Question

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 \(\frac{1}{3}\), the value of X is______.

• A. 0.25
• B. 0.33
• C. 0.67
• D. 0.75

Answer 1

The Correct Option is D

Charge Neutrality and Fraction of Ions Calculation 

Consider the compound \( \text{M}_x \text{Y}_2 \text{O}_4 \), where the M ions exist in two oxidation states: \( \text{M}^{2+} \) and \( \text{M}^{3+} \). The fraction of \( \text{M}^{2+} \) ions is given as \( \frac{1}{3} \). Thus:

\[ \text{M}^{2+} = \frac{x}{3}, \quad \text{M}^{3+} = \frac{2x}{3} \]

Step 1: Apply Charge Neutrality

For the compound to be neutral, the total positive and negative charges must balance. This gives us the equation: \[ \frac{2x}{3} \cdot (+3) + \frac{x}{3} \cdot (+2) + 2 \cdot (+3) + 4 \cdot (-2) = 0 \] Simplify the terms: \[ \frac{6x}{3} + \frac{2x}{3} + 6 - 8 = 0 \]

Step 2: Solve for \( x \)

Combine terms: \[ \frac{6x}{3} + \frac{2x}{3} = 2 \] Multiply through by 3 to eliminate the denominators: \[ 6x + 2x = 6 \] Simplify: \[ 8x = 6 \quad \Rightarrow \quad x = \frac{6}{8} = 0.75 \]

Final Answer:

The value of \( x \) is \( x = 0.75 \).

Answer 2

To solve the problem, we use charge neutrality to find the value of \(X\) in the formula \(M_X Y_2 O_4\), given the oxidation states and fraction of \(M^{2+}\) ions.

Given:
- Formula: \(M_X Y_2 O_4\)
- \(M\) exists in \(+2\) and \(+3\) oxidation states
- Fraction of \(M^{2+}\) ions = \(\frac{1}{3}\)
- \(Y\) is in \(+3\) oxidation state
- Oxygen is in \(-2\) oxidation state

Step 1: Define variables for moles of \(M^{2+}\) and \(M^{3+}\):
Let total moles of \(M\) be \(X\). Then:
\[ \text{Moles of } M^{2+} = \frac{1}{3} X \] \[ \text{Moles of } M^{3+} = X - \frac{1}{3} X = \frac{2}{3} X \]

Step 2: Calculate total positive charge from \(M\) and \(Y\):
\[ \text{Charge from } M = \left(\frac{1}{3} X\right) \times (+2) + \left(\frac{2}{3} X\right) \times (+3) = \frac{2}{3} X + 2 X = \frac{2}{3} X + 2 X = \frac{8}{3} X \] (Notice corrected addition: \(\frac{1}{3}X \times 2 = \frac{2}{3} X\), and \(\frac{2}{3} X \times 3 = 2 X\), sum is \(\frac{2}{3} X + 2 X = \frac{8}{3} X\)) Charge from \(Y_2\): \[ 2 \times (+3) = +6 \]

Step 3: Total negative charge from oxygen:
\[ 4 \times (-2) = -8 \]

Step 4: Apply charge neutrality:
\[ \text{Total positive charge} + \text{Total negative charge} = 0 \] \[ \frac{8}{3} X + 6 - 8 = 0 \implies \frac{8}{3} X - 2 = 0 \] \[ \frac{8}{3} X = 2 \implies X = \frac{2 \times 3}{8} = \frac{6}{8} = 0.75 \]

Final Answer:
\[ \boxed{0.75} \]