Number of right angle bonds in `PCl_(5)` is _______.

Answer

To determine the number of right angle bonds in PCl₅, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the central atom and its valence electrons**:
- The central atom in PCl₅ is phosphorus (P). Phosphorus has 5 valence electrons.

2. **Count the number of surrounding atoms**:
- In PCl₅, there are 5 chlorine (Cl) atoms surrounding the phosphorus atom.

3. **Determine the hybridization**:
- The hybridization can be calculated using the formula:
\[ \text{Hybridization} = \frac{(n + m)}{2} \]
where \(n\) is the number of valence electrons of the central atom and \(m\) is the number of surrounding atoms. Here, \(n = 5\) (from P) and \(m = 5\) (from 5 Cl atoms):
\[ \text{Hybridization} = \frac{(5 + 5)}{2} = \frac{10}{2} = 5 \]
- This corresponds to \(sp^3d\) hybridization.

4. **Visualize the molecular geometry**:
- PCl₅ has a trigonal bipyramidal geometry. In this structure, there are two types of bonds:
- **Axial bonds**: These are the bonds that are oriented vertically (up and down).
- **Equatorial bonds**: These are the bonds that are in the horizontal plane.

5. **Identify the angles between bonds**:
- The axial bonds are at 90 degrees to the equatorial bonds.
- Specifically, the angle between the two axial bonds is 180 degrees, while the angle between an axial bond and an equatorial bond is 90 degrees.

6. **Count the right angle bonds**:
- In the PCl₅ structure, there are 2 axial bonds that are at right angles (90 degrees) to the 3 equatorial bonds. Therefore, there are 2 right angle bonds in total.

### Final Answer:
The number of right angle bonds in PCl₅ is **2**.

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