The interionic distance for Cesium chloride ( | vedclass

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(C) In a $CsCl$ crystal,the structure is body-centered cubic $(BCC)$.In this structure,the $Cs^+$ ion is at the body center and $Cl^-$ ions are at the

Vedclass · Accepted Answer

$\frac{\sqrt{3} a}{2}$

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(C) In a $CsCl$ crystal,the structure is body-centered cubic $(BCC)$.In this structure,the $Cs^+$ ion is at the body center and $Cl^-$ ions are at the corners.The body diagonal of the cube is equal to $\sqrt{3} a$.The distance between the body center and a corner is half of the body diagonal,which is $\frac{\sqrt{3} a}{2}$.Thus,the interionic distance $(d)$ is $\frac{\sqrt{3} a}{2}$.

The interionic distance for $Cesium$ $chloride$ $(CsCl)$ crystal will be (where $a$ is the edge length of the unit cell):

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