangement right 8 is to 8 arrangement okay next write down structure of ZLS zinc sulfide zinc blend also it has two types of structure the first one write down zinc blend structure the first type of structure in this we have zinc blend zinc plate and oreside two different structure we have the first time to write down right on this it has cct arrangement it has cct arrangement s-2 next line s-2 this I am ZNS gives what s-2 and ZN plus 2 right again you see the structure is ZNS it means the number of ZN plus 2 should be equals to the number of s-2 okay s-2 present at corners and face center at corners and face center ZN plus 2 occupy half of the tetrahedral void half of the tetrahedral void okay how many s-2 will there how it is 2 ZN plus 2 S minus 2 one by it so 80 to one by it face center half so six into half four corner and face center simply ahead s-2 is forming cct arrangement or fcc arrangement the number will be 4 okay ZN plus 2 now present half of the tetrahedral how many times are there 8 2 and we know so 4 tetrahedral voids are occupied by the one of which is same table ZN plus 2 okay so what is the arrangement here it is 4 is to 4 arrangement the arrangement is 4 is to 4 ZNS coordination number of both iron is 4 and here the arrangement is 4 is 2 4 okay this way r of ZN plus 2 by r of s minus 2 the ratio is 0.40 it is in that range only 0.2 to 5 to 0.414 e of the ratio is 1 bar they have asked this question only once ratio of this