KBr has fcc str The density of KBr is 2 75 g/cm cube Find the distance b/w K+ and Br- - Chemistry - The Solid State

Question

KBr has fcc str The density of KBr is 2 75 g/cm cube Find the
distance b/w K+ and Br-

Pankaj Singh · Accepted Answer

For PbS M= 239g Z= 4 Let distance between ions = (R++R-) d= 2 75g/cm NA=6 023x1023 Br- is present at corners while K+ is present at edge centers so (2R++2R-)=a or 2(R++R-)=a (1) we know that $a^{3} = \frac{z \times m}{N_{A} \times d}$ Form e q (1) ${(R^{+} + R^{-})}^{3} = \frac{1}{8} \frac{z \times m}{N_{A} \times d}$ on Substituting the values and calculating you will get the value of (R++R-) Good luck

KBr has fcc str.The density of KBr is 2.75 g/cm cube.Find the distance b/w K+ and Br- .