To draw the Lewis dot structure for SOF4 (sulfur oxyfluoride), we first need to determine the total number of valence electrons present in the molecule.
Sulfur (S) has 6 valence electrons, oxygen (O) has 6, and each fluorine (F) has 7. Since there are four fluorine atoms, we calculate:
- S: 6
- O: 6
- F: 4 x 7 = 28
Adding these together gives us a total of:
Total valence electrons = 6 + 6 + 28 = 40 valence electrons
Next, we place the sulfur atom in the center because it is the least electronegative atom, and will be able to form bonds with the surrounding atoms. Then we connect each of the four fluorine atoms to the sulfur atom with single bonds, and also attach the oxygen atom through a single bond.
This arrangement uses up:
- 4 bonds to F (4 x 2 = 8 electrons)
- 1 bond to O (1 x 2 = 2 electrons)
So far, we have used 10 electrons out of the 40 total. Now we need to account for the remaining electrons:
Remaining electrons = 40 – 10 = 30 electrons
We can distribute the remaining electrons by placing three lone pairs on each fluorine atom (since each fluorine needs 8 electrons total) and two lone pairs on the oxygen atom (to fulfill its octet). Each fluorine atom now has a complete octet:
- Each F has 6 electrons left as lone pairs + 2 from the bond with S = 8
The lone pairs on the oxygen also complete its octet, as follows:
- 2 lone pairs (4 electrons) + 2 from the bond with S = 6
Finally, the Lewis dot structure can be visualized with sulfur at the center, single bonds connecting to one oxygen and four fluorine atoms, and each fluorine atom having three lone pairs:
F:
..|..
F:..S..O:
..|..
F:
..|..
F:
This structure conveys the distribution of electrons in the sulfur oxyfluoride molecule.