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Posted

In the lecture I have just attended the professor talked about molecular orbital (MO) theory. He said a bond orbital, even for an individual orbital contribute so little that the contribution tends to none, should have an energy lower than all the contributing constituent orbitals. I have been thinking about this statement, and got some confusion in the following circumstance.

 

If, now, we consider oxygen molecule, each oxygen has its 2s and three 2p considered.

1) The two 2s would form a bonding and an anti-bonding orbitals (1sigma and 2sigma*).

2) The two 2px would form a bonding and an anti-bonding orbitals (3sigma and 4sigma*).

3) The 2py and 2pz in each oxygen atom interact to form 4 MO in total, two bonding and two anti-bonding, all with pi nature (1pi, 2pi, 3pi* and 4pi*).

 

Here we have 12e- in consideration (6 in each):

2 to 1sigma ---a

2 to 2sigma ---b

(a and b cancel one another out)

2 to 3sigma

2 to 1pi ---c

2 to 2pi ---d

1 to 3pi* ---e

1 to 4pi* ---f

(#due to half-cancelling capacity of 3pi* and 4pi*, c to f totally give just 1 pi bonding)

The result would be 1 sigma bonding and 1 pi bonding.

 

The above is what is usually interpreted. Now, actually 2s orbital of either oxygen should have, at least some, contribution to 3sigma (that formed between two 2px). But 3sigma certainly has an energy higher than the two 2s.

 

Another thing is that I don't know if the statement in # is exactly the explanation or not, maybe there is other interpretations?

 

So this is the case I want to discuss, please give some opinions. Thanks.

Posted

The s orbitals will not have any infulence on the sigma bond formed between the two p-orbitals. Remember the orbitals are centred at the atom and if you then try to overlap an s-orbital with a p-orbital you'll find that any contrustive interferance exactly cancels all the destructive interferance.

 

Regarding the # section, its probably easier to count the number of electrons in both the bonding and anti-bonding orbitals. If you do that, only counting orbitals from the 2s and 2p, you should find that you have 8 bonding electrons and 4 antibonding electrons, so overal there are 4 electrons involved in a stable bond. As you need to electrons for a bond, dioxygen therefore has a double bond. But yes, your statment is also correct.

 

Incidently, the two unpaired electrons that occupy the two antibonding orbitals are what make oxygen paramagnetic (it sticks to a magnetic) and what gives it its colour (liquid oxygen is very visibly blue).

Posted

Its due to the presence of two antibonding orbitals which make oxygen to behave paramagnetically.

Posted
The s orbitals will not have any infulence on the sigma bond formed between the two p-orbitals. Remember the orbitals are centred at the atom and if you then try to overlap an s-orbital with a p-orbital you'll find that any contrustive interferance exactly cancels all the destructive interferance.

But if s-orbital is just overlapping side by side, there should be constructive interference, because it is not s- and px orbital of the same atom overlapping, there should not be cancelled interference, I think...

  • 2 weeks later...
Posted

Yes your are right that an s-orbtial and one of the p-orbitals (depends on wht u label the axis) will interect with each other. But if I remember my MO lectures correctly, it is too low in energy to make any reasonable contribution

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