Experimentally it was found that a metal oxide has formula ` M_(0.98)O`. Metal M is present as ` M^(2+) and M^(3+)` in iyts oxide. Fraction of the metal which exists as ` M^(3+)` would be

Options

A. 0.0508
B. 0.0701
C. 0.0408
D. 0.0605

Explanation

To solve the problem, we need to determine the fraction of the metal M that exists as \( M^{3+} \) in the oxide \( M_{0.98}O \).

### Step-by-Step Solution:

1. **Identify the Formula and Oxidation States**:
The given formula of the metal oxide is \( M_{0.98}O \). In this oxide, metal M can exist in two oxidation states: \( M^{2+} \) and \( M^{3+} \).

2. **Determine the Total Charge from Oxygen**:
Oxygen in the oxide has an oxidation state of \(-2\). Since there is one oxygen atom, the total negative charge contributed by oxygen is \(-2\).

3. **Set Up the Charge Balance Equation**:
Let \( x \) be the fraction of metal M that is in the \( M^{3+} \) state. Therefore, the fraction of metal M in the \( M^{2+} \) state will be \( 0.98 - x \). The total positive charge contributed by the metal can be expressed as:
\[ 3x + 2(0.98 - x) = 2 \]
Here, \( 3x \) is the contribution from \( M^{3+} \) and \( 2(0.98 - x) \) is from \( M^{2+} \).

4. **Simplify the Equation**:
Expanding the equation gives:
\[ 3x + 1.96 - 2x = 2 \]
Simplifying this results in:
\[ x + 1.96 = 2 \]

5. **Solve for \( x \)**:
Subtract \( 1.96 \) from both sides:
\[ x = 2 - 1.96 = 0.04 \]

6. **Calculate the Fraction of \( M^{3+} \)**:
The fraction of metal M that exists as \( M^{3+} \) is \( x = 0.04 \).

7. **Calculate the Total Fraction of Metal**:
To find the fraction of the total metal that is \( M^{3+} \), we divide by the total amount of metal:
\[ \text{Fraction of } M^{3+} = \frac{x}{0.98} = \frac{0.04}{0.98} \approx 0.0408 \]

### Final Answer:
The fraction of the metal which exists as \( M^{3+} \) is approximately \( 0.0408 \) or \( 4.08\% \).