20 ml of 0.1 M solution of metal ion reacted with 20 ml to 0.1 M `SO_(2)` solution . `SO_(2)` reacted according to the equation `SO_(2) + 2H_(2)O rarr SO_(4)^(2-) + 4H^(+) + 2e^(-)`. If the oxidation no. of metal ion was `+ 3`, the new oxidation number of the metal would be `:`

Options

A. 0
B. `+1`
C. `+2`
D. None of these

Answer

To solve the problem step by step, we will analyze the reaction and calculate the new oxidation number of the metal ion after it reacts with sulfur dioxide (SO₂).

### Step 1: Understand the Reaction
The reaction given is:
\[ \text{SO}_2 + 2 \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4 \text{H}^+ + 2 e^- \]

From this equation, we can see that sulfur in SO₂ is oxidized to sulfate (SO₄²⁻), which involves the loss of 2 electrons. This indicates that SO₂ acts as a reducing agent.

### Step 2: Determine the n-factor for SO₂
The n-factor for SO₂ can be determined from the reaction:
- The oxidation state of sulfur changes from +4 in SO₂ to +6 in SO₄²⁻.
- The change in oxidation state is \( +6 - (+4) = +2 \).
Thus, the n-factor for SO₂ is 2.

### Step 3: Calculate the Number of Milliequivalents
We need to find the number of milliequivalents for both SO₂ and the metal ion.

1. **For SO₂:**
\[ \text{Molarity} = 0.1 \, \text{M}, \quad \text{Volume} = 20 \, \text{mL} \]
\[ \text{Number of moles of SO}_2 = \text{Molarity} \times \text{Volume (L)} = 0.1 \times 0.020 = 0.002 \, \text{moles} \]
\[ \text{Milliequivalents of SO}_2 = \text{Number of moles} \times \text{n-factor} = 0.002 \times 2 = 0.004 \, \text{equivalents} = 4 \, \text{milliequivalents} \]

2. **For the Metal Ion:**
\[ \text{Molarity} = 0.1 \, \text{M}, \quad \text{Volume} = 20 \, \text{mL} \]
\[ \text{Number of moles of Metal Ion} = 0.1 \times 0.020 = 0.002 \, \text{moles} \]
Let the n-factor for the metal ion be \( x \):
\[ \text{Milliequivalents of Metal Ion} = 0.002 \times x \]

### Step 4: Set the Milliequivalents Equal
Since the milliequivalents of SO₂ and the metal ion must be equal:
\[ 4 = 0.002 \times x \]
Solving for \( x \):
\[ x = \frac{4}{0.002} = 2000 \]

### Step 5: Determine the New Oxidation Number
The initial oxidation number of the metal ion is +3. The n-factor indicates how many electrons are involved in the reaction. Since the metal ion is reacting and losing electrons, we can find the new oxidation number by subtracting the n-factor from the initial oxidation number:
\[ \text{New Oxidation Number} = +3 - x \]
Substituting \( x = 2 \) (since the metal ion is losing 2 electrons):
\[ \text{New Oxidation Number} = +3 - 2 = +1 \]

### Final Answer
The new oxidation number of the metal ion after the reaction is **+1**.

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