Oxyacids of phosphorous and the starting materials for their preparation are given below.
`{:(,"Ocyacid", , "Matereifal for preparation"),((A), H_(3)PO_(2), (i), "Red P + alpali"),((B), H_(3)P_(3), (ii) , P_(4)O_(10)+H_(2)O),((C), H_(3)PO_(4), (iii), P_(2)O_(3)+H_(2)O),((D), H_(4)P_(2)O_(6),(iv), "White P+ alkali"):}`
Choose the correct answer from the codes given below:

Options

A. (A) - (iv), (B) - (iii), (C) - (ii), (D) - (i)
B. (A) - (i), (B) - (iii), (C) - (ii), (D) - (iv)
C. (A) - (iv), (B) - (iii), (C) - (i), (D) - (ii)
D. (A) - (ii), (B) - (iii), (C) - (i), (D) - (iv)

Answer

To solve the problem of matching the oxyacids of phosphorus with their respective starting materials for preparation, we will analyze each oxyacid and determine the correct starting material.

### Step-by-Step Solution:

1. **Identify the Oxyacids and Their Formulas:**
- A: H₃PO₂ (Phosphorous acid)
- B: H₃PO₃ (Phosphoric acid)
- C: H₃PO₄ (Phosphoric acid)
- D: H₄P₂O₆ (Hypophosphorous acid)

2. **Analyze the Starting Materials:**
- (i) Red P + alkali
- (ii) P₄O₁₀ + H₂O
- (iii) P₂O₃ + H₂O
- (iv) White P + alkali

3. **Match Each Oxyacid with Its Preparation:**
- **For H₃PO₂ (A):**
- H₃PO₂ can be prepared from white phosphorus (P) reacting with an alkali.
- Therefore, A matches with (iv) White P + alkali.

- **For H₃PO₃ (B):**
- H₃PO₃ can be prepared from P₂O₃ reacting with water (H₂O).
- Therefore, B matches with (iii) P₂O₃ + H₂O.

- **For H₃PO₄ (C):**
- H₃PO₄ can be prepared from P₄O₁₀ reacting with water (H₂O).
- Therefore, C matches with (ii) P₄O₁₀ + H₂O.

- **For H₄P₂O₆ (D):**
- H₄P₂O₆ can be prepared from red phosphorus reacting with an alkali.
- Therefore, D matches with (i) Red P + alkali.

4. **Final Matching:**
- A → (iv) White P + alkali
- B → (iii) P₂O₃ + H₂O
- C → (ii) P₄O₁₀ + H₂O
- D → (i) Red P + alkali

### Conclusion:
The correct matches are:
- A: (iv)
- B: (iii)
- C: (ii)
- D: (i)

Thus, the answer is **A → (iv), B → (iii), C → (ii), D → (i)**.