Draw the Lewis structure for the SF5 ion

To draw the Lewis structure for the SF₅^– ion, we need to follow these steps:

1. Count the total valence electrons: Sulfur (S) is in group 16 and has 6 valence electrons, and fluorine (F) is in group 17 with 7 valence electrons. For SF₅, we have:
• 1 sulfur = 6 electrons
• 5 fluorines = 5 x 7 = 35 electrons
2. Total = 6 + 35 = 41 electrons
3. Account for the charge: Since the ion has a -1 charge, we add an additional electron to the total, giving us 41 + 1 = 42 electrons to use for the structure.
4. Draw the basic skeleton: Place the sulfur atom in the center and surround it with five fluorine atoms. This arrangement ensures that sulfur can accommodate more than an octet.
5. Add bonds: Connect each fluorine atom to the sulfur with a single bond. This uses 10 electrons (2 electrons per bond).
6. Distribute the remaining electrons: Since we started with 42 electrons and used 10 for bonding, we have 32 electrons left. Each fluorine needs 6 electrons to complete its octet (total of 30 electrons for the 5 fluorines). This leaves us with 2 electrons.
7. Place the remaining electrons: The 2 leftover electrons represent a lone pair on the sulfur atom, which allows it to expand its octet.

The final Lewis structure for the SF₅^– ion shows sulfur in the center bonded to five fluorine atoms, with the sulfur holding one lone pair of electrons. This structure illustrates the expanded octet of the sulfur atom, which is unique for elements in the third period.