Question

Calculate lone pairs for PCL2F

Answer 1

To calculate the number of lone pairs in the molecule phosphorus dichlorofluoride (PCl₂F), we need to follow a few steps:

1. **Determine the total number of valence electrons:**
- Phosphorus (P) has 5 valence electrons.
- Each chlorine (Cl) has 7 valence electrons, and there are 2 chlorine atoms: \(2 \times 7 = 14\) valence electrons.
- Fluorine (F) has 7 valence electrons.
- Total: \(5 + 14 + 7 = 26\) valence electrons.

2. **Construct the Lewis structure:**
- Phosphorus is the central atom. It is bonded to two chlorines and one fluorine.
- Start by forming single bonds between P and each of the three surrounding atoms (Cl and F). Each bond uses 2 electrons, so 3 bonds use \(3 \times 2 = 6\) electrons.
- Subtract the used electrons from the total: \(26 - 6 = 20\) electrons remaining.
- Distribute the remaining electrons to satisfy the octet rule for the surrounding atoms. Each Cl and F requires 8 electrons to complete their octet.
- Place 6 electrons around each of the two Cl atoms and the F atom. So, each surrounding atom has 3 lone pairs: \(3 \times 2 + 3 = 9\) electrons (for both Cl and F).

3. **Calculate the lone pairs on the central atom:**
- After placing electrons around Cl and F, we have \(20 - 18 = 2\) electrons remaining.
- These 2 electrons can form one lone pair on the phosphorus atom.

4. **Final count of lone pairs:**
- Phosphorus has 1 lone pair.
- Chlorine atoms and fluorine do not have lone pairs on P, since they are already completing their octet by forming bonds.

**In summary:**
In the molecule PCl₂F, there is 1 lone pair of electrons on the phosphorus atom.