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A solid AB has CsCl-type structure. The ...

A solid `AB` has `CsCl`-type structure. The edge length of the unit cell is `404` pm. Calculate the distance of closest approach between `A^(o+)` and `B^(Θ)` ions.

Text Solution

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The distance of closest approach is equal to the distance between the nearest neighbour `(d)`. As `CsCl` has `b c c` lacttice,
`d = (sqrt3)/(2) a = (1.732)/(2) xx 400 "pm" = 349.9` pm
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