The structure of `PF_5` molecule is

Options

A. Tetrahedral
B. Trigonal bipyramidal
C. Square planar
D. Pentagonal bipyramidal

Explanation

To determine the structure of the `PF_5` molecule, we will follow these steps:

### Step 1: Identify the central atom and its valence electrons
- The central atom in `PF_5` is phosphorus (P). Phosphorus is in group 15 of the periodic table and has 5 valence electrons.

### Step 2: Determine the number of fluorine atoms and their contribution
- There are 5 fluorine (F) atoms in `PF_5`. Each fluorine atom contributes 1 electron for bonding. Therefore, the total number of bonding electrons contributed by fluorine is 5.

### Step 3: Calculate the total number of electrons around phosphorus
- The total number of electrons around phosphorus in `PF_5` is the sum of its own valence electrons and the electrons contributed by the fluorine atoms:
\[ \text{Total electrons} = 5 \text{ (from P)} + 5 \text{ (from 5 F)} = 10 \text{ electrons} \]

### Step 4: Determine the steric number
- The steric number is calculated as the number of sigma bonds plus the number of lone pairs around the central atom. In `PF_5`, there are 5 sigma bonds and no lone pairs:
\[ \text{Steric number} = 5 \text{ (sigma bonds)} + 0 \text{ (lone pairs)} = 5 \]

### Step 5: Identify the hybridization
- The steric number of 5 indicates that the hybridization of phosphorus in `PF_5` is `sp^3d`. This involves one s orbital, three p orbitals, and one d orbital.

### Step 6: Determine the molecular geometry
- With a steric number of 5 and no lone pairs, the molecular geometry of `PF_5` is trigonal bipyramidal. In this geometry, there are three equatorial bonds at 120-degree angles and two axial bonds at 90-degree angles.

### Step 7: Conclude the structure
- Therefore, the structure of the `PF_5` molecule is trigonal bipyramidal.

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