1.

CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance is ………(a) 400pm (b) 800pm(c)  \(\sqrt{3}\)x100pm(d)  g\(\Big(\frac{\sqrt{3}}{2}\Big)\times 400\)pm

Answer»

(d)  g\(\Big(\frac{\sqrt{3}}{2}\Big)\times 400\)pm

g\(\sqrt{3a}\) = rCs+ + 2rCs- + rCs+

\(\Big(\frac{\sqrt{3}}{2}\Big)\)a  =  (rCs+ + rCs- )

\(\Big(\frac{\sqrt3}{2}\Big)\)400 = inter ionic distance



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