Question

Paper overlay C C. Zinc Blende, ZnS Use the plastic black template marked phere on a metal get the pink sphere e firmly. Note that It takes a little patience to build this structure. Rather than placing a small pink sphere on rod, you are balancing it in one of the vacancies of the lattice. Have faith, if you get the nint in the correct location it will not fall out, the colorless spheres will hold it in place firmly. N you are using the same template and base as you did for NaCl. 1. If the sulfide ions are on the lattice points of a face-centered cubic unit cell, how would describe the location of the zinc ions? 2. How many zinc cations surround each sulfide? How many sulfide ions surround each zinc cation? Does this match the empirical formula of zinc sulfide? OS 3. A certain form of silver iodide adopts the zinc blende structure. The density of the salt is 5.69 g/cmº. Calculate the length of a side of the unit cell. Calculate an approximate distance between the silver cations and iodide anions in the structure. (You can approximate the location of silver ion because it resides in a pocket 4 of the way down the body diagonal.)

Answer 1

solution In zus, sulphide ims formed cubic close parking arrangement and zinc ions are lies at 50% of tetra- - hedral voids in the alternative manner. ie only & tetrahedons voids are to occupied by Zu2t in tw alternative manner. FFFFFFFFF O O=52 1 e 11 = volds. 5 0 ᎫᎫ Ꭻ ᎫᏇ Ꭻ ᎫᎫᎯ If the sulfide ions are on the lothic points of a face centered cubic mit cell, then zine ion will lie at 50% of tetrahedsol voids in the alternative manner,

from the figure , we can imagine that zinc cations will be surrounded by 4 Sulfide ions and one sulfides ions will be surrounded by 4 zinc cation. So, the empirical formula of 2lnc sulfide (zus) mat because coordination no. t for both anion is same. I ③ we know the formula, of density ( 2 x molecular mass of Agl. 6.022*1023 x 93 11 z = no. of particle in mit cell. total No. of 22 in mit cell - 4 tatal No. of 52 in unit cell Notas + Bx + 6x so 42u2t & 4 5 2 4 zus so, 2= 4 (Also for Agc1) molecular mass of Agel = 143.32

a= side of unit cell. so will et 4 x 143.32 5.69 = 6,022 X1033 x 93 bobo tavan a3 = od 4 x 143.32 6.022 x 1023x5.69 93 - 16.73 X10-23 93 - 167.3. X10-24 a = 5.5x108 cm from the figure distance se. the anion and Cotion the inter sum of is given lonic radius by of Vzn2t + rs - = ava dy - 5.5*108 xV3 4 = 2.38 x 18 8 can

? and So the chloride distance im is b/w silver in 2.38 x 108 can.