The Vespiary

Just one of many chem questions that plague me due to my unfortunate ignorance and predisposition for the humanities and not the natural sciences. 

OK, SO2:  S has 6 valence e- and same with O.  So wouldn't it make sense for S to form a double bond with EACH O atom? (As in O=S=O)  Yet we know that SO2 is O=S-O [leaving the S with 3 unpaired valence shell electrons?]; well, acutally it looks a bit different because of delocalization.  So, obviously I have a poor grasp of this very simple octet rule.  If one of my fellow bees, so wise in the ways of chemistry could help this little dolt, it woiuld be appreciated.

OK, SO2:  S has 6 valence e- and same with O.  So wouldn't it make sense for S to form a double bond with EACH O atom? (As in O=S=O)

Sulphur can have -2,0,+2,+4 and +6 oxidation states and in the case of SO₂ it's +4 ie it shares four unpaired electrons and therefore is O=S=O

Find out about s,p,d orbitals etc.

Just learn about the s,p,d orbitals.  OK, in SO2, then, there are 6 shared electrons, as there are 3 non-H atoms and 18 available valence electrons.  Next question, though:

In diagramming electron geometric configurations according to molecular orbital theory, how do I know when to use a pi bond or a pi* bond?  Ditto for sigma, of course.  I appreciate the help.

Thanks a lot for the help acid_egg

Try going here and check out orbitals,

http://www.colby.edu/chemistry/OChem/demoindex.html (http://www.colby.edu/chemistry/OChem/demoindex.html)

for all your help

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OK, SO2:  S has 6 valence e- and same with O.  So wouldn't it make sense for S to form a double bond with EACH O atom? (As in O=S=O)  Yet we know that SO2 is O=S-O [leaving the S with 3 unpaired valence shell electrons?]; well, acutally it looks a bit different because of delocalization.  So, obviously I have a poor grasp of this very simple octet rule.  If one of my fellow bees, so wise in the ways of chemistry could help this little dolt, it woiuld be appreciated.

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Your only thinking 1/2 way through the problem, sure it seems to make sense to have the 4 bonds, but now count how many electrons that takes up, and how many you have left? now distribute the ones that you have left (unbonded) and keep the octet rule from being broken... well guess what, there's nowhere to put those extra 10 electrons, now take one of the bonds away to leave O=S-O and now distribute the remaining 12 electrons that aren't forming bonds, and hey it's now filling the desire of the elements to reach noble gases (full shells) SO2 is neither O=S-O or O=S=O it's really "O=S-O <-> O-S=O"

AFAIK
Regards

This question didn't really make sense from the start and the thread is becoming increasingly non-sensical.

This is the electronic configuration of sulphur in the ground state

http://www.webelements.com/webelements/elements/text/S/econ.html (http://www.webelements.com/webelements/elements/text/S/econ.html)

. In that state you can see sulphur has only 2 unpaired electrons.
Sulphur however has easily accessable 3d orbitals which means it's 3s and 3p electrons can move to the 3d orbitals giving a greater number of unpaired electrons. So in the case of SO₂ picture one of the 3p_x electrons having moved to the 3d_x orbital.
Thus there are now four unpaired electrons, 2 for each oxygen in the case of SO₂.

In SF₆ the sulphur nucleus actually has a compliment of 12 valence electrons. ;)