Which among the following molecules have `sp^(3)d` hybridization with one lone pair of electrons on the central atom?
(P) `SF_(4)`
(Q) `[PCl_(4)]^(+)`
(R) `XeO_(2)F_(2)`
(S) `ClOF_(3)`

Options

A. P, Q and R only
B. P, R and S only
C. P and R only
D. R and S only

Explanation

To determine which of the given molecules have `sp^(3)d` hybridization with one lone pair of electrons on the central atom, we will analyze each molecule step by step.

### Step 1: Analyze SF₄
1. **Identify the central atom**: Sulfur (S).
2. **Count the valence electrons**: Sulfur has 6 valence electrons, and each fluorine (F) has 7 valence electrons. Thus, for 4 fluorine atoms, we have:
\[ 6 + (4 \times 7) = 6 + 28 = 34 \text{ total valence electrons.} \]
3. **Bonding**: Sulfur forms 4 bonds with fluorine, using 4 of its electrons. This leaves 2 electrons (1 lone pair) on sulfur.
4. **Hybridization**: The number of bond pairs (4) + lone pairs (1) = 5. Therefore, the hybridization is `sp^(3)d`.
5. **Conclusion**: SF₄ has `sp^(3)d` hybridization with one lone pair.

### Step 2: Analyze [PCl₄]⁺
1. **Identify the central atom**: Phosphorus (P).
2. **Count the valence electrons**: Phosphorus has 5 valence electrons, and each chlorine (Cl) has 7 valence electrons. For 4 chlorines:
\[ 5 + (4 \times 7) - 1 = 5 + 28 - 1 = 32 \text{ total valence electrons.} \]
(We subtract 1 for the positive charge.)
3. **Bonding**: Phosphorus forms 4 bonds with chlorine, using all 4 of its available electrons. There are no lone pairs left on phosphorus.
4. **Hybridization**: The number of bond pairs (4) + lone pairs (0) = 4. Therefore, the hybridization is `sp³`.
5. **Conclusion**: [PCl₄]⁺ does not have `sp^(3)d` hybridization with a lone pair.

### Step 3: Analyze XeO₂F₂
1. **Identify the central atom**: Xenon (Xe).
2. **Count the valence electrons**: Xenon has 8 valence electrons, each oxygen (O) has 6, and each fluorine (F) has 7. For 2 oxygens and 2 fluorines:
\[ 8 + (2 \times 6) + (2 \times 7) = 8 + 12 + 14 = 34 \text{ total valence electrons.} \]
3. **Bonding**: Xenon forms 4 bonds (2 with oxygen and 2 with fluorine), using 4 electrons. This leaves 4 electrons (2 lone pairs) on xenon.
4. **Hybridization**: The number of bond pairs (4) + lone pairs (2) = 6. Therefore, the hybridization is `sp^(3)d^(2)`.
5. **Conclusion**: XeO₂F₂ does not have `sp^(3)d` hybridization with one lone pair.

### Step 4: Analyze ClOF₃
1. **Identify the central atom**: Chlorine (Cl).
2. **Count the valence electrons**: Chlorine has 7 valence electrons, oxygen (O) has 6, and each fluorine (F) has 7. For 3 fluorines:
\[ 7 + 6 + (3 \times 7) = 7 + 6 + 21 = 34 \text{ total valence electrons.} \]
3. **Bonding**: Chlorine forms 4 bonds (1 with oxygen and 3 with fluorine), using 4 electrons. This leaves 3 electrons (1 lone pair) on chlorine.
4. **Hybridization**: The number of bond pairs (4) + lone pairs (1) = 5. Therefore, the hybridization is `sp^(3)d`.
5. **Conclusion**: ClOF₃ has `sp^(3)d` hybridization with one lone pair.

### Final Conclusion
The molecules that have `sp^(3)d` hybridization with one lone pair of electrons on the central atom are:
- **SF₄**
- **ClOF₃**

### Summary of Answers:
- **Correct Answers**: (P) SF₄ and (S) ClOF₃.

---