The hybridization of central atoms of compounds A, B, C and D are `sp^(3)d, sp^(3), sp^(2) and sp` respectively. If compounds A and D have same shape like `I_(3)^(-)` and compounds B and C have same shape like water structure. Then calculate value of "P+Q+R+S", where P, Q, R, and S are number of lone pairs on central atoms of compounds A, B, C and D respectively.

Explanation

To solve the problem, we need to determine the number of lone pairs on the central atoms of compounds A, B, C, and D based on their hybridization and the shapes provided.

### Step-by-Step Solution:

1. **Identify the Hybridization and Shapes:**
- Compound A: Hybridization = sp³d (Shape: I₃⁻)
- Compound B: Hybridization = sp³ (Shape: Water)
- Compound C: Hybridization = sp² (Shape: Water)
- Compound D: Hybridization = sp (Shape: I₃⁻)

2. **Determine the Lone Pairs for Compound A (sp³d):**
- The I₃⁻ ion has a linear shape with three lone pairs on the central atom (Iodine).
- Therefore, for compound A, P (the number of lone pairs) = 3.

3. **Determine the Lone Pairs for Compound B (sp³):**
- Water (H₂O) has a bent shape with two lone pairs on the central atom (Oxygen).
- Therefore, for compound B, Q (the number of lone pairs) = 2.

4. **Determine the Lone Pairs for Compound C (sp²):**
- The sp² hybridized atom in a water-like structure has one lone pair (for example, in a molecule like SO₂).
- Therefore, for compound C, R (the number of lone pairs) = 1.

5. **Determine the Lone Pairs for Compound D (sp):**
- The sp hybridized atom in a linear structure (like in I₃⁻) has no lone pairs.
- Therefore, for compound D, S (the number of lone pairs) = 0.

6. **Calculate the Total:**
- Now we add the values of P, Q, R, and S:
\[ P + Q + R + S = 3 + 2 + 1 + 0 = 6 \]

### Final Answer:
The value of \( P + Q + R + S \) is **6**.