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 analyze the hybridization and the shapes of the compounds A, B, C, and D, and then determine the number of lone pairs on the central atoms of each compound.

### Step 1: Identify the hybridization and shapes of the compounds
- **Compound A**: Hybridization is `sp^3d` (trigonal bipyramidal shape). Since it has the same shape as `I3^-`, we can infer that it has 2 lone pairs.
- **Compound B**: Hybridization is `sp^3` (tetrahedral shape). The water structure indicates that it has 2 lone pairs.
- **Compound C**: Hybridization is `sp^2` (trigonal planar shape). The water structure indicates that it has 1 lone pair.
- **Compound D**: Hybridization is `sp` (linear shape). Since it has the same shape as `I3^-`, it has 0 lone pairs.

### Step 2: Count the lone pairs for each compound
- **P (Compound A)**: 2 lone pairs
- **Q (Compound B)**: 2 lone pairs
- **R (Compound C)**: 1 lone pair
- **S (Compound D)**: 0 lone pairs

### Step 3: Calculate the sum of P, Q, R, and S
Now, we can calculate the total:
- P + Q + R + S = 2 + 2 + 1 + 0 = 5

### Final Answer
The value of "P + Q + R + S" is **5**.

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