Question

Draw the dominant Lewis structure for the phosphorus trifluoride molecule, PF3₃.

Answer 1

To draw the dominant Lewis structure for phosphorus trifluoride (PF3), count the total number of valence electrons, place the least electronegative element (phosphorus) in the center, and distribute the remaining electrons to satisfy the octet rule, resulting in a structure with a lone pair of electrons on phosphorus and three fluorine atoms attached to it.

Step-by-step explanation:

The Lewis structure for phosphorus trifluoride (PF3) can be determined by following these steps:

1. Count the total number of valence electrons. Phosphorus has 5 valence electrons and each fluorine atom contributes 7 valence electrons, so the total is 26.
2. Place the least electronegative element (phosphorus) in the center. Connect the fluorine atoms to the phosphorus atom with single bonds.
3. Place the remaining electrons around the atoms to satisfy the octet rule. Start with the fluorine atoms, which each need 8 electrons (2 already accounted for in the bond), and then distribute the remaining electrons on the phosphorus atom.
4. Check if all atoms have satisfied the octet rule. If any atoms have fewer than 8 electrons, move lone pairs from atoms with more than 8 electrons to form multiple bonds.

The final Lewis structure for PF3 has a lone pair of electrons on phosphorus with three fluorine atoms attached to it.