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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.

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The correct Answer is:
349.9 pm
Distance of closest approach is equal to the distance between the nearest neighbours (d). As CsCl has BCC lattice,
`d=sqrt3/2a=1.732/2xx404` pm =349.9 pm
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