To find the value of $ X $ in the metal deficient oxide sample $ M_x Y_2 O_4 $, where $ M $ is present in both +2 and +3 oxidation states and $ Y $ is in the +3 oxidation state, and given that the fraction of $ M^{2+} $ ions present in $ M $ is $\frac{1}{3}$, follow these steps:

1. Charge Balance Equation:

The total charge contributed by the cations ($ M $ and $ Y $) must balance the total negative charge contributed by the oxide ions $ O $:

$ x(M^{2+} + M^{3+}) + 2(Y^{3+}) = 4(-2) $

1. Oxidation States and Charges:

• Let the total number of $ M $ ions be $ x $.


• Given $\frac{1}{3}$ of $ M $ ions are $ M^{2+} $, the remaining $\frac{2}{3}$ are $ M^{3+} $.

1. Number of Ions:

• Number of $ M^{2+} $ ions = $\frac{x}{3}$.


• Number of $ M^{3+} $ ions = $\frac{2x}{3}$.

1. Total Positive Charge:

• Charge from $ M^{2+} $ ions = $ \left(\frac{x}{3}\right) \times 2 = \frac{2x}{3} $.


• Charge from $ M^{3+} $ ions = $ \left(\frac{2x}{3}\right) \times 3 = 2x $.


• Charge from $ Y^{3+} $ ions = $ 2 \times 3 = 6 $.

1. Total Positive Charge Calculation:

$ \frac{2x}{3} + 2x + 6 $

1. Total Negative Charge:

• Charge from 4 oxygen ions $ (O^{2-}) $:

$ 4 \times (-2) = -8 $

1. Setting Up the Charge Balance:

$ \frac{2x}{3} + 2x + 6 = 8 $

1. Solving for $ x $:

$ \frac{2x}{3} + 2x + 6 = 8 $

$ \frac{2x}{3} + 2x = 2 $

$ \frac{2x + 6x}{3} = 2 $

$ \frac{8x}{3} = 2 $

$ 8x = 6 $

$ x = \frac{6}{8} $

$ x = 0.75 $

Thus, the value of $ X $ is $ \boxed{0.75} $.

Option D: 0.75