(Hive Addict)
07-13-03 22:09
No 446880
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.
Rainbows & Butterflies Forever :)
(Hive Addict)
07-14-03 02:03
No 446924
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.
(Hive Addict)
07-14-03 16:31
No 447076
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
Rainbows & Butterflies Forever :)
(Hive Bee)
07-14-03 17:37
No 447101
Try going here and check out orbitals,
http://www.colby.edu/chemistry/OChem/dem
We're all in this world together,
http://www.chiapaslink.ukgateway.net/
(Hive Addict)
07-14-03 21:00
No 447162
for all your help
Rainbows & Butterflies Forever :)
(Stranger)
07-17-03 18:41
No 448060
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.
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
Signature? I'm not signing anything. You prove it.
(Hive Addict)
07-17-03 19:59
No 448069
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/e
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 3px electrons having moved to the 3dx 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.
