Which of the following compounds have the same no. of lone pairs with their central atom?
(I) `XeF_(5)^(-)` (II)`BrF_(3)` (III) `XeF_(2)` (IV)`H_(3)S^(+)` (V) Triplet methylene

Options

A. (IV) and (V)
B. (I) and (III)
C. (I) and (II)
D. II, IV and V

Explanation

To determine which of the given compounds have the same number of lone pairs on their central atom, we will analyze each compound one by one using the concept of hybridization and the number of lone pairs.

### Step 1: Analyze `XeF5^(-)`

1. **Valence Electrons Calculation**:
- Xenon (Xe) has 8 valence electrons.
- There are 5 fluorine (F) atoms, each contributing 1 electron.
- The negative charge adds 1 electron.
- Total = 8 (Xe) + 5 (F) + 1 (charge) = 14 electrons.

2. **Hybridization**:
- Using the formula: \( \text{Steric Number} = \frac{(V + M + \text{Charge})}{2} \)
- \( V = 8 \) (from Xe), \( M = 5 \) (from F), and charge = +1.
- \( \text{Steric Number} = \frac{(8 + 5 + 1)}{2} = \frac{14}{2} = 7 \).
- Hybridization is \( sp^3d^3 \).

3. **Lone Pairs**:
- Total pairs = 7 (steric number).
- Bond pairs = 5 (from 5 F).
- Lone pairs = 7 - 5 = 2.

### Step 2: Analyze `BrF3`

1. **Valence Electrons Calculation**:
- Bromine (Br) has 7 valence electrons.
- There are 3 fluorine (F) atoms.
- Total = 7 (Br) + 3 (F) = 10 electrons.

2. **Hybridization**:
- \( V = 7 \) (from Br), \( M = 3 \) (from F).
- \( \text{Steric Number} = \frac{(7 + 3)}{2} = \frac{10}{2} = 5 \).
- Hybridization is \( sp^3d \).

3. **Lone Pairs**:
- Total pairs = 5 (steric number).
- Bond pairs = 3 (from 3 F).
- Lone pairs = 5 - 3 = 2.

### Step 3: Analyze `XeF2`

1. **Valence Electrons Calculation**:
- Xenon (Xe) has 8 valence electrons.
- There are 2 fluorine (F) atoms.
- Total = 8 (Xe) + 2 (F) = 10 electrons.

2. **Hybridization**:
- \( V = 8 \) (from Xe), \( M = 2 \) (from F).
- \( \text{Steric Number} = \frac{(8 + 2)}{2} = \frac{10}{2} = 5 \).
- Hybridization is \( sp^3d \).

3. **Lone Pairs**:
- Total pairs = 5 (steric number).
- Bond pairs = 2 (from 2 F).
- Lone pairs = 5 - 2 = 3.

### Step 4: Analyze `H3S^(+)`

1. **Valence Electrons Calculation**:
- Sulfur (S) has 6 valence electrons.
- There are 3 hydrogen (H) atoms.
- The positive charge removes 1 electron.
- Total = 6 (S) + 3 (H) - 1 (charge) = 8 electrons.

2. **Hybridization**:
- \( V = 6 \) (from S), \( M = 3 \) (from H).
- \( \text{Steric Number} = \frac{(6 + 3 - 1)}{2} = \frac{8}{2} = 4 \).
- Hybridization is \( sp^3 \).

3. **Lone Pairs**:
- Total pairs = 4 (steric number).
- Bond pairs = 3 (from 3 H).
- Lone pairs = 4 - 3 = 1.

### Step 5: Analyze Triplet Methylene

1. **Valence Electrons Calculation**:
- Carbon (C) has 4 valence electrons.
- In triplet methylene (CH2), there are 2 H atoms.
- Total = 4 (C) + 2 (H) = 6 electrons.

2. **Hybridization**:
- \( V = 4 \) (from C), \( M = 2 \) (from H).
- \( \text{Steric Number} = \frac{(4 + 2)}{2} = \frac{6}{2} = 3 \).
- Hybridization is \( sp^2 \).

3. **Lone Pairs**:
- Total pairs = 3 (steric number).
- Bond pairs = 2 (from 2 H).
- Lone pairs = 3 - 2 = 1.

### Conclusion

After analyzing all compounds, we find:

- `XeF5^(-)` has **2 lone pairs**.
- `BrF3` has **2 lone pairs**.
- `XeF2` has **3 lone pairs**.
- `H3S^(+)` has **1 lone pair**.
- Triplet methylene has **0 lone pairs**.

Thus, the compounds that have the same number of lone pairs with their central atom are **(I) `XeF5^(-)` and (II) `BrF3`**.

### Final Answer:
**(I) and (II)** have the same number of lone pairs (2).