CsCl crystallises in a body-centred cubic (BC | vedclass

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Question

(C) In a $BCC$ unit cell of $CsCl$,the $Cs^{+}$ ion is at the body center and $Cl^{-}$ ions are at the corners.The body diagonal of the cube is equal

Vedclass · Accepted Answer

$r_{Cs^{+}} + r_{Cl^{-}} = \frac{\sqrt{3}}{2} a$

Vedclass · Answer

(C) In a $BCC$ unit cell of $CsCl$,the $Cs^{+}$ ion is at the body center and $Cl^{-}$ ions are at the corners.The body diagonal of the cube is equal to $\sqrt{3} a$.Along the body diagonal,the ions are in contact such that the distance is $r_{Cl^{-}} + 2r_{Cs^{+}} + r_{Cl^{-}} = \sqrt{3} a$.This simplifies to $2(r_{Cs^{+}} + r_{Cl^{-}}) = \sqrt{3} a$.Therefore,the correct expression is $r_{Cs^{+}} + r_{Cl^{-}} = \frac{\sqrt{3}}{2} a$.

$CsCl$ crystallises in a body-centred cubic $(BCC)$ lattice. If '$a$' is its edge length,then which of the following expressions is correct?

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