Questions: Draw the Lewis structure of PF3 and then determine the number of nonbonding electron pairs on the central atom.

Transcript text: Draw the Lewis structure of $\mathrm{PF}_{3}$ and then determine the number of nonbonding electron pairs on the central atom.

Solution

Solution Steps

Step 1: Determine the total number of valence electrons

Phosphorus (P) is in group 15 and has 5 valence electrons. Each fluorine (F) atom is in group 17 and has 7 valence electrons. Since there are three fluorine atoms, the total number of valence electrons is: \[ 5 + 3 \times 7 = 5 + 21 = 26 \]

Step 2: Draw the skeletal structure

The central atom is phosphorus (P), and the three fluorine (F) atoms are bonded to it. The initial skeletal structure is: \[ \mathrm{F} - \mathrm{P} - \mathrm{F} \] with the third fluorine also bonded to phosphorus.

Step 3: Distribute electrons to form bonds

Each P-F bond requires 2 electrons. With three P-F bonds, we use: \[ 3 \times 2 = 6 \text{ electrons} \] Subtracting these from the total valence electrons: \[ 26 - 6 = 20 \text{ electrons remaining} \]

Step 4: Distribute remaining electrons to satisfy the octet rule

First, place the remaining electrons around the fluorine atoms to complete their octets. Each fluorine needs 6 more electrons (since each already has 2 from the P-F bond): \[ 3 \times 6 = 18 \text{ electrons} \] Subtracting these from the remaining electrons: \[ 20 - 18 = 2 \text{ electrons remaining} \]

Step 5: Place remaining electrons on the central atom

The remaining 2 electrons are placed on the phosphorus atom as a lone pair.

Step 6: Verify the structure

Phosphorus now has 3 bonding pairs and 1 lone pair, making a total of 4 electron pairs around it, which is consistent with the valence shell electron pair repulsion (VSEPR) theory.