The shapes of `PCl_4^+ , PCl_4^-` and `AsCl_5` are respectively

Options

A. square planer, tetrahedral and see-saw
B. tetrahedral, see-saw and trigonal bipyramidal
C. tetrahedral, square planer and pentagonal bipyramidal
D. trigonal bipyramidal, tetrahedral and square pyramidal

Answer

To determine the shapes of \( PCl_4^+ \), \( PCl_4^- \), and \( AsCl_5 \), we will analyze each compound step by step.

### Step 1: Determine the shape of \( PCl_4^+ \)

1. **Count the valence electrons**:
- Phosphorus (P) has 5 valence electrons.
- Each Chlorine (Cl) has 7 valence electrons, and there are 4 Chlorine atoms, contributing \( 4 \times 7 = 28 \) electrons.
- The positive charge (\(+1\)) indicates that we lose one electron.
- Total valence electrons = \( 5 + 28 - 1 = 32 \).

2. **Determine bonding and lone pairs**:
- Phosphorus uses 4 of its valence electrons to form bonds with the 4 Chlorine atoms.
- This results in 4 bond pairs and 0 lone pairs.

3. **Calculate the steric number**:
- Steric number = number of bond pairs + number of lone pairs = \( 4 + 0 = 4 \).

4. **Determine hybridization and shape**:
- Hybridization for a steric number of 4 is \( sp^3 \).
- The geometry is tetrahedral, and since there are no lone pairs, the shape is also tetrahedral.

### Step 2: Determine the shape of \( PCl_4^- \)

1. **Count the valence electrons**:
- Phosphorus has 5 valence electrons.
- Each Chlorine has 7 valence electrons, contributing \( 4 \times 7 = 28 \).
- The negative charge (\(-1\)) indicates that we gain one electron.
- Total valence electrons = \( 5 + 28 + 1 = 34 \).

2. **Determine bonding and lone pairs**:
- Phosphorus uses 4 electrons to form bonds with the 4 Chlorine atoms.
- This results in 4 bond pairs and 2 electrons remaining, which form 1 lone pair.

3. **Calculate the steric number**:
- Steric number = number of bond pairs + number of lone pairs = \( 4 + 1 = 5 \).

4. **Determine hybridization and shape**:
- Hybridization for a steric number of 5 is \( sp^3d \).
- The geometry is trigonal bipyramidal. However, since there is 1 lone pair, the shape is based on the positions of the bond pairs, which results in a seesaw shape.

### Step 3: Determine the shape of \( AsCl_5 \)

1. **Count the valence electrons**:
- Arsenic (As) has 5 valence electrons.
- Each Chlorine has 7 valence electrons, contributing \( 5 \times 7 = 35 \).
- Total valence electrons = \( 5 + 35 = 40 \).

2. **Determine bonding and lone pairs**:
- Arsenic uses all 5 of its valence electrons to form bonds with the 5 Chlorine atoms.
- This results in 5 bond pairs and 0 lone pairs.

3. **Calculate the steric number**:
- Steric number = number of bond pairs + number of lone pairs = \( 5 + 0 = 5 \).

4. **Determine hybridization and shape**:
- Hybridization for a steric number of 5 is \( sp^3d \).
- The geometry is trigonal bipyramidal, and since there are no lone pairs, the shape is also trigonal bipyramidal.

### Final Summary of Shapes:
- The shape of \( PCl_4^+ \) is **tetrahedral**.
- The shape of \( PCl_4^- \) is **seesaw**.
- The shape of \( AsCl_5 \) is **trigonal bipyramidal**.