In `BrF_(3)` molecule, the lone pair occupies equatorial position minimize

Options

A. Bond pair – bond pair repulsion only
B. Lone pair – lone pair repulsion and lone pair– bond pair repulsion
C. Lone pair–lone pair repulsion only
D. Lone pair–bond pair repulsion ony

Answer

To solve the question regarding the molecular structure of BrF₃ and the position of the lone pairs, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Central Atom and its Valence Electrons:**
- In BrF₃, bromine (Br) is the central atom.
- Bromine belongs to group 17 (halogens) and has 7 valence electrons.

2. **Determine the Number of Bonds:**
- BrF₃ has three fluorine (F) atoms bonded to the bromine atom.
- Each F atom forms a single bond with Br, using up 3 of the 7 valence electrons.

3. **Calculate Remaining Electrons:**
- After forming 3 bonds (3 electrons used), there are 4 valence electrons left (7 - 3 = 4).

4. **Assign Lone Pairs:**
- The remaining 4 electrons will be assigned as lone pairs on the bromine atom.
- Since 4 electrons can form 2 lone pairs, Br will have 2 lone pairs.

5. **Hybridization of the Central Atom:**
- The hybridization of Br in BrF₃ can be determined by the formula:
\[ \text{Hybridization} = \text{Number of Bonding Pairs} + \text{Number of Lone Pairs} \]
- Here, there are 3 bonding pairs (from F atoms) and 2 lone pairs.
- Thus, the hybridization is \(3 + 2 = 5\), which corresponds to \(sp^3d\) hybridization.

6. **Determine the Molecular Geometry:**
- With \(sp^3d\) hybridization, the electron pair geometry is trigonal bipyramidal.
- In this geometry, the lone pairs will occupy positions to minimize repulsion.

7. **Position of Lone Pairs:**
- The two lone pairs will occupy the equatorial positions to minimize the lone pair-lone pair and lone pair-bond pair repulsions.
- This results in a T-shaped molecular geometry for BrF₃.

### Conclusion:
- The lone pairs in BrF₃ occupy the equatorial positions to minimize repulsions, resulting in a T-shaped structure.