A solid $A B$ has $C s C l$ type structure. The edge length of the unit cell is $404 p m$ . The distance of closest approach between $A^{+}$ and $B^{-}$ ions is:

250 p m

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349.9 p m

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349.4 p m

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450 p m

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Solution

The correct option is C $349.9 p m$
Distance of closest approach in BCC unit cell

$= \frac{a \sqrt{3}}{2} = \frac{404 \times 1.732}{2}$

$= 349.9 p m$

Option C is correct.

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