For cesium chloride structure, the interionic distance (in terms of edge length, a) is equal to $\frac{\sqrt{3 a}}{2}$ .

True

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False

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Solution

The correct option is B False
In $C S C L, C l^{-}$ is present at body center and $C s^{+}$ are present at corner
(diagram)
$I n △ A B D c^{2} = a^{2} + b^{2} c = \sqrt{a^{2} + {2 a}^{2}} = \sqrt{3} a S i n c e △ A B D i n e q u i l a t e r a l \sqrt{3} a = 4 r r = \frac{\sqrt{3} a}{4} w h e r e a = e d g e l e n g t h r = i n t e r i o n i c d i s t a n c e$

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