Formula of metal oxide with metal deficiency defect in its crystal is A_0.8O. The crystal contains A²⁺ and A³⁺ ions. The fraction of metal existing as A²⁺ ions in the crystal is

Thus, if there are 100 O atoms, then number of A atoms = 80

$\text{ Charge on }100{\mathrm{O}}^{2-}\text{ ions }=100\times (-2)=-200$

Suppose number of A atoms present as A2^x = x

$\therefore$   Number of A atoms present as $A^{3+}=80-x$

Total charge on $A^{2+}\text{ and }{A}^{3+}=x\times (+2)+(80-x)\times (+3)$

$=2x+240-3x=240-x$

As metal oxide is neutral,
Total charge on cations = total charge on anions

$\therefore 240-x=200\text{ or }x=40$

$\therefore$   Fraction of metal existing as A²⁺ ions in the crystal

$=\frac{40}{80}=0.5$