Question

In a metal-deficient oxide sample, \(M_xY_2O_4\) (\(M\) 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 \(M^{2+}\) ions present in \(M\) is \(\frac{1}{3}\), the value of \(X\) is \_\_\_\_.

• A. \(0.25\)
• B. \(0.33\)
• C. \(0.67\)
• D. \(0.75\)

Hint:

In metal-deficient oxides, balance the charges of all ions using the oxidation states and stoichiometry.

Answer 1

The Correct Option is D

Let \(M_xY_2O_4\) contain \(M^{2+}\) ions in a fraction of \(\frac{1}{3}\).
So, \(M^{2+} = \frac{x}{3}\) and \(M^{3+} = \frac{2x}{3}\). From charge neutrality: \[ \frac{2x}{3} \cdot (+3) + \frac{x}{3} \cdot (+2) + 2(+3) + 4(-2) = 0. \] Simplify: \begin{align*} \frac{2x}{3} + 2x - 2 &= 0
\frac{8x}{3} &= 2
x &= \frac{6}{8} = \frac{3}{4} = 0.75 \end{align*} Thus, the value of \(X\) is \(0.75\).