Which of the following can react with `K_(2)Cr_(2)O_(7)`

Options

A. a. `SO_(3)^(-2)`
B. b.`CO_(3)^(-2)`
C. c. `SO_(4)^(-2)`
D. d. `NO_(3)^(-)`

Answer

To determine which of the given compounds can react with potassium dichromate (K₂Cr₂O₇), we need to analyze the oxidation states of the central atoms in each compound. K₂Cr₂O₇ is a strong oxidizing agent, and it can oxidize species that have lower oxidation states to higher oxidation states.

Let's analyze each option step by step:

### Step 1: Analyze Option A - SO₃²⁻ (Sulfite Ion)
1. Let the oxidation state of sulfur (S) be \( X \).
2. The formula for sulfite is SO₃²⁻, which has 3 oxygen atoms each with an oxidation state of -2.
3. The equation for the oxidation state is:
\[ X + 3(-2) = -2 \]
\[ X - 6 = -2 \implies X = +4 \]
4. The oxidation state of sulfur in SO₃²⁻ is +4.

### Step 2: Analyze Option B - CO₃²⁻ (Carbonate Ion)
1. Let the oxidation state of carbon (C) be \( X \).
2. The formula for carbonate is CO₃²⁻, which has 3 oxygen atoms each with an oxidation state of -2.
3. The equation for the oxidation state is:
\[ X + 3(-2) = -2 \]
\[ X - 6 = -2 \implies X = +4 \]
4. The oxidation state of carbon in CO₃²⁻ is +4.

### Step 3: Analyze Option C - SO₄²⁻ (Sulfate Ion)
1. Let the oxidation state of sulfur (S) be \( X \).
2. The formula for sulfate is SO₄²⁻, which has 4 oxygen atoms each with an oxidation state of -2.
3. The equation for the oxidation state is:
\[ X + 4(-2) = -2 \]
\[ X - 8 = -2 \implies X = +6 \]
4. The oxidation state of sulfur in SO₄²⁻ is +6.

### Step 4: Analyze Option D - NO₃⁻ (Nitrate Ion)
1. Let the oxidation state of nitrogen (N) be \( X \).
2. The formula for nitrate is NO₃⁻, which has 3 oxygen atoms each with an oxidation state of -2.
3. The equation for the oxidation state is:
\[ X + 3(-2) = -1 \]
\[ X - 6 = -1 \implies X = +5 \]
4. The oxidation state of nitrogen in NO₃⁻ is +5.

### Step 5: Determine Reactivity with K₂Cr₂O₇
- K₂Cr₂O₇ can oxidize species that have lower oxidation states to higher oxidation states.
- The oxidation states we found are:
- SO₃²⁻: +4
- CO₃²⁻: +4
- SO₄²⁻: +6
- NO₃⁻: +5

### Conclusion
- The only species that can be oxidized by K₂Cr₂O₇ is SO₃²⁻, as it has a lower oxidation state (+4) compared to the maximum oxidation state of sulfur (+6) in SO₄²⁻.
- Therefore, the correct answer is **Option A: SO₃²⁻**.