Which of the following molecule have same structure and shape ?

Options

A. `IF_(7)`
B. `[AlF_(6)]^(3-)`
C. `BrF_(5)`
D. `XeOF_(4)`

Explanation

To determine which of the given molecules have the same structure and shape, we need to analyze each molecule by calculating its steric number, hybridization, and then identifying its shape and geometry. Let's go through the molecules step by step.

### Step 1: Analyze IF7
1. **Valence Electrons**: Iodine (I) has 7 valence electrons.
2. **Sigma Bonds**: IF7 forms 7 sigma bonds with fluorine (F).
3. **Steric Number Calculation**:
\[ \text{Steric Number} = \frac{1}{2} \times (\text{Valence Electrons} + \text{Number of Sigma Bonds}) = \frac{1}{2} \times (7 + 7) = 7 \]
4. **Hybridization**: With a steric number of 7, the hybridization is \( sp^3d^3 \).
5. **Shape and Geometry**: The shape is pentagonal bipyramidal (PPP), and since there are no lone pairs, the geometry is also pentagonal bipyramidal.

### Step 2: Analyze AlF6^3-
1. **Valence Electrons**: Aluminium (Al) has 3 valence electrons.
2. **Sigma Bonds**: AlF6^3- forms 6 sigma bonds with fluorine and has a charge of -3.
3. **Steric Number Calculation**:
\[ \text{Steric Number} = \frac{1}{2} \times (\text{Valence Electrons} + \text{Number of Sigma Bonds} + \text{Charge}) = \frac{1}{2} \times (3 + 6 + 3) = 6 \]
4. **Hybridization**: With a steric number of 6, the hybridization is \( sp^3d^2 \).
5. **Shape and Geometry**: The shape is octahedral, and since there are no lone pairs, the geometry is also octahedral.

### Step 3: Analyze BrF5
1. **Valence Electrons**: Bromine (Br) has 7 valence electrons.
2. **Sigma Bonds**: BrF5 forms 5 sigma bonds with fluorine.
3. **Steric Number Calculation**:
\[ \text{Steric Number} = \frac{1}{2} \times (\text{Valence Electrons} + \text{Number of Sigma Bonds}) = \frac{1}{2} \times (7 + 5) = 6 \]
4. **Hybridization**: With a steric number of 6, the hybridization is \( sp^3d^2 \).
5. **Shape and Geometry**: The shape is square pyramidal (due to one lone pair), while the geometry is octahedral.

### Step 4: Analyze XeOF4
1. **Valence Electrons**: Xenon (Xe) has 8 valence electrons.
2. **Sigma Bonds**: XeOF4 forms 4 sigma bonds (4 with F and 1 with O).
3. **Steric Number Calculation**:
\[ \text{Steric Number} = \frac{1}{2} \times (\text{Valence Electrons} + \text{Number of Sigma Bonds}) = \frac{1}{2} \times (8 + 5) = 6.5 \]
(Note: This should be adjusted for lone pairs.)
4. **Hybridization**: With a steric number of 6, the hybridization is \( sp^3d^2 \).
5. **Shape and Geometry**: The shape is square planar (due to two lone pairs), while the geometry is octahedral.

### Conclusion
After analyzing all four compounds:
- **IF7**: Pentagonal bipyramidal
- **AlF6^3-**: Octahedral
- **BrF5**: Square pyramidal (octahedral geometry)
- **XeOF4**: Square planar (octahedral geometry)

The compounds **BrF5** and **XeOF4** have the same geometry (octahedral) but different shapes.

Thus, the answer to the question is that **BrF5 and XeOF4 have the same structure and shape**.