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# A binary solid has zinc blend structure with $B^+$ ions constituting the lattice and $A^+$ ions occupying 25% tetrahedral voids. The formula of solid is

$\begin{array}{1 1}(A)\;AB&(B)\;A_2B\\(C)\;AB_2&(D)\;AB_4\end{array}$

No. of B− ions in unit cell =$8\times \large\frac{1}{8}+\frac{1}{2}$$\times 6=4 Now A^+ ion occupies 25% of tetrahedral voids \therefore No.of A^+=\large\frac{8\times 25}{100}$$=2$
Thus ratio of $A^+$ to B is 1: 2
Hence formula is $AB_2$
Hence (C) is the correct answer.