CsBr has bcc structure with edge length 4.3 | vedclass

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(A) In a $bcc$ 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 body

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} 	imes 4.3$

Vedclass · Answer

(A) In a $bcc$ 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 body diagonal consists of two radii of the cation $(r^+)$ and two radii of the anion $(r^-)$,i.e.,$2r^+ + 2r^- = \sqrt{3}a$.The shortest interionic distance between the cation and anion is the sum of their radii,$d = r^+ + r^-$.Therefore,$d = \frac{\sqrt{3}a}{2}$.Given $a = 4.3 \ \mathring{A}$,the distance $d = \frac{\sqrt{3}}{2} 	imes 4.3 \ \mathring{A}$.

$CsBr$ has $bcc$ structure with edge length $4.3 \ \mathring{A}$. The shortest interionic distance in between $Cs^{+}$ and $Br^{-}$ (in $\mathring{A}$) is :

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