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 its oxide ,Fraction of the metal which exists as `M^(3+)` would be

Options

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

Answer

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

### Step 1: Understand the composition of the oxide
The formula \( M_{0.98}O \) indicates that there are 0.98 moles of metal \( M \) for every mole of oxygen \( O \). Since oxygen typically has a charge of \( -2 \), we will use this information to balance the charges in the compound.

### Step 2: Define variables for the oxidation states
Let:
- \( x \) = fraction of metal \( M \) that is in the \( M^{3+} \) state.
- Therefore, the fraction of metal \( M \) that is in the \( M^{2+} \) state will be \( 0.98 - x \).

### Step 3: Set up the charge balance equation
The total positive charge contributed by the metal must equal the total negative charge from the oxygen. The equation can be set up as follows:

\[ 3x + 2(0.98 - x) = 2 \]

### Step 4: Simplify the equation
Expanding the equation gives:

\[ 3x + 1.96 - 2x = 2 \]

This simplifies to:

\[ x + 1.96 = 2 \]

### Step 5: Solve for \( x \)
Now, isolate \( x \):

\[ x = 2 - 1.96 = 0.04 \]

### Step 6: Calculate the fraction of \( M^{3+} \)
To find the fraction of the metal that exists as \( M^{3+} \), we divide \( x \) by the total amount of metal:

\[ \text{Fraction of } M^{3+} = \frac{x}{0.98} = \frac{0.04}{0.98} \]

### Step 7: Simplify the fraction
Calculating this gives:

\[ \frac{0.04}{0.98} = \frac{4}{98} = \frac{2}{49} \approx 0.0408 \]

### Final Answer
The fraction of the metal \( M \) that exists as \( M^{3+} \) is approximately \( 0.0408 \).

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