In the cubic crystal of CsCl (d=3.97gcm⁻³) the eight corners are occupied by Cl⁻ with a Cs⁺ at the centre and vice-versa. Calculate the distance between the neighboring Cs⁺ and Cl⁻ ions. What is the approximate radius ratio of the two ions? (At. wt. of Cs=132.91 and Cl=35.45)

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$0.51 A °, 0.732$

$3.46 A °, 1.2$

$3.57 A °, 0.732$

$3.57 A °, 0.54$

answer is C.

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Detailed Solution

Edge length of CsCl is:

$\begin{matrix} ρ & = & \frac{nM}{N_{a} a^{3}} \\ 3.97 & = & \frac{1 \times 168.35}{a^{3} \times 6.02 \times 10^{23}} \\ a & = & 4.13 \overset{°}{A} \end{matrix}$

For BCC:

$r_{+} + r_{-} = \frac{a \sqrt{3}}{2} = 3.57 A °$

The coordination number of CsCl is 8, therefore its radius ratio falls under range of 0.732-1.0.

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