What is the geometry of `BrF_(3)`?

Options

A. `sp^(3)d` and T-shaped
B. `sp^(3)d^(2)` and tetrahedral
C. `sp^(3)d` and bent
D. None of these

Explanation

To determine the geometry of BrF₃, we will follow these steps:

### Step 1: Determine the Valence Electrons of Bromine
Bromine (Br) is the central atom in BrF₃. It is in Group 17 of the periodic table, so it has 7 valence electrons.

### Step 2: Count the Monovalent Atoms
In BrF₃, there are 3 fluorine (F) atoms. Each fluorine atom is monovalent, meaning it forms one bond.

### Step 3: Calculate the Value of X
Using the formula:
\[ X = \frac{1}{2} \left( \text{Number of valence electrons of central atom} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]
For BrF₃:
- Number of valence electrons of Br = 7
- Number of monovalent atoms = 3 (from 3 F atoms)
- Charge on cation = 0 (neutral molecule)
- Charge on anion = 0 (neutral molecule)

Substituting these values:
\[ X = \frac{1}{2} \left( 7 + 3 + 0 + 0 \right) = \frac{1}{2} \times 10 = 5 \]

### Step 4: Determine the Hybridization
From the value of X:
- If \( X = 5 \), the hybridization is \( sp^3d \).

### Step 5: Predict the Geometry
When the hybridization is \( sp^3d \), the geometry is trigonal bipyramidal. However, we need to account for lone pairs.

### Step 6: Account for Lone Pairs
Bromine has 7 valence electrons, and it forms 3 bonds with the 3 fluorine atoms, using 3 of its electrons. This leaves 4 electrons, which will form 2 lone pairs.

### Step 7: Determine the Molecular Shape
In a trigonal bipyramidal arrangement:
- The 3 bond pairs (Br-F) will occupy the equatorial positions.
- The 2 lone pairs will occupy the axial positions.

This arrangement results in a T-shaped molecular geometry for BrF₃.

### Final Answer
The geometry of BrF₃ is T-shaped.

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