Question

A solid AB crystallises in $NaCl$ structure. The anion $B^{-}$form fcc lattice, while the cation $A^{+}$ occupy all the octahedral holes. If all the particles as shown in the figure are removed, then the formula of solid becomes :-

• A. $AB$
• B. $A_5{B}_7$
• C. $A_7{B}_5$
• D. $A{B}_2$

Answer 1

Answer: (A)

$A^{+}$
$4\times \frac{1}{8}+4\frac{1}{2}$
$\frac{1}{2}+2$
$2.5$
$B^{-}$
$10\times \frac{1}{4}$
$2.5$
$AB$