Covalent Bonding

dnorric

Ok I understand that O2 wants to have a stable number of electrons for its period which is 8 and so it shares two pairs of electrons. But what about S8. I thought it would form S2. Why is it forming S8, what is the logic.
Cheers
Damian

Borek

I am afraid there is no simple logic behind. For some reasons S₈ has the lowest energy in a given circumstances. Quite often that's just the way it is and it is not possible to give a systematic explanation other than "if you calculate the energy of different molecules, there is a minimum here".

dnorric

This is why chemistry drives me crazy. There are so many exceptions to the rule, it hardly seems a science in comparison to physics. Thanks heaps for your help though

Ygggdrasil

That's what happens when you study real systems instead of idealized models :p

Borek

It depends on the selection of the rule. If the rule is "whichever system has the lowest energy, is the one observed in nature", this is not an exception.

But then you can't explain the system in the simple terms like "stable octet"...

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warenzeichen

I think Mathematics and Physics are the ultimate weapons in dealing with Chemistry...

pzona

Since there are so many exceptions to the rule, perhaps it's just that we have yet to find the true and complete conditions of the rule. Just a thought...

Borek

There are many exceptions because the rule is oversimplified.

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DrDu

There is a logic. Take some classes in quantum chemistry.
In the case of oxigen and sulfur the problem has to do with electron correlation.
Loosely speaking, in Oxigen the electrons are already so densely packed, that O2 tries to avoid contact with other oxigen atoms or molecules and instead develops a very strange double bond. In sulfur the orbitals are so much larger that electronic correlation is not a relevant topic any more. So it forms rather simplistic single bonds. The energetic differences between S4 S6 S8 to S_infinity are rather minute so that it is quite difficult to predict which one is actually most stable.