Formula of metal oxide with metal deficiency defect in its crystal...

$A: \mathrm{O}=0.8: 1=80: 100$ Thus, if there are $100 \mathrm{O}$ atoms, then number of $A$ atoms $=80$ Charge on $100 \mathrm{O}^{2-}$ ions $=100 \times(-2)=-200$ Suppose number of $A$ atoms present as $A^{2+}=x$ $\therefore \quad$ Number of $\mathrm{A}$ atoms present as $A^{3+}=80-x$ Total charge on $A^{2+}$ and $A^{3+}=x \times(+2)$ $+(80-x) \times(+3)$ $=2 x+240-3 x=240-x$ As metal oxide is neutral, Total charge on cations = total charge on anions $\therefore \quad 240-x=200$ or $x=40$ $\therefore \quad$ Fraction of metal existing as $A^{2+}$ ions in the crystal $=40 / 80=0.5$