$C s B r$ has B.C.C structure with edge length $4.3 ˚ A$ . The shortest inter-ionic distance (in $˚ A$ ) between $C s^{+}$ and $B r^{-}$ is :

3.72

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1.86

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7.44

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4.3

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Solution

The correct option is D $3.72$
As we know, distance of closest approach in bcc lattice

$= \frac{1}{2}$ of body diagonal $= \frac{1}{2} \times \sqrt{3} a$

$= \frac{\sqrt{3}}{2} \times 4.3 = 3.72 ˚ A$

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Similar questions

$C s B r$ has a BCC structure where the edge length is 4.3. The shortest inter ionic distance between $C s^{+}$ and $B r^{-}$ is:

CsBr crystal has bcc structure. It has an edge length of The shortest interionic distance between is

[IIT 1995]

CsBr crystal has bcc structure. It has an edge length of $4.3 \circ A$ . The shortest interionic distance between $C s^{+}$ and $B r^{-}$ ions is

C s B r

4.3 pm

C s^{+}

B r^{-}

C s B r

4.3

(in pm)

C s^{+}

B r^{-}

\sqrt{3} = 1.73

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