What is the difference between bond angles in cationic species of
`PCl_(5)` and `PBr_(5)` in solid state.

Options

A. `60^(@)`
B. `109^(@)28'`
C. `0^(@)`
D. `90^(@)`

Answer

To determine the difference between the bond angles in the cationic species of \( PCl_5 \) and \( PBr_5 \) in the solid state, we will follow these steps:

### Step 1: Identify the Cationic Species
Both \( PCl_5 \) and \( PBr_5 \) can exist in cationic forms. The relevant cationic species are:
- \( PCl_4^+ \) for \( PCl_5 \)
- \( PBr_4^+ \) for \( PBr_5 \)

### Step 2: Determine the Hybridization
To find the hybridization of these cationic species, we can use the formula:
\[ x = \frac{1}{2} \left( \text{valence electrons} + \text{monovalent atoms} - \text{charge on cation} + \text{charge on anion} \right) \]

#### For \( PCl_4^+ \):
- Valence electrons of phosphorus = 5 (Group 15)
- Monovalent atoms (Cl) = 4
- Charge on cation = +1
- Charge on anion = 0

Substituting these values:
\[ x = \frac{1}{2} \left( 5 + 4 - 1 + 0 \right) = \frac{1}{2} \times 8 = 4 \]
This indicates \( sp^3 \) hybridization, which corresponds to a tetrahedral shape.

#### For \( PBr_4^+ \):
Using similar values:
- Valence electrons of phosphorus = 5
- Monovalent atoms (Br) = 4
- Charge on cation = +1
- Charge on anion = 0

Substituting these values:
\[ x = \frac{1}{2} \left( 5 + 4 - 1 + 0 \right) = \frac{1}{2} \times 8 = 4 \]
This also indicates \( sp^3 \) hybridization, corresponding to a tetrahedral shape.

### Step 3: Determine the Bond Angles
In a tetrahedral geometry, the bond angles are approximately \( 109.5^\circ \). Both \( PCl_4^+ \) and \( PBr_4^+ \) have the same tetrahedral shape, leading to the same bond angles.

### Step 4: Calculate the Difference in Bond Angles
Since both species have the same bond angles:
\[ \text{Difference in bond angles} = 109.5^\circ - 109.5^\circ = 0^\circ \]

### Final Answer
The difference between the bond angles in the cationic species of \( PCl_5 \) and \( PBr_5 \) in the solid state is \( 0^\circ \).

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