The shapes of `PCI_(4)^(o+),PCI_(4)^(Θ)` and `AsCI_(5)` and are respectively .

Options

A. Squar planar, tetrahedral and see-saw
B. Tetrahedral,see-saw and trigonal bipyramidal
C. Tetrahedarl ,square planar and pentagonal bipyramidal
D. Trigonal bipyramidal, tetrahedral and square pyramidal

Explanation

To determine the shapes of the molecules \( \text{PCl}_4^+ \), \( \text{PCl}_4^- \), and \( \text{AsCl}_5 \), we will follow a systematic approach to calculate the number of valence electrons, determine the hybridization, and then deduce the molecular shapes.

### Step 1: Determine the shape of \( \text{PCl}_4^+ \)

1. **Calculate the number of valence electrons (V)**:
- Phosphorus (P) has 5 valence electrons.
- Each Chlorine (Cl) has 7 valence electrons, and there are 4 Cl atoms.
- Since there is a positive charge, we subtract 1 electron.
\[ V = 5 + (4 \times 7) - 1 = 5 + 28 - 1 = 32 \]

2. **Determine the number of bond pairs and lone pairs**:
- Divide \( V \) by 8:
\[ \text{Quotient} = \frac{32}{8} = 4 \quad (\text{number of bond pairs}) \]
- Remainder is 0, which means:
\[ \text{Number of lone pairs} = 0 \]

3. **Determine hybridization**:
- Since there are 4 bond pairs and 0 lone pairs, the hybridization is \( \text{sp}^3 \).

4. **Determine the shape**:
- With \( \text{sp}^3 \) hybridization and no lone pairs, the shape is **tetrahedral**.

### Step 2: Determine the shape of \( \text{PCl}_4^- \)

1. **Calculate the number of valence electrons (V)**:
- Phosphorus (P) has 5 valence electrons.
- Each Chlorine (Cl) has 7 valence electrons, and there are 4 Cl atoms.
- Since there is a negative charge, we add 1 electron.
\[ V = 5 + (4 \times 7) + 1 = 5 + 28 + 1 = 34 \]

2. **Determine the number of bond pairs and lone pairs**:
- Divide \( V \) by 8:
\[ \text{Quotient} = \frac{34}{8} = 4 \quad (\text{number of bond pairs}) \]
- Remainder is 2, which means:
\[ \text{Number of lone pairs} = \frac{2}{2} = 1 \]

3. **Determine hybridization**:
- With 4 bond pairs and 1 lone pair, the hybridization is \( \text{sp}^3\text{d} \).

4. **Determine the shape**:
- With \( \text{sp}^3\text{d} \) hybridization and one lone pair, the shape is **seesaw**.

### Step 3: Determine the shape of \( \text{AsCl}_5 \)

1. **Calculate the number of valence electrons (V)**:
- Arsenic (As) has 5 valence electrons.
- Each Chlorine (Cl) has 7 valence electrons, and there are 5 Cl atoms.
\[ V = 5 + (5 \times 7) = 5 + 35 = 40 \]

2. **Determine the number of bond pairs and lone pairs**:
- Divide \( V \) by 8:
\[ \text{Quotient} = \frac{40}{8} = 5 \quad (\text{number of bond pairs}) \]
- Remainder is 0, which means:
\[ \text{Number of lone pairs} = 0 \]

3. **Determine hybridization**:
- With 5 bond pairs and 0 lone pairs, the hybridization is \( \text{sp}^3\text{d} \).

4. **Determine the shape**:
- With \( \text{sp}^3\text{d} \) hybridization and no lone pairs, the shape is **trigonal bipyramidal**.

### Final Answer:
- The shapes of \( \text{PCl}_4^+ \), \( \text{PCl}_4^- \), and \( \text{AsCl}_5 \) are respectively **tetrahedral**, **seesaw**, and **trigonal bipyramidal**.