A metal oxide has empirical formula `M_(0.96)O_(1.00)` What will be the percentage of `M^(2+)` ion in the crystal ?

Options

A. `90.67`
B. `91.67`
C. `8.33`
D. `9.33`

Explanation

To find the percentage of \( M^{2+} \) ions in the crystal with the empirical formula \( M_{0.96}O_{1.00} \), we can follow these steps:

### Step 1: Understand the Composition
The empirical formula indicates that there are 0.96 moles of metal \( M \) for every 1 mole of oxygen \( O \). This suggests that the metal can exist in different oxidation states, specifically \( M^{2+} \) and \( M^{3+} \).

### Step 2: Define Variables
Let:
- \( x \) = moles of \( M^{2+} \)
- \( (0.96 - x) \) = moles of \( M^{3+} \)

### Step 3: Charge Balance
The total positive charge from the metal ions must equal the total negative charge from the oxygen ions. Since there is 1 mole of \( O^{2-} \), the total negative charge is \( -2 \).

The total positive charge can be expressed as:
\[ 2x + 3(0.96 - x) = 2 \]
This equation accounts for the charge contribution from both \( M^{2+} \) and \( M^{3+} \).

### Step 4: Simplify the Equation
Expanding the equation gives:
\[ 2x + 2.88 - 3x = 2 \]
Combining like terms results in:
\[ -1x + 2.88 = 2 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ -1x = 2 - 2.88 \]
\[ -1x = -0.88 \]
Thus,
\[ x = 0.88 \]
This means there are 0.88 moles of \( M^{2+} \).

### Step 6: Calculate the Percentage of \( M^{2+} \)
To find the percentage of \( M^{2+} \) ions in the crystal:
\[ \text{Percentage of } M^{2+} = \left( \frac{0.88}{0.96} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage of } M^{2+} = 91.67\% \]

### Final Answer
The percentage of \( M^{2+} \) ions in the crystal is **91.67%**.

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