CsBr has b.c.c. structure with edge length 4. | vedclass

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(A) In a $b.c.c.$ structure,the atoms touch along the body diagonal.The length of the body diagonal is $\sqrt{3}a$,where $a$ is the edge length.The di

Vedclass · Accepted Answer

$3.72$

Vedclass · Answer

(A) In a $b.c.c.$ structure,the atoms touch along the body diagonal.The length of the body diagonal is $\sqrt{3}a$,where $a$ is the edge length.The distance between the center of the body-centered ion and the corner ion is half of the body diagonal,which is $\frac{\sqrt{3}a}{2}$.Given $a = 4.3 \ \mathring{A}$,the shortest inter-ionic distance $d = \frac{\sqrt{3} 	imes 4.3}{2}$.$d = \frac{1.732 	imes 4.3}{2} = \frac{7.4476}{2} = 3.7238 \ \mathring{A}$.Thus,the shortest inter-ionic distance is approximately $3.72 \ \mathring{A}$.

$CsBr$ has $b.c.c.$ structure with edge length $4.3 \ \mathring{A}$. The shortest inter-ionic distance between $Cs^{+}$ and $Br^{-}$ is .............. $\mathring{A}$.

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