`PCl_5` has trigonal pyramidal geometry with `sp^3 d` hybridisation in gases and liquid state but in solid state it exist as ionic compound.
In crystalline state `PCl_5` exists as.

Options

A. `[PCl_3]^(2+) + 2 Cl^(Ө)`
B. `[PCl_4)^(oplus) [PCl_6]^(Ө)`
C. `[PCl_4]^(oplus) Cl^(Ө)`
D. `[PCl_6]^(oplus) [PCl_4)^(Ө)`

Answer

To solve the question regarding the crystalline state of \( PCl_5 \), we can break down the solution into clear steps:

### Step 1: Understand the Hybridization and Geometry of \( PCl_5 \)
- In the gaseous and liquid states, \( PCl_5 \) exhibits **trigonal bipyramidal geometry** due to **\( sp^3d \)** hybridization. This means that phosphorus forms five bonds with chlorine atoms, resulting in a specific arrangement.

### Step 2: Analyze the Solid State of \( PCl_5 \)
- When \( PCl_5 \) transitions to the solid state, it does not retain the same molecular structure. Instead, it forms an ionic compound. This is due to the interactions between the molecules in the solid state.

### Step 3: Identify the Ionic Forms of \( PCl_5 \)
- In the solid state, \( PCl_5 \) can be represented as \( PCl_4^+ \) and \( PCl_6^- \). This indicates that the phosphorus atom can lose one chloride ion to form a positively charged ion \( PCl_4^+ \) and gain an extra chloride ion to form a negatively charged ion \( PCl_6^- \).

### Step 4: Determine the Hybridization in the Ionic Forms
- In \( PCl_4^+ \), the phosphorus atom has **\( sp^3 \)** hybridization, leading to a **tetrahedral shape**. In \( PCl_6^- \), the phosphorus atom has **\( sp^3d^2 \)** hybridization, resulting in an **octahedral shape**.

### Step 5: Conclusion
- Therefore, in the crystalline state, \( PCl_5 \) exists as the ionic compound formed by \( PCl_4^+ \) and \( PCl_6^- \).

### Final Answer
In crystalline state, \( PCl_5 \) exists as \( PCl_4^+ \) and \( PCl_6^- \).

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