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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 $Cl^{-}$ if the ionic radius of $Cl^{-}$ 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 :175.8pm" role="presentation" style="position: relative;">175.8pm175.8pm 175.8\;pm
Now Since Cs+" role="presentation" style="position: relative;">Cs+Cs+Cs^{+} ion touch the two chlorides along the cross diagonal of the cube, we will have
2rc+2ra=3a" role="presentation" style="position: relative;">2rc+2ra=3a2rc+2ra=3a2r_c+2 r_a=\sqrt {3}a
or rc=(32)" role="presentation" style="position: relative;">rc=(32)rc=(32)r_c =\bigg( \large\frac{\sqrt 3}{2} \bigg)a−ra" role="presentation" style="position: relative;">araa−raa-r_a
=(32)" role="presentation" style="position: relative;">=(32)=(32)\qquad= \bigg(\large\frac{\sqrt 3}{2}\bigg)(412pm)−(181pm)" role="presentation" style="position: relative;">(412pm)(181pm)(412pm)−(181pm)\normalsize (412\;pm) - (181\;pm)
=175.8pm" role="presentation" style="position: relative;">=175.8pm=175.8pm\qquad= 175.8\;pm
answered Aug 8, 2014 by meena.p
edited May 9 by meena.p
 

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