Question

The radius ratio of $C{s}^{+}$ to $C{l}^{-}$ is 0.90. If the radius of $C{l}^{-}$ ion is $1.80\stackrel{\circ }{A}$, the false statement regarding the CsCl structure is ?

• A. The distance between the centres of $C{l}^{-}$ ions is greater than $3.60\stackrel{\circ }{A}$
• B. The distance between $C{s}^{+}$ ion and $C{l}^{-}$ is $3.60\stackrel{\circ }{A}$
• C. The distance between $C{s}^{+}$ ion and $C{l}^{-}$ is $3.42\stackrel{\circ }{A}$
• D. The distance between two $C{s}^{+}$ ions is greater than $3.24\stackrel{\circ }{A}$

Answer 1

The correct option is B

The distance between $C{s}^{+}$ ion and $C{l}^{-}$ is $3.60\stackrel{\circ }{A}$

$\frac{r_{C{s}^{+}}}{r_{C{l}^{-}}}=0.90\stackrel{\circ }{A}\Rightarrow r_{C{s}^{+}}=0.90\times r_{C{l}^{-}}=0.90\times 1.80=1.62\stackrel{\circ }{A}$
For precise fitting of the cubic voids, $\frac{r_{C{s}^{+}}}{r_{C{l}^{-}}}$ must be 0.732 which is smaller than the given value of 0.90. Hence cubic packing of $C{l}^{-}$ ions is loose to some extent, i.e., two $C{l}^{-}$ ions are not in touch with each other and the distance between two $C{l}^{-}$ ions will be greater than $3.60\stackrel{\circ }{A}$.
The $C{s}^{+}$ ion at the centre of the body is in touch with $2C{l}^{-}$ ions at the opposite corners of the cube. Hence distance between $C{s}^{+}$ and $C{l}^{-}=r_{C{S}^{+}}+r_{C{l}^{-}}=1.62\stackrel{\circ }{A}+1.80\stackrel{\circ }{A}=3.42\stackrel{\circ }{A}$