Question 2

In the oxide $M_X Y_2 O_4$: Oxidation state of $Y = +3$. Total charge from $Y = 2 \times (+3) = +6$. Oxidation state of $O = -2$. Total charge from $O = 4 \times (-2) = -8$. Let the total number of $M$ ions be $X$. Given: Fraction of $M^{2+}$ is $\frac{1}{3}$, so fraction of $M^{3+}$ is $1 - \frac{1}{3} = \frac{2}{3}$. Number of $M^{2+}$ ions $= \frac{1}{3}X$. Number of $M^{3+}$ ions $= \frac{2}{3}X$. For the compound to be electrically neutral, the sum of charges must be zero: $[\frac{1}{3}X \times (+2) + \frac{2}{3}X \times (+3)] + [2 \times (+3)] + [4 \times (-2)] = 0$ $\frac{2}{3}X + 2X + 6 - 8 = 0$ $\frac{8}{3}X = 2$ $X = \frac{6}{8} = 0.75$ Thus, the value of $X$ is 0.75. Correct option is (D).