Question

A solid made up of ions of A and B which possesses edge length of unit cell equal to $0.564$ nm has four formula units. Among the two ions, the smaller one occupies the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass $9.712\times {10}^{-23}$ g.
The ionic radius (in $\mathring{A}$) for $B^{-}$ ion assuming anion - anion contact, is :

• A. $1.815$
• B. $1.994$
• C. $1.682$
• D. $1.562$

Answer 1

Answer: (B)

$1.994$

Here, $B^{-}$ ions occupy the space lattice with ccp type of arrangement.

So, $d=\sqrt{2}a=1.41\times 5.64\mathring{A}$ and $d=4r^{-}$

So, $r^{-}=1.994\mathring{A}$

Option B is correct.