In solid state `PCl_(5)` exists as ionic solid i.e., `[X]^(+)[Y]^(-)`, shapes of `X^(+) and Y^(-)` are respectively

Options

A. Tetrahedral, Pyramidal
B. Tetrahedral, Octahedral
C. Octahedral, Linear
D. Octahedral, Trigonal bipyramidal

Answer

To solve the question regarding the shapes of the ions in the solid state of \( PCl_5 \), we can break down the problem into a series of steps:

### Step 1: Understand the Ionic Form of \( PCl_5 \)
In the solid state, \( PCl_5 \) exists as an ionic solid consisting of \( PCl_4^+ \) and \( PCl_6^- \).

### Step 2: Determine the Structure of \( PCl_4^+ \)
- Phosphorus (P) has 5 valence electrons.
- In \( PCl_4^+ \), phosphorus loses one electron, leaving it with 4 valence electrons.
- These 4 electrons are used to form bonds with 4 chlorine (Cl) atoms.
- The bonding involves sharing electrons, leading to the formation of 4 sigma bonds.

### Step 3: Identify the Hybridization of \( PCl_4^+ \)
- With 4 sigma bonds and no lone pairs, the hybridization of phosphorus in \( PCl_4^+ \) is \( sp^3 \).
- The geometry corresponding to \( sp^3 \) hybridization is tetrahedral.

### Step 4: Determine the Structure of \( PCl_6^- \)
- In \( PCl_6^- \), phosphorus gains an extra electron, resulting in 6 valence electrons.
- These 6 electrons are used to form bonds with 6 chlorine atoms.
- This results in the formation of 6 sigma bonds.

### Step 5: Identify the Hybridization of \( PCl_6^- \)
- With 6 sigma bonds and no lone pairs, the hybridization of phosphorus in \( PCl_6^- \) is \( sp^3d^2 \).
- The geometry corresponding to \( sp^3d^2 \) hybridization is octahedral.

### Step 6: Conclusion
- Therefore, the shapes of \( X^+ \) (which is \( PCl_4^+ \)) and \( Y^- \) (which is \( PCl_6^- \)) are tetrahedral and octahedral, respectively.

### Final Answer
The shapes of \( X^+ \) and \( Y^- \) are tetrahedral and octahedral, respectively.

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