Hybridisation and shape of `BrF_(5)` is :

Options

A. `sp^(3)d` (Trigonal bipyramidal)
B. `sp^(3)d^(2)` (Octahedral)
C. `dsp^(2)` (Square planar)
D. `sp^(3)d^(2)` (Square pyramidal)

Explanation

To determine the hybridization and shape of BrF₅ (Bromine Pentafluoride), we can follow these steps:

### Step 1: Determine the Valence Electrons
Bromine (Br) is a halogen and has 7 valence electrons. Each Fluorine (F) atom has 7 valence electrons as well. In BrF₅, there are 5 Fluorine atoms.

- Total valence electrons from Br = 7
- Total valence electrons from 5 F = 5 × 7 = 35
- Total valence electrons = 7 (Br) + 35 (5 F) = 42

### Step 2: Draw the Lewis Structure
In the Lewis structure of BrF₅, Bromine will form 5 single bonds with 5 Fluorine atoms. After forming these bonds, Bromine will have 1 lone pair of electrons remaining.

- Bond pairs: 5 (Br-F bonds)
- Lone pairs: 1 (on Br)

### Step 3: Determine the Hybridization
To find the hybridization, we can use the formula:
\[ \text{Hybridization} = \text{Number of bond pairs} + \text{Number of lone pairs} \]

In BrF₅:
- Number of bond pairs = 5
- Number of lone pairs = 1

So, the total is:
\[ 5 + 1 = 6 \]

The hybridization corresponding to 6 regions of electron density is sp³d².

### Step 4: Determine the Molecular Shape
With 5 bond pairs and 1 lone pair, the molecular geometry is determined by the arrangement of the bond pairs. The shape of BrF₅ is square pyramidal.

### Conclusion
- **Hybridization of BrF₅**: sp³d²
- **Shape of BrF₅**: Square pyramidal