An X^{+}Y^{-} ionic compound has a bcc struct | vedclass

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(A) In a $bcc$ (body-centered cubic) structure, the nearest ions are located at the body diagonal.For a $bcc$ unit cell, the distance between the near

Vedclass · Accepted Answer

$200 \ pm$

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(A) In a $bcc$ (body-centered cubic) structure, the nearest ions are located at the body diagonal.For a $bcc$ unit cell, the distance between the nearest ions (which is the distance between the corner atom and the body-centered atom) is given by $d = \frac{\sqrt{3}}{2} a$, where $a$ is the edge length.Given $d = 1.73 \ \mathring{A}$.Since $1 \ \mathring{A} = 100 \ pm$, $d = 1.73 	imes 100 = 173 \ pm$.Substituting the values: $173 = \frac{\sqrt{3}}{2} a$.$a = \frac{173 	imes 2}{\sqrt{3}}$.Using $\sqrt{3} \approx 1.732$, $a = \frac{173 	imes 2}{1.732} \approx 200 \ pm$.

An $X^{+}Y^{-}$ ionic compound has a $bcc$ structure. The distance between two nearest ions is $1.73 \ \mathring{A}$. What is the edge length of the unit cell?

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