The answer to each of the following questions is a non- negative integer.
Find the maximum number equal of P-F bond lengths in `PF_5`

Answer

To solve the problem of finding the maximum number of equal P-F bond lengths in PF₅ (phosphorus pentafluoride), we can follow these steps:

### Step 1: Understand the Structure of PF₅
PF₅ has a trigonal bipyramidal structure. In this structure, phosphorus (P) is at the center, and it is surrounded by five fluorine (F) atoms.

### Step 2: Identify the Types of Bonds
In the trigonal bipyramidal geometry of PF₅:
- There are three equatorial P-F bonds.
- There are two axial P-F bonds.

### Step 3: Analyze the Bond Lengths
- The three equatorial P-F bonds are of equal length because they are symmetrically arranged in the equatorial plane.
- The two axial P-F bonds are longer than the equatorial bonds due to the geometry of the molecule.

### Step 4: Count the Maximum Number of Equal Bond Lengths
Since the three equatorial bonds are equal, we can conclude that:
- The maximum number of equal P-F bond lengths in PF₅ is **3** (the three equatorial bonds).

### Final Answer
The maximum number of equal P-F bond lengths in PF₅ is **3**.

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