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

(i) 329.1 pm
(ii) 154.1 pm

Explanation

In a bcc lattice,
body diagonal = `sqrt(3)a`
Now, `2 (r_(A^+) + r_(B^-)) = sqrt(3)a`
`r_(A^+) + r_(B^-) = (sqrt2)/2 xx a = (1.732 xx 380)/(2) "pm " = 329.1 ` pm
`r_(A^+) = 329.1 - r_(B^-) = 154.1 ` pm