Select in which both have sea-saw shape-

Options

A. `XeO_(2)F_(2), SiF_(4)`
B. `XeO_(2)F_(2), IO_(2)F_(2)^(-)`
C. `TeCl_(4), I Cl_(4)^(-)`
D. `BrO_(3)F, XeOF_(2)`

Answer

To determine which compounds have a seesaw shape, we will follow these steps:

### Step 1: Identify the compounds and their central atoms.
We need to analyze the given compounds to find their central atoms and the number of surrounding atoms. The compounds provided are:
1. SiF4
2. XeO2F2
3. I2F2^(-)
4. TeCl4
5. ICl4^(-)
6. BrO3F

### Step 2: Calculate the hybridization using the formula.
The formula for calculating the hybridization (X) is:
\[ X = \frac{1}{2} \left( \text{Valence electrons of central atom} + \text{Number of surrounding monovalent atoms} + \text{Charge} \right) \]

### Step 3: Analyze each compound.

#### Compound 1: SiF4
- Central atom: Silicon (Si) has 4 valence electrons.
- Surrounding atoms: 4 Fluorine (F) atoms.
- Charge: 0.
- Calculation:
\[ X = \frac{1}{2} (4 + 4 + 0) = \frac{1}{2} \times 8 = 4 \]
- Lone pairs: 4 - 4 = 0.
- Shape: Tetrahedral (not a seesaw).

#### Compound 2: XeO2F2
- Central atom: Xenon (Xe) has 8 valence electrons.
- Surrounding atoms: 2 Oxygen (O) and 2 Fluorine (F).
- Charge: 0.
- Calculation:
\[ X = \frac{1}{2} (8 + 4 + 0) = \frac{1}{2} \times 12 = 6 \]
- Lone pairs: 6 - 4 = 2.
- Shape: Seesaw (4 bond pairs, 1 lone pair).

#### Compound 3: I2F2^(-)
- Central atom: Iodine (I) has 7 valence electrons.
- Surrounding atoms: 2 Fluorine (F) atoms.
- Charge: -1 (adds 1).
- Calculation:
\[ X = \frac{1}{2} (7 + 2 + 1) = \frac{1}{2} \times 10 = 5 \]
- Lone pairs: 5 - 4 = 1.
- Shape: Seesaw (4 bond pairs, 1 lone pair).

#### Compound 4: TeCl4
- Central atom: Tellurium (Te) has 6 valence electrons.
- Surrounding atoms: 4 Chlorine (Cl) atoms.
- Charge: 0.
- Calculation:
\[ X = \frac{1}{2} (6 + 4 + 0) = \frac{1}{2} \times 10 = 5 \]
- Lone pairs: 5 - 4 = 1.
- Shape: Seesaw (4 bond pairs, 1 lone pair).

#### Compound 5: ICl4^(-)
- Central atom: Iodine (I) has 7 valence electrons.
- Surrounding atoms: 4 Chlorine (Cl) atoms.
- Charge: -1 (adds 1).
- Calculation:
\[ X = \frac{1}{2} (7 + 4 + 1) = \frac{1}{2} \times 12 = 6 \]
- Lone pairs: 6 - 4 = 2.
- Shape: Square planar (not a seesaw).

#### Compound 6: BrO3F
- Central atom: Bromine (Br) has 7 valence electrons.
- Surrounding atoms: 3 Oxygen (O) and 1 Fluorine (F).
- Charge: 0.
- Calculation:
\[ X = \frac{1}{2} (7 + 4 + 0) = \frac{1}{2} \times 11 = 5.5 \]
- Lone pairs: 5.5 - 4 = 1.5 (not applicable).
- Shape: Not determined as seesaw.

### Step 4: Conclusion
The compounds that have a seesaw shape are:
- XeO2F2
- I2F2^(-)
- TeCl4

However, the question specifically asks for pairs that both have a seesaw shape. The only pair that meets this criterion is:
- XeO2F2 and I2F2^(-)

### Final Answer
The answer is **option number B: XeO2F2 and I2F2^(-)**.

---