How many right angle bonds are present in `BrF_(5)`.

Answer

To determine the number of right angle bonds present in the compound \( \text{BrF}_5 \), we can follow these steps:

### Step 1: Determine the Hybridization of Bromine
1. **Count the Valence Electrons**: Bromine (Br) has 7 valence electrons. Each fluorine (F) contributes 1 valence electron, and there are 5 fluorine atoms.
\[ \text{Total valence electrons} = 7 + (5 \times 1) = 12 \]

2. **Calculate Hybridization**: The hybridization can be determined using the formula:
\[ \text{Hybridization} = \frac{\text{Total valence electrons}}{2} \]
Therefore,
\[ \text{Hybridization} = \frac{12}{2} = 6 \]
This corresponds to \( \text{sp}^3\text{d}^2 \) hybridization.

### Step 2: Determine the Molecular Geometry
1. **Identify Lone Pairs**: In \( \text{BrF}_5 \), there is 1 lone pair on the bromine atom and 5 bond pairs with fluorine atoms.

2. **Draw the Structure**: The molecular geometry of \( \text{BrF}_5 \) is based on the arrangement of the bond pairs and lone pairs. The structure is a square pyramidal shape, where the lone pair occupies an axial position.

### Step 3: Analyze Bond Angles
1. **Bond Angles**: In a perfect octahedral geometry, the bond angles would be 90 degrees. However, due to the presence of the lone pair, the bond angles are slightly distorted.

2. **Calculate Actual Bond Angles**: According to VSEPR theory, the presence of a lone pair causes repulsion that slightly reduces the bond angles. In \( \text{BrF}_5 \), the bond angles are approximately 88.4 degrees instead of the ideal 90 degrees.

### Step 4: Count Right Angle Bonds
1. **Determine Right Angle Bonds**: Since the bond angles are not exactly 90 degrees due to the lone pair's influence, there are no right angle bonds in \( \text{BrF}_5 \).

### Conclusion
Thus, the number of right angle bonds in \( \text{BrF}_5 \) is **0**.

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