Formula of a metal oxide with metal deficiency defect in its crystal is  ${\mathrm{M}}_{0.96}\mathrm{O}.$If the crystal contains ${\mathrm{M}}^{+2}$and ${\mathrm{M}}^{+3}$ ions, the fraction of metal existing as ${\mathrm{M}}^{+3}$ions in the crystal is

               Let number of oxide ions = 100

Net charge of oxide ions =  $2\times 100$

$\mathrm{Let}\mathrm{the}\mathrm{sum}\mathrm{of}\left({\mathrm{M}}^{+2}+{\mathrm{M}}^{+3}\right)=96$

$\mathrm{Number}\mathrm{of}{\mathrm{M}}^{+2}\mathrm{ions}=\text{\hspace{0.17em}X}$

$\mathrm{Number}\mathrm{of}{\mathrm{M}}^{+3}\mathrm{ions}=(96-\text{\hspace{0.17em}X})$

Net charge of cation = 2X+3(96- X) = 288 - X
                                                       
Even though it is non-stoichiometric, electrical neutrality is maintained.

Net charge of cations = Net charge of anions

                          288- X   =      200
                                      X = 88

$$\begin{gathered} \therefore \mathrm{Number}\mathrm{of}{\mathrm{M}}^{+2}\mathrm{ions}=88 \\ \mathrm{Number}\mathrm{of}{\mathrm{M}}^{+3}\mathrm{ions}=(96-88)=8 \end{gathered}$$
 

Fraction of   ${\mathrm{M}}^{+3}$ ions = $\frac{8}{96} =0.0833$