$C s B r$ has BCC structure. The length of its one side is $4.3 o A$ . The minimum distance between $C s^{+}$ and $B r^{-}$ ion will be:

0.897 o A

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3.72 o A

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1.749 o A

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none of the above

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Solution

The correct option is C $3.72 o A$
$C s^{+}$ is at body centre and $B r^{-}$ forms the lattice. Along the body diagonal minimum distance is $\frac{a \sqrt{3}}{2}$ between $C s^{+}$ and $B r^{-}$ .
$d = \frac{4.3 \sqrt{3}}{2} = 3.72 ˚ A$

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