CsBr ha cubic structure with edge length $4.4 o A$ . The shortest interionic distance between $C s^{+}$ and $B r^{-}$ is:

3.82 o A

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1.86 o A

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7.44 o A

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4.3 o A

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Solution

The correct option is A $3.82 o A$
$C s B r$ has body centered cubic (BCC) structure. The edge length of unit cell is given by $4.4 ˚ A$

Now, for a cube of edge $a$ units, the length of the diagonal is $\sqrt{3} a$ units.

The shortest ionic distance between $C s^{+}$ and $B r^{-} = \frac{\sqrt{3} a}{2}$

$= \frac{\sqrt{3} \times 4.4}{2} = 3.81 ˚ A$

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