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A solid AB has $CsCl −$ structure edge length of the unit cell is $4 .04 \overset{\circ}{A}$ , the closest distance between $A +$ and $B −$ is:

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$3.5 \overset{\circ}{A}$

$7 \overset{\circ}{A}$

$4 \overset{\circ}{A}$

$6.4 \overset{\circ}{A}$

answer is C.

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Detailed Solution

For

CsCl

3 a = 2 r + + r −

r + + r − = 3 × 4.04 2 = 3.5 A 0

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