Question

An ionic compound AB has $ZnS$ (Zinc blende) type structure. If the radius $A^{+}$ is $22.5pm$, then the ideal radius of $B^{-}$ would be

• A. $54.35pm$
• B. none of these
• C. $100pm$
• D. $145.16pm$

Answer 1

The correct option is C $100pm$

The given ionic compound $AB$ has Zinc blende type structure.



In Zinc blende,
$S^{2-}$ ions forms the fcc lattice. Thus, it occupies the corners and face centre of the cubic unit cell.
Thus lattice form has 8 tetrahedral voids and 4 octahedral void for an unit cell.
$Z{n}^{2+}$ ion occupy only alternate (non adjacent) four tetrahedral void.
Each $Z{n}^{2+}$ in terahedral void is surrounded tetrahedrally by four $S^{2-}$ ions.
Each $S^{2-}$ ion is surrounded tetrahedrally by four $Z{n}^{2+}$ ions.
Thus, the coordination number of $ZnS$ is 4 : 4
Hence, by radius ratio,
$\frac{r^{A+}}{r^{B^{-}}}=0.225$

$r^{B^{-}}=\frac{22.5}{0.225}=100pm$