Value of X in Metal Deficient Oxide Formula

To find the value of $X$ in the metal deficient oxide formula $\mathbf{M}_X\mathbf{Y}_2\mathbf{O}_4$, we use the principle of electrical neutrality, which states that the sum of the positive and negative charges in a neutral compound must equal zero.

Step 1: Calculate the total negative charge

The oxide contains $4$ oxygen atoms ($\text{O}^{2-}$), each having an oxidation state of $-2$. $\text{Total negative charge from }\text{O}^{2-} = 4 \times (-2) = -8$

Step 2: Calculate the charge from metal $\mathbf{Y}$

There are $2$ atoms of metal $\mathbf{Y}$, each in the $+3$ oxidation state. $\text{Total positive charge from }\mathbf{Y}^{3+} = 2 \times (+3) = +6$

Step 3: Calculate the average oxidation state of metal $\mathbf{M}$

$\mathbf{M}$ exists in two oxidation states, $+2$ and $+3$.

• Fraction of $\mathbf{M}^{2+}$ ions $= \frac{1}{3}$
• Fraction of $\mathbf{M}^{3+}$ ions $= 1 - \frac{1}{3} = \frac{2}{3}$

The average oxidation state of $\mathbf{M}$, denoted as $\overline{Z}_{\mathbf{M}}$, is calculated as: $\overline{Z}_{\mathbf{M}} = \left(\frac{1}{3} \times (+2)\right) + \left(\frac{2}{3} \times (+3)\right)$ $\overline{Z}_{\mathbf{M}} = \frac{2}{3} + 2 = \frac{8}{3}$

Therefore, the total positive charge contributed by $X$ atoms of metal $\mathbf{M}$ is: $\text{Total charge from }\mathbf{M} = X \times \left(\frac{8}{3}\right) = \frac{8}{3}X$

Step 4: Apply the Principle of Electrical Neutrality

Setting the sum of all charges equal to zero: $\left(\text{Total charge from }\mathbf{M}\right) + \left(\text{Total charge from }\mathbf{Y}\right) + \left(\text{Total charge from }\mathbf{O}\right) = 0$

$\frac{8}{3}X + (+6) + (-8) = 0$ $\frac{8}{3}X - 2 = 0$ $\frac{8}{3}X = 2$ $X = \frac{2 \times 3}{8} = \frac{6}{8} = 0.75$

Thus, the value of $X$ is 0.75.

Correct Option: (D) 0.75