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CsBr has b.c.c structure with edge length $4.3 A^{\large\circ}$. The shortest inter ionic distance between $Cs^+$ and $Br^-$ is

$(a)\;3.72A^{\large\circ}\qquad(b)\;1.86A^{\large\circ}\qquad(c)\;7.44A^{\large\circ}\qquad(d)\;4.3A^{\large\circ}$

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$r_c + r_a = \large\frac{\sqrt3}{2}a$
$\therefore r_c + r_a = \large\frac{\sqrt3\times4.3}{2}$
$=3.72 A^{\large\circ}$
Hence answer is (a)
answered Feb 27, 2014 by sharmaaparna1
 

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