Question

In a cubic close packed structure of mixed oxides, the lattice is made up of oxides ions one eighth tetrahedral voids are occupied by divalent ions (A^2+) while one half of octahedral voids are occupied by trivalent ions (B^3+). what is the formula of the oxide.

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Answer 1

${\bold{\underline{\underline{Answer:}}}}$

Solution : In a closed packed arrangement there is one octahedral and two tetrahedral voids corresponding to each atom constituting the lattice. Therefore, if

$\implies$ Number of oxide ions (O) per unit cell = 1

$\implies$ Number if tetrahedral voids per oxide ion in lattice = 1 × 2 = 2

$\implies$ Number of divalent ions

$(A^{ +2})$ions = $\frac{1}{8} \times2 = \frac{1}{4}$

$\implies$ Number of Octahedral voids per oxide ion in lattice = 1 × 1 = 1

$\implies$ Number of trivalent

$(B^{3 + })\:ions = 1 \times \frac{1}{2} = \frac{1}{2}$

$\implies$ Now, Formula of the compound = $\sf \:A_{\tiny{}^{1} / {}_{4}}\sf B_{\tiny{}^{1} / {}_{2}}O$

$\implies$ $\therefore$ Whole Number Formula = $\sf\:AB_2O_4.$