i) CsCl is simple cubic cell. Cesium ion is surrounded by eight chloride ion which are also disposed towards the corner of a cube therefore both type ions are in equivalent positions and the stoichiometry is 1:1 . the coordination of CsCl is 8:8.
ii) Zinc sulfide is a FCC unit cell. The net number of zinc cation per unit cell is four, and the net number of sulfide anions per unit cell is four therefore, the ratio of ZnS ion in the cell is 1:1.
iii)Calcium fluoride is a FCC unit cell. The net number of Calcium cation per unit cell is four and the net number of fluoride anion is eight. Therefore the ratio of CaF2 ion in cell is 1:2
iv) Na2O has the structure opposite to CaF2. In
this case coordination number of Na+ ions is 4 and that of O2- ion is 8. Thus Na2O has 4:8 coordination.
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3rd of tetrahedral voids?
An element with molar mass 27 g mol-1 forms a cubic unit cell with edge length 4.05 x 10-8 cm . If its density is 2.7 g cm-3 , what is the nature of the cubic unit cell?
An element with density 11.2 g cm-3 forms a f.c.c. lattice with edge length of 4 x10-8
Calculate the atomic mass of the element. (Given: NA = 6.022x 10-23 (mol-1)