Step-by-Step Solution

1. Apply the Principle of Electroneutrality: The given compound, $M_x Y_2 O_4$, is electrically neutral. This implies that the sum of the positive charges from the cations (M and Y) must balance the sum of the negative charges from the anions (O). $\text{Total Positive Charge} + \text{Total Negative Charge} = 0$

2. Calculate the Total Negative Charge: The formula unit contains 4 oxide ions ($O^{2-}$). The charge of each oxide ion is -2. $\text{Total Negative Charge} = 4 \times (-2) = -8$

3. Calculate the Total Positive Charge: For the compound to be neutral, the total positive charge must be +8 to balance the -8 charge from the oxide ions. $\text{Total Positive Charge} = +8$

4. Calculate the Charge Contribution from Y Ions: The formula unit contains 2 Y ions, and it is given that the oxidation state of Y is +3 ($Y^{3+}$). $\text{Charge from Y ions} = 2 \times (+3) = +6$

5. Determine the Required Charge Contribution from M Ions: The total positive charge of +8 is contributed by both M and Y ions. The charge required from the $x$ M ions is the difference between the total positive charge and the charge from Y ions. $\text{Charge from M ions} = (\text{Total positive charge}) - (\text{Charge from Y ions})$ $\text{Charge from M ions} = +8 - (+6) = +2$ Thus, the $x$ moles of M ions in one formula unit must contribute a total charge of +2.

6. Express the Number of M²⁺ and M³⁺ Ions in terms of x: The total number of M ions in the formula unit is $x$. We are given that the fraction of $M^{2+}$ ions among all M ions is $\frac{1}{3}$.

   • Number of $M^{2+}$ ions = $(\text{fraction of } M^{2+}) \times (\text{total M ions}) = \frac{1}{3} \times x = \frac{x}{3}$ The remaining fraction of M ions must be in the +3 state.
   • Fraction of $M^{3+}$ ions = $1 - \frac{1}{3} = \frac{2}{3}$
   • Number of $M^{3+}$ ions = $(\text{fraction of } M^{3+}) \times (\text{total M ions}) = \frac{2}{3} \times x = \frac{2x}{3}$

7. Set up an Equation and Solve for x: The total charge from M ions is the sum of the charges from $M^{2+}$ and $M^{3+}$ ions. We equate this to the required charge of +2 calculated in Step 5. $(\text{Number of } M^{2+} \times \text{Charge}) + (\text{Number of } M^{3+} \times \text{Charge}) = +2$ $\left(\frac{x}{3} \times (+2)\right) + \left(\frac{2x}{3} \times (+3)\right) = 2$ $\frac{2x}{3} + \frac{6x}{3} = 2$ $\frac{8x}{3} = 2$ $8x = 6$ $x = \frac{6}{8} = \frac{3}{4}$

8. Convert to Decimal and Select the Correct Option: $x = 0.75$ This value corresponds to option D.