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

Options

A. `4.08%`
B. `6.05%`
C. `5.08%`
D. `7.01%`

Explanation

To solve the problem, we need to determine the fraction of the metal \( M \) that exists in the \( M^{3+} \) oxidation state in the metal oxide \( M_{0.98}O \). Here’s a step-by-step breakdown of the solution:

### Step 1: Define the variables
Let:
- \( x \) = number of moles of \( M^{3+} \)
- \( 98 - x \) = number of moles of \( M^{2+} \)

### Step 2: Write the charge balance equation
The total charge contributed by the metal ions and oxygen must equal zero for the compound to be electrically neutral. The charge from oxygen (which is always \( -2 \)) for \( 100 \) moles of the oxide is:
\[ \text{Total charge from oxygen} = -2 \times 100 = -200 \]

The charge contributed by \( M^{3+} \) and \( M^{2+} \) can be expressed as:
\[ \text{Total charge from metal} = 3x + 2(98 - x) \]

### Step 3: Set up the equation for charge neutrality
Setting the total charge from the metal equal to the total charge from oxygen gives us:
\[ 3x + 2(98 - x) - 200 = 0 \]

### Step 4: Simplify the equation
Expanding and simplifying the equation:
\[ 3x + 196 - 2x - 200 = 0 \]
\[ x - 4 = 0 \]
\[ x = 4 \]

### Step 5: Calculate the fraction of \( M^{3+} \)
Now that we have \( x = 4 \), we can find the fraction of \( M^{3+} \) in the total amount of metal:
\[ \text{Total amount of metal} = 0.98 \text{ moles} \]
The fraction of \( M^{3+} \) is:
\[ \text{Fraction of } M^{3+} = \frac{x}{0.98} = \frac{4}{0.98} \]

### Step 6: Calculate the percentage
To find the percentage of \( M^{3+} \):
\[ \text{Percentage of } M^{3+} = \left( \frac{4}{0.98} \right) \times 100 \approx 4.08\% \]

### Final Answer
The fraction of metal \( M \) that exists as \( M^{3+} \) is approximately **4.08%**.

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