`{:("Column I","Column II"),((A)PbO_(2)+HNO_(3)to,(p)"One of the products has bond order of two." ),((B)Cr_(2)O_(7)^(2-)+H^(+)+H_(2)O_(2)to,(q)"One of the products has peroxide linkages(s)."),((C)H_(2)O_(2)+CIO_(2)+OH^(-)to,(r)"One of the products is a hydride and is liquid at room temperature."),((D)XeF_(2)+NaOH to,(s)"One of the products has bent shape with two lone pairs of electrons on central atom".):}`

Explanation

To solve the given question, we will analyze each reaction in Column I and determine which statements from Column II correspond to the products formed in each reaction.

### Step 1: Analyze the first reaction (A)
**Reaction:** PbO₂ + HNO₃ → Products

**Products formed:**
When lead dioxide (PbO₂) reacts with nitric acid (HNO₃), the products are lead nitrate (Pb(NO₃)₂), water (H₂O), and oxygen (O₂).

**Analysis of products:**
- **O₂** has a bond order of 2 (double bond).
- **H₂O** is a hydride and is liquid at room temperature.
- **H₂O** has a bent shape with two lone pairs on the oxygen atom.

**Conclusion for A:**
- P: True (O₂ has bond order of 2)
- R: True (H₂O is a hydride and liquid at room temperature)
- S: True (H₂O has bent shape with two lone pairs)
- Q: False (No peroxide linkages)

**Answer for A:** P, R, S

### Step 2: Analyze the second reaction (B)
**Reaction:** Cr₂O₇²⁻ + H⁺ + H₂O₂ → Products

**Products formed:**
The reaction produces chromium pentoxide (CrO₅) and water (H₂O).

**Analysis of products:**
- **CrO₅** contains peroxide linkages (O-O bonds).
- **H₂O** is a hydride and is liquid at room temperature.
- **H₂O** has a bent shape with two lone pairs on the oxygen atom.

**Conclusion for B:**
- P: False (No product has bond order of 2)
- Q: True (CrO₅ has peroxide linkages)
- R: True (H₂O is a hydride and liquid at room temperature)
- S: True (H₂O has bent shape with two lone pairs)

**Answer for B:** Q, R, S

### Step 3: Analyze the third reaction (C)
**Reaction:** H₂O₂ + ClO₂ + OH⁻ → Products

**Products formed:**
The reaction produces ClO₂⁻, water (H₂O), and oxygen (O₂).

**Analysis of products:**
- **O₂** has a bond order of 2.
- **H₂O** is a hydride and is liquid at room temperature.
- **H₂O** has a bent shape with two lone pairs on the oxygen atom.

**Conclusion for C:**
- P: True (O₂ has bond order of 2)
- Q: False (No peroxide linkages)
- R: True (H₂O is a hydride and liquid at room temperature)
- S: True (H₂O has bent shape with two lone pairs)

**Answer for C:** P, R, S

### Step 4: Analyze the fourth reaction (D)
**Reaction:** XeF₂ + NaOH → Products

**Products formed:**
The reaction produces xenon (Xe), water (H₂O), and sodium fluoride (NaF).

**Analysis of products:**
- **O₂** is not produced, but water is produced.
- **H₂O** is a hydride and is liquid at room temperature.
- **H₂O** has a bent shape with two lone pairs on the oxygen atom.

**Conclusion for D:**
- P: True (O₂ is released)
- Q: False (No peroxide linkages)
- R: True (H₂O is a hydride and liquid at room temperature)
- S: True (H₂O has bent shape with two lone pairs)

**Answer for D:** P, R, S

### Final Summary of Answers:
- A: P, R, S
- B: Q, R, S
- C: P, R, S
- D: P, R, S