In a metal deficient oxide sample, M_XY₂O₄ (M and Y are metals), M is present in both +2 and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M²⁺ ions present in M is 1/3 , the value of X is _____.

To find the value of $X$ in the metal-deficient oxide sample $M_X{Y}_2{O}_4$, we use the principle of overall electrical neutrality. The sum of the positive charges of the metal cations must equal the total negative charge of the oxide anions.

Let us analyze the charges of the constituents in the formula $M_X{Y}_2{O}_4$:
1. Oxygen is present as oxide ions, $O^{2-}$. There are 4 oxide ions, so the total negative charge is:
$\text{Total negative charge}=4\times (-2)=-8$
Therefore, the total positive charge of the cations $M_X$ and $Y_2$ must be $+8$ to maintain electrical neutrality.

2. The metal $Y$ is present in the $+3$ oxidation state as $Y^{3+}$. There are 2 ions of $Y$, so the total charge contributed by $Y$ is:
$\text{Charge from}Y=2\times (+3)=+6$

3. Let the total charge contributed by the metal $M$ be $Q_M$. Using the charge balance equation:
$Q_M+6=8$
$Q_M=+2$

4. The metal $M$ is present in both $+2$ and $+3$ oxidation states ($M^{2+}$ and $M^{3+}$). The total number of moles/atoms of $M$ per formula unit is $X$.
We are given that the fraction of $M^{2+}$ ions in $M$ is $\frac{1}{3}$. This means:
$\text{Number of}{M}^{2+}\text{ions}=\frac{1}{3}X$
The remaining fraction of $M$ must be in the $+3$ oxidation state:
$\text{Number of}{M}^{3+}\text{ions}=(1-\frac{1}{3})X=\frac{2}{3}X$

5. Now we express the total positive charge contributed by $M$ in terms of $X$:
$Q_M=(\text{Number of}{M}^{2+}\times 2)+(\text{Number of}{M}^{3+}\times 3)$
Substitute the expressions into the equation:
$2=(\frac{1}{3}X\times 2)+(\frac{2}{3}X\times 3)$
$2=\frac{2}{3}$
$2=(\frac{2}{3}+\frac{6}{3})X$
$2=\frac{8}{3}$X
Solving for $X$:
$X=2\times \frac{3}{8}$
$X=\frac{3}{4}$
$X=0.75$

Thus, the value of $X$ is $0.75$, which corresponds to the correct option of 0.75.