# Why is there a lone pair in thionyl fluoride?

Why is there a lone pair in $$\ce{SOF2}$$? I drew its structure, which according to me should look like this:

Why is there a lone pair on sulfur? Isn't its octet complete? If yes, why should it expand its octet and gain more electrons?

• How many valence electrons did S have in the first place? Apr 15 '19 at 5:25
• @Ivan Oh you're right! S has 6. Apr 15 '19 at 5:47
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When you want to write a Lewis structure, I suggest you start from considering how many valence electrons each atom has. In your case, you would have:

S = 6
F = 7; total = 14
O = 6

The final total would be 14 + 6 + 6 = 26. The structure you guessed for is correct, sulfur is the central atom. You start by drawing a single bond per each atom, getting something like this:

F
|
S–O
|
F


Then, you can draw a double bond for oxygen:

F
|
S=O
|
F


If you do the math, by counting the electrons that you put in the previous structure, you would have 2 + 2 + 4 (two S-F bonds and a S=O) = 8. Then we will add in the lone pairs: three on each halogen, two for oxygen, and one for sulfur. Therefore we add 3 × 2 × 2 = 12 electrons (three lone pairs on each fluorine) plus 4 electrons on O (the two lone pairs), and we have a total of 24. The final 2 , adding up to 26, come from the lone pair on sulfur.

This makes sense because sulfur is in period 3, so it is possible for it to have more than 8 electrons: in other words, the octet rule applies not only for 8 electrons, if that makes sense.

• A much, much better description would use partial charges at the sulfur and oxygen and a single bond instead. A double bond would require d-orbitals from sulfur, a theory that hat been disproved. This has been discussed on our platform a couple of times. In any case, following the octet rule is usually safer than expecting 'octet expansion' (because the latter is wrong). Apr 15 '19 at 13:44
• You are right, thank you for catching that! Do you have any references about what you said? :)
– Pier
Apr 15 '19 at 20:03
• See for example chemistry.stackexchange.com/questions/29101/… Apr 16 '19 at 20:36