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CsCl crystallises in a cubic and that has a Cl– at each corner and $Cs^+$ at the centre of unit cell. If $r_{Cs^+} = 1.69A^{\large\circ}$ and $r_{Cl^-}= 1.81A^{\large\circ}$, what is the value of edge-length ‘a’ of the cube? Compare this value of ‘a’ (calculated) from the observed density of CsCl, $3.97 gm/cm^3$ [M(CsCl) = 168.5]

1 Answer

Hence (A) is the correct answer
answered Jun 4, 2014 by sreemathi.v