$M^{+} X^{-}$ crystallizes as FCC structure. Edge length is 3.2 $A^{0}$ . Radius of anion is 1.4 A° . Assuming $M^{+} - X^{-}$ contact along the cell edge, radius of $M^{+}$ ion is . The value of is $x \times 10^{- 1} A^{0}$ . The value of ‘x’ is

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answer is 2.

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Detailed Solution

Given,

Edge length, a = 3.2 A°

Radius of anion, $r^{-} = 1.4 A^{0}$

Radius of cation, $r^{+} = ?$

$\begin{array}{l} \Rightarrow 2 (r^{+} + r^{-}) = 3.2 \\ \Rightarrow r^{+} + r^{-} = 1.6 \\ \Rightarrow r^{+} = 1.6 - 1.4 \\ \Rightarrow r^{+} = 0.2 A^{0} \\ \Rightarrow r^{+} = \underset{x}{\underset{⏟}{2}} \times 10^{- 1} A^{0} \end{array}$

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