In which of the following pairs. the oxidation states, of sulphuric and chromium are same?

Options

A. `SO_(3)^(2-) , CrO_(4)^(2-)`
B. `SO_(3), CrO_(4)^(2-)`
C. `SO_(2), CrO_(4)^(2-)`
D. `SO_(2), Cr_(2)O_(7)^(2-)`

Answer

To determine in which of the given pairs the oxidation states of sulfur and chromium are the same, we will analyze each pair step by step.

### Step 1: Analyze the first pair - SO3²⁻ and CrO4²⁻
1. **Calculate the oxidation state of sulfur in SO3²⁻:**
- Let the oxidation state of sulfur be \( x \).
- The oxidation state of oxygen is \(-2\).
- The equation for the ion is:
\[ x + 3(-2) = -2 \]
- Simplifying gives:
\[ x - 6 = -2 \implies x = +4 \]
- Therefore, the oxidation state of sulfur in SO3²⁻ is **+4**.

2. **Calculate the oxidation state of chromium in CrO4²⁻:**
- Let the oxidation state of chromium be \( y \).
- The equation for the ion is:
\[ y + 4(-2) = -2 \]
- Simplifying gives:
\[ y - 8 = -2 \implies y = +6 \]
- Therefore, the oxidation state of chromium in CrO4²⁻ is **+6**.

### Step 2: Analyze the second pair - SO3 and CrO4²⁻
1. **Calculate the oxidation state of sulfur in SO3:**
- Let the oxidation state of sulfur be \( x \).
- The equation is:
\[ x + 3(-2) = 0 \]
- Simplifying gives:
\[ x - 6 = 0 \implies x = +6 \]
- Therefore, the oxidation state of sulfur in SO3 is **+6**.

2. **The oxidation state of chromium in CrO4²⁻ is already calculated as +6.**

### Step 3: Analyze the third pair - SO2 and CrO4²⁻
1. **Calculate the oxidation state of sulfur in SO2:**
- Let the oxidation state of sulfur be \( x \).
- The equation is:
\[ x + 2(-2) = 0 \]
- Simplifying gives:
\[ x - 4 = 0 \implies x = +4 \]
- Therefore, the oxidation state of sulfur in SO2 is **+4**.

2. **The oxidation state of chromium in CrO4²⁻ is still +6.**

### Step 4: Analyze the fourth pair - SO2 and Cr2O7²⁻
1. **The oxidation state of sulfur in SO2 is +4 (already calculated).**

2. **Calculate the oxidation state of chromium in Cr2O7²⁻:**
- Let the oxidation state of chromium be \( y \).
- The equation is:
\[ 2y + 7(-2) = -2 \]
- Simplifying gives:
\[ 2y - 14 = -2 \implies 2y = 12 \implies y = +6 \]
- Therefore, the oxidation state of chromium in Cr2O7²⁻ is **+6**.

### Conclusion
- The pairs with oxidation states:
- SO3²⁻ (S: +4) and CrO4²⁻ (Cr: +6) - Not the same
- SO3 (S: +6) and CrO4²⁻ (Cr: +6) - Same
- SO2 (S: +4) and CrO4²⁻ (Cr: +6) - Not the same
- SO2 (S: +4) and Cr2O7²⁻ (Cr: +6) - Not the same

The correct answer is **the second pair: SO3 and CrO4²⁻**, where both sulfur and chromium have an oxidation state of **+6**.