`{:("List-I"," ""List-II"),(""," ""In the aqueous medium the ion can be"),((A)HPO_(4)^(-2),P."Arrhenius acid"),((B)CO_(3)^(-2),Q."Bronstead base"),((C)H_(2)PO_(3)^(-1),R."Amphoteric"),((D) H_(2)PO_(2)^(-1),S."Bronstead acid"):}`

Answer

To solve the problem, we need to classify each ion from List-I (A, B, C, D) according to their behavior in aqueous medium as described in List-II (P, Q, R, S). Here’s a step-by-step breakdown:

### Step 1: Analyze Ion A (HPO₄²⁻)
- **Identification**: HPO₄²⁻ is derived from H₃PO₄ (phosphoric acid) after losing 2 H⁺ ions.
- **Behavior**:
- It can donate a proton (H⁺) making it a **Bronsted acid**.
- It can accept a proton due to the presence of O⁻, making it a **Bronsted base**.
- Since it can act as both an acid and a base, it is **amphoteric**.
- **Conclusion**: A corresponds to P (Bronsted acid), Q (Bronsted base), and R (Amphoteric).

### Step 2: Analyze Ion B (CO₃²⁻)
- **Identification**: CO₃²⁻ is the carbonate ion.
- **Behavior**:
- It can accept a proton (H⁺) to form HCO₃⁻ (bicarbonate), making it a **Bronsted base**.
- **Conclusion**: B corresponds to Q (Bronsted base).

### Step 3: Analyze Ion C (H₂PO₃⁻)
- **Identification**: H₂PO₃⁻ is derived from phosphorous acid.
- **Behavior**:
- It has one ionizable hydrogen, making it a **Bronsted acid**.
- It can accept a proton due to the presence of O⁻, making it a **Bronsted base**.
- It can act as both an acid and a base, hence it is **amphoteric**.
- **Conclusion**: C corresponds to P (Bronsted acid), Q (Bronsted base), and R (Amphoteric).

### Step 4: Analyze Ion D (H₂PO₂⁻)
- **Identification**: H₂PO₂⁻ is derived from hypophosphorous acid.
- **Behavior**:
- It has no ionizable hydrogen, so it cannot act as an acid.
- However, it has O⁻, which allows it to accept a proton, making it a **Bronsted base**.
- **Conclusion**: D corresponds to Q (Bronsted base).

### Final Summary:
- A corresponds to P, Q, R (Amphoteric).
- B corresponds to Q (Bronsted base).
- C corresponds to P, Q, R (Amphoteric).
- D corresponds to Q (Bronsted base).