CsCl crystallises in a cubic structure that h | vedclass

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(A) In a $CsCl$ crystal structure,the $Cs^{+}$ ion is located at the body center and $Cl^{-}$ ions are at the corners of the cube.For this body-center

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$4.04$

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(A) In a $CsCl$ crystal structure,the $Cs^{+}$ ion is located at the body center and $Cl^{-}$ ions are at the corners of the cube.For this body-centered cubic arrangement,the body diagonal is given by $\sqrt{3} \ a = 2(r_{Cs^{+}} + r_{Cl^{-}})$.Substituting the given values: $\sqrt{3} \ a = 2(1.69 \ \mathring{A} + 1.81 \ \mathring{A})$.$\sqrt{3} \ a = 2(3.50 \ \mathring{A}) = 7.00 \ \mathring{A}$.$a = \frac{7.00}{1.732} \ \mathring{A} \approx 4.04 \ \mathring{A}$.

$CsCl$ crystallises in a cubic structure that has $Cl^{-}$ at each corner and $Cs^{+}$ at the centre of the unit cell. If $r_{Cs^{+}} = 1.69 \ \mathring{A}$ and $r_{Cl^{-}} = 1.81 \ \mathring{A}$,what is the value of the edge length of the cube in $\mathring{A}$?

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