What is the structure of `BrF_5` :

Options

A. Pentagonal bipyramidal
B. Octahedral
C. Square pyramidal
D. Bent T-shape

Explanation

To determine the structure of \( \text{BrF}_5 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Bromine (Br) is in Group 17 (halogens) and has 7 valence electrons. Each fluorine (F) atom has 7 valence electrons, and since there are 5 fluorine atoms, the total contribution from fluorine is \( 5 \times 7 = 35 \) electrons.

**Total valence electrons:**
\[ \text{Valence electrons from Br} + \text{Valence electrons from 5 F} = 7 + 35 = 42 \text{ electrons} \]

### Step 2: Calculate the Electron Pairs
To find the number of electron pairs, we divide the total number of valence electrons by 2:
\[ \text{Total electron pairs} = \frac{42}{2} = 21 \text{ pairs} \]

### Step 3: Determine the Hybridization
Since bromine is the central atom and it forms 5 bonds with fluorine, we need to consider the hybridization. The number of bonding pairs is 5 and the number of lone pairs can be calculated as follows:
\[ \text{Lone pairs} = \text{Total pairs} - \text{Bonding pairs} = 21 - 5 = 1 \text{ lone pair} \]

The hybridization can be determined by the formula:
- 5 bonding pairs + 1 lone pair = 6 electron pairs → \( \text{SP}^3\text{D}^2 \)

### Step 4: Determine the Geometry
The geometry corresponding to \( \text{SP}^3\text{D}^2 \) hybridization is octahedral. However, since there is 1 lone pair, the molecular shape will be different from the geometry.

### Step 5: Identify the Molecular Shape
With 5 bonding pairs and 1 lone pair, the molecular shape is derived from the octahedral arrangement but with one position occupied by a lone pair. This results in a square pyramidal shape.

### Final Structure
Thus, the structure of \( \text{BrF}_5 \) is square pyramidal.

### Summary
- **Valence Electrons**: 42
- **Electron Pairs**: 21
- **Hybridization**: \( \text{SP}^3\text{D}^2 \)
- **Geometry**: Octahedral
- **Molecular Shape**: Square Pyramidal