Which of the following reaction depicts the oxidsing behaviour of `H_(2)SO_(4)`?

Options

A. `2HI + H_2 SO_4 rarr I_2 + SO_2 + 2H_2 O`
B. `Ca(OH)_2 + H_2 SO_4 rarr CaSO_4 + 2H_2 O`
C. `NaCl + H_2 SO_4 rarr NaHSO_4 + HCl`
D. `2PCl_5 + H_2 SO_4 rarr 2POCl_3 + 2HCl + SO_2 Cl_2`

Explanation

To determine which reaction depicts the oxidizing behavior of \( H_2SO_4 \), we need to analyze the oxidation states of the elements involved in each reaction. An oxidizing agent is a substance that oxidizes another substance and is itself reduced in the process. This means that the oxidizing agent will undergo a decrease in its oxidation state.

Let's go through the options step by step:

### Step 1: Analyze the first reaction with HI
1. **Identify oxidation states**:
- In \( HI \):
- \( H \) has an oxidation state of +1
- \( I \) has an oxidation state of -1
- In \( H_2SO_4 \):
- \( H \) has +1
- \( S \) has +6
- \( O \) has -2
2. **Reaction products**:
- When \( HI \) reacts with \( H_2SO_4 \), iodine is oxidized from -1 to 0.
3. **Conclusion**:
- Since iodine is losing electrons (oxidation), \( H_2SO_4 \) is acting as an oxidizing agent.

### Step 2: Analyze the second reaction with calcium hydroxide
1. **Identify oxidation states**:
- In \( Ca(OH)_2 \):
- \( Ca \) has +2
- \( O \) has -2
- \( H \) has +1
- In \( H_2SO_4 \):
- Same as above, \( S \) is +6.
2. **Reaction products**:
- The products are \( CaSO_4 \) and water.
3. **Conclusion**:
- There is no change in oxidation states for calcium or sulfur, hence \( H_2SO_4 \) is not acting as an oxidizing agent.

### Step 3: Analyze the third reaction with NaCl
1. **Identify oxidation states**:
- In \( NaCl \):
- \( Na \) has +1
- \( Cl \) has -1
- In \( H_2SO_4 \):
- Same as above.
2. **Reaction products**:
- The products are \( HCl \) and \( Na_2SO_4 \).
3. **Conclusion**:
- There is no change in oxidation states, so \( H_2SO_4 \) is not acting as an oxidizing agent.

### Step 4: Analyze the fourth reaction with PCl5
1. **Identify oxidation states**:
- In \( PCl_5 \):
- \( P \) has +5
- \( Cl \) has -1
- In \( H_2SO_4 \):
- Same as above.
2. **Reaction products**:
- The products are \( POCl_3 \) and \( HCl \).
3. **Conclusion**:
- There is no change in oxidation states for phosphorus or chlorine, so \( H_2SO_4 \) is not acting as an oxidizing agent.

### Final Conclusion
The only reaction where \( H_2SO_4 \) is acting as an oxidizing agent is the first one with \( HI \). Thus, the answer is:

**Answer: The first reaction with HI depicts the oxidizing behavior of \( H_2SO_4 \).**