A compound AB crystallises in the b.c.c lattice with unit cell edge length of 390 pm. Calculate
(a) the distance between the oppsitely charged ions in the lattice.
(b) the radius of `A^(+)` ion if radius of `B^(-)` ion is 175 pm.

Answer

(a) 195 pm (b) 128.1 pm

Explanation

(a) Distance between the oppositely charged ions in the lattice `=(("390 pm"))/(2)` = 195 pm.
(b) For a cubic unit cell, `(r_(A^(+)))/(r_(B^(-)))=0.732`
`r_(A^(+))=0.732xxr_(B^(-))=0.723xx("175 pm")=128.1 "pm"`