Question

Analysis shows that a transition metal forms an oxide with the formula \[ \mathrm{M}_x\mathrm{O}_{1.00} \] (where \(x<1\)). The ratio of \(\mathrm{M}^{3+}\) and \(\mathrm{M}^{2+}\) ions in the oxide is:

• A. \(\dfrac{x-1}{3x-2}\)
• B. \(\dfrac{3x-2}{x-1}\)
• C. \(\dfrac{2(x-1)}{3x-2}\)
• D. \(\dfrac{2(1-x)}{3x-2}\)

Hint:

For non-stoichiometric oxides, assume the amount of oxygen as unity and apply charge neutrality. This directly gives the amount of ions in different oxidation states.

Answer 1

The Correct Option is D

Concept: This question is based on metal deficiency defects in transition metal oxides. Due to variable oxidation states, some \(\mathrm{M}^{2+}\) ions are oxidized to \(\mathrm{M}^{3+}\), maintaining overall electrical neutrality.

Step 1: Determine the number of ions Given oxide: \[ \mathrm{M}_x\mathrm{O} \] Assume: \[ \text{Number of } \mathrm{M}^{3+}\text{ ions}=n \] Then \[ \text{Number of } \mathrm{M}^{2+}\text{ ions}=x-n \]

Step 2: Apply charge neutrality Total positive charge: \[ 3n+2(x-n) \] Total negative charge from one oxide ion: \[ 2 \] Hence, \[ 3n+2(x-n)=2 \] \[ 3n+2x-2n=2 \] \[ n+2x=2 \] \[ n=2-2x=2(1-x) \]

Step 3: Find the number of \(\mathrm{M}^{2+}\) ions \[ x-n=x-\bigl(2-2x\bigr) \] \[ =3x-2 \]

Step 4: Calculate the required ratio \[ \frac{\mathrm{M}^{3+}}{\mathrm{M}^{2+}} = \frac{n}{x-n} \] \[ = \frac{2(1-x)}{3x-2} \] Hence, \[ \boxed{\frac{2(1-x)}{3x-2}} \] Therefore, the correct answer is Option (D).