Which of these doesn’t exist?

Options

A. `PH _(3)`
B. `PH _(5)`
C. `LuH _(3)`
D. `PF _(5)`

Answer

To determine which of the given compounds does not exist, we need to analyze each of them: PH3, PH5, LuH3, and PF5.

### Step-by-Step Solution:

1. **Analyze PH3 (Phosphine)**:
- PH3 is a well-known compound where phosphorus (P) forms bonds with three hydrogen (H) atoms.
- The bonding involves the sharing of electrons and is stable.
- **Conclusion**: PH3 exists.

2. **Analyze PH5 (Phosphorus Pentahydride)**:
- PH5 would require phosphorus to form five bonds with hydrogen.
- Phosphorus typically has a maximum of three bonds due to its valence shell configuration (3s² 3p³).
- To form five bonds, phosphorus would need to utilize its d orbitals, but the energy difference between the d orbitals and the 1s orbitals of hydrogen is significant.
- This large energy difference makes the bond formation unfavorable.
- **Conclusion**: PH5 does not exist.

3. **Analyze LuH3 (Lutetium Hydride)**:
- LuH3 involves lutetium (Lu) bonding with three hydrogen atoms.
- Lutetium can form stable bonds with hydrogen through electron sharing.
- **Conclusion**: LuH3 exists.

4. **Analyze PF5 (Phosphorus Pentafluoride)**:
- PF5 involves phosphorus bonding with five fluorine (F) atoms.
- In this case, phosphorus can utilize its d orbitals to form bonds with fluorine, which uses its p orbitals.
- The energy difference between the d orbitals of phosphorus and the p orbitals of fluorine is much smaller than that in PH5, making bond formation favorable.
- **Conclusion**: PF5 exists.

### Final Conclusion:
Among the compounds analyzed, PH5 does not exist due to the unfavorable energy difference in orbital hybridization.

### Answer:
**PH5 does not exist.**

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