In a solid, $\mathrm{B}^{-}$ions occupy corners of a cube forming ccp structure. If $A^{+}$ion occupy half…
In a solid, $\mathrm{B}^{-}$ions occupy corners of a cube forming ccp structure. If $A^{+}$ion occupy half the tetrahedral voids, formula of the solid is
$\mathrm{A}_2 \mathrm{~B}$
$\mathrm{AB}_2$
$\quad A_2 B_3$
AB
Solution
$\mathrm{B}^{-}$ions occupy corners of cube and forms ccp structure.
The number of tetrahedral voids generated is twice the number of $\mathrm{B}^{-}$ions.
Thus, number of tetrahedral voids $=2 \mathrm{~B}$
$\mathrm{A}^{+}$ions occupy ( $1 / 2$ ) of these tetrahedral voids.
Hence, number of $\mathrm{A}^{+}$ions $=2 \mathrm{~B} \times 1 / 2=1 \mathrm{~B}$
Ratio of $\mathrm{A}^{+}$and $\mathrm{B}^{-}$ions $=1 \mathrm{~B}: 1 \mathrm{~B}=1: 1$
$\therefore \quad$ Formula of compound $=\mathrm{AB}$