0
$\begingroup$

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

structure of thionyl fluoride

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?

$\endgroup$
  • 1
    $\begingroup$ How many valence electrons did S have in the first place? $\endgroup$ – Ivan Neretin Apr 15 at 5:25
  • 1
    $\begingroup$ @Ivan Oh you're right! S has 6. $\endgroup$ – PhysicsMonster_01 Apr 15 at 5:47
  • 1
    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ – Martin - マーチン Apr 15 at 13:37
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Martin - マーチン Apr 15 at 13:44
  • $\begingroup$ You are right, thank you for catching that! Do you have any references about what you said? :) $\endgroup$ – Pier Apr 15 at 20:03
  • $\begingroup$ See for example chemistry.stackexchange.com/questions/29101/… $\endgroup$ – Mithoron Apr 16 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.