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Caesium chloride may be considered to form interpenetrating simple primitive cubic crystals. The edge length of unit cell is $412\;pm$. Determine the ionic radius of $Cs^{+}$ if the ionic radius of $CI^{-}$ if the ionic radius of $CI^{-}$ is $181$ P.m Given $M(C_s)=133\;g\;mol^{-1}$

$\begin{array}{1 1} 175.8\;pm \\ 7 \;pm \\ 34.8\;pm \\ 9.6\;pm \end{array} $

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Answer :$ 175.8\;pm$
Now Since $Cs^{+}$ ion touch the two chlorides along the cross diagonal of the cube, we will have
$2r_c+2 r_a=\sqrt {3}a$
or $r_c =\bigg( \large\frac{\sqrt 3}{2} \bigg)$$a-r_a$
$\qquad= \bigg(\large\frac{\sqrt 3}{2}\bigg)$$\normalsize (412\;pm) - (181\;pm)$
$\qquad= 175.8\;pm$
answered Aug 8, 2014 by meena.p

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