Question

A compound AB crystallizes in bcc lattice with unit cell edge length of 480pm.if radius of B- is 225pm then the radius of A+ is?

Answer 1

Bro hope it is helpful.

Answer 2

Explanation:

It is given that unit cell edge length is 480 pm and radius of $B^{-}$ is 225 pm.

Let the radius of $A^{+}$ is x.

Formula to calculate BCC edge length is as follows.

                a = $\frac{2}{\sqrt3} \times (R_{A^{+}} + R_{B^{-}})$

               480 = $\frac{2}{\sqrt3} \times (x + 225 pm)$

                   x = 190.68 pm

Thus, we can conclude that the radius of $A^{+}$ for given BCC is 190.68 pm.