QJEEAdv24Ch4253: JEE Advanced 2024 Chemistry - Class 12 JEEAdv24Ch4253 Solution: In a metal deficient oxide sample, ( and are metals), is…

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