Question

Ab has nacl type of structure .if edge length of crystal lattice is141A.the radius of anion ​

Answer 1

as you know, NaCl is face centered cubic lattice in which Na+ is placed at midpoint of edge corner and Cl^- ion is placed at corner and face center of lattice.so, AB is also a face - centered cubic lattice. Let A^+ is cation and Cl^- is anion. then, A^+ ion will be place at midpoint of edge corner and B^- ion will be place at corner and face center of lattice AB.

from structure of fcc , $\frac{r^+}{r^-}=0.414$

or, $\frac{r^-+r^+}{r^-}=0.41+1=1.41$

or, $\frac{2r^++2r^-}{r^-}=2.82$

we know, $2r^-+2r^+$ = edge length of fcc lattice

so, edge length/radius of anion = 2.82

or, 141/radius of anion = 2.828

or, radius of anion = 141/2.82 =50A°

hence, radius of anion = 50A°