Hydridisation of the central atom in `BrF_(5)` molecule is :

Options

A. `sp^(3)`
B. `dsp^(2)`
C. `sp^(3)d^(2)`
D. `d^(2)sp^(2)`

Explanation

To determine the hybridization of the central atom in the `BrF_(5)` molecule, we can follow these steps:

### Step 1: Determine the Valence Electrons
- Bromine (Br) is in Group 17 of the periodic table and has 7 valence electrons.
- Each fluorine (F) atom is also in Group 17 and has 7 valence electrons. Since there are 5 fluorine atoms, they contribute a total of 5 × 7 = 35 valence electrons.

### Step 2: Calculate the Total Valence Electrons
- Total valence electrons = Valence electrons from Br + Valence electrons from 5 F
- Total = 7 (from Br) + 35 (from 5 F) = 42 valence electrons.

### Step 3: Determine the Number of Bond Pairs and Lone Pairs
- In `BrF_(5)`, bromine forms 5 bonds with the 5 fluorine atoms.
- Each bond uses 2 electrons, so 5 bonds will use 10 electrons.
- Remaining electrons = Total valence electrons - Electrons used in bonds
- Remaining = 42 - 10 = 32 electrons.
- Since each lone pair consists of 2 electrons, we can calculate the number of lone pairs on bromine.
- Lone pairs on Br = Remaining electrons / 2 = 32 / 2 = 16 lone pairs.

### Step 4: Determine the Steric Number
- The steric number is the sum of the number of bond pairs and lone pairs around the central atom.
- In `BrF_(5)`, there are 5 bond pairs and 1 lone pair (since we have 5 bonds and 1 lone pair).
- Steric number = Number of bond pairs + Number of lone pairs = 5 + 1 = 6.

### Step 5: Determine the Hybridization
- The hybridization can be determined using the steric number.
- A steric number of 6 corresponds to `sp^3d^2` hybridization.

### Conclusion
- Therefore, the hybridization of the central atom in `BrF_(5)` is `sp^3d^2`.

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