A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of B^- if the radius of A^+ is 190 pm

A compound AB crystallises in bcc lattice with the unit cell edge leng

1.	A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of B^- if the radius of A^+ is 190 pm
Answer» Solution :(i)For BCC, distance between `A^+` and `B^- =sqrt3/2a` , i.e., it is half of BODY diagonal (ii)As the CATIONS and the ANIONS touch each other , `r_(A^+)+r_(B^-)`329 pm `therefore r_(B^-)`=329-190=139 pm